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All Textbook Solutions for STATISTICS F/BUSINESS+ECONOMICS-TEXT
Delta Airlines quotes a flight time of 2 hours, 5 minutes for its flights from Cincinnatito Tampa. Suppose we believe that actual flight times are uniformly distributed between2 hours and 2 hours, 20 minutes. a. Show the graph of the probability density function for flight time. b. What is the probability that the flight will be no more than 5 minutes late? c. What is the probability that the flight will be more than 10 minutes late? d. What is the expected flight time?4EIn October 2012, Apple introduced a much smaller variant of the Apple iPad, known as theiPad Mini. Weighing less than 11 ounces, it was about 50% lighter than the standard iPad.Battery tests for the iPad Mini showed a mean life of 10.25 hours (The Wall Street Journal,October 31, 2012). Assume that battery life of the iPad Mini is uniformly distributed between8.5 and 12 hours. a. Give a mathematical expression for the probability density function of battery life. b. What is the probability that the battery life for an iPad Mini will be 10 hours or less? c. What is the probability that the battery life for an iPad Mini will be at least 11 hours? d. What is the probability that the battery life for an iPad Mini will be between 9.5 and11.5 hours? e. In a shipment of 100 iPad Minis, how many should have a battery life of at least9 hours?A Gallup Daily Tracking Survey found that the mean daily discretionary spending byAmericans earning over 90,000 per year was 136 per day (USA Today, July 30, 2012).The discretionary spending excluded home purchases, vehicle purchases, and regularmonthly bills. Let x = the discretionary spending per day and assume that a uniform probability density function applies with f (x) = .00625 for a x b. a. Find the values of a and b for the probability density function. b. What is the probability that consumers in this group have daily discretionary spendingbetween 100 and 200? c. What is the probability that consumers in this group have daily discretionary spendingof 150 or more? d. What is the probability that consumers in this group have daily discretionary spendingof 80 or less?Suppose we are interested in bidding on a piece of land and we know one other bidder isinterested.1 The seller announced that the highest bid in excess of 10,000 will be accepted. Assume that the competitors bid x is a random variable that is uniformly distributedbetween 10,000 and 15,000. a. Suppose you bid 12,000. What is the probability that your bid will be accepted? b. Suppose you bid 14,000. What is the probability that your bid will be accepted? c. What amount should you bid to maximize the probability that you get the property? d. Suppose you know someone who is willing to pay you 16,000 for the property.Would you consider bidding less than the amount in part (c)? Why or why not?Using Figure 6.4 as a guide, sketch a normal curve for a random variable x that has a meanof m = 100 and a standard deviation of = 10. Label the horizontal axis with values of70, 80, 90, 100, 110, 120, and 130.A random variable is normally distributed with a mean of m = 50 and a standard deviation of = 5. a. Sketch a normal curve for the probability density function. Label the horizontal axiswith values of 35, 40, 45, 50, 55, 60, and 65. Figure 6.4 shows that the normal curvealmost touches the horizontal axis at three standard deviations below and at threestandard deviations above the mean (in this case at 35 and 65). b. What is the probability the random variable will assume a value between 45 and 55? c. what is the probability the random variable will assume a value between 40 and 60?Draw a graph for the standard normal distribution. Label the horizontal axis at values of3, 2, 1, 0, 1, 2, and 3. Then use the table of probabilities for the standard normaldistribution inside the front cover of the text to compute the following probabilities. a. P(z 1.5) b. P(z 1) c. P(1 z 1.5) d. P(0 z 2.5)Given that z is a standard normal random variable, compute the following probabilities. a. P(z 1.0) b. P(z 1) c. P(z 1.5) d. P(2.5 z) e. P(3 z 0)Given that z is a standard normal random variable, compute the following probabilities. a. P(0 z 83) b. P(1.57 z 0) c. P(z .44) d. P(z .23) e. P(z 1.20) f. P(z .71)Given that z is a standard normal random variable, compute the following probabilities. a. P(1.98 z .49) b. P(.52 z 1.22) c. P(1.75 z 1.04)Given that z is a standard normal random variable, find z for each situation. a. The area to the left of z is .9750. b. The area between 0 and z is .4750. c. The area to the left of z is .7291. d. The area to the right of z is .1314. e. The area to the left of z is .6700. f. The area to the right of z is .3300.Given that z is a standard normal random variable, find z for each situation. a. The area to the left of z is .2119. b. The area between z and z is .9030. c. The area between z and z is .2052. d. The area to the left of z is .9948. e. The area to the right of z is .6915.Given that z is a standard normal random variable, find z for each situation. a. The area to the right of z is .01. b. The area to the right of z is .025. c. The area to the right of z is .05. d. The area to the right of z is .10.The mean cost of domestic airfares in the United States rose to an all-time high of 385per ticket (Bureau of Transportation Statistics website, November 2, 2012). Airfares werebased on the total ticket value, which consisted of the price charged by the airlines plusany additional taxes and fees. Assume domestic airfares are normally distributed with astandard deviation of 110. a. What is the probability that a domestic airfare is 550 or more? b. What is the probability that a domestic airfare is 250 or less? c. What is the probability that a domestic airfare is between 300 and 500? d. What is the cost for the 3% highest domestic airfares?The average return for large-cap domestic stock funds over the three years 20092011was 14.4% (AAII Journal, February, 2012). Assume the three-year returns were normallydistributed across funds with a standard deviation of 4.4%. a. What is the probability an individual large-cap domestic stock fund had a three-yearreturn of at least 20%? b. What is the probability an individual large-cap domestic stock fund had a three-yearreturn of 10% or less? c. How big does the return have to be to put a domestic stock fund in the top 10% for thethree-year period?Automobile repair costs continue to rise with the average cost now at 367 per repair (U.S.News World Report website, January 5, 2015). Assume that the cost for an automobilerepair is normally distributed with a standard deviation of 88. Answer the followingquestions about the cost of automobile repairs. a. What is the probability that the cost will be more than 450? b. What is the probability that the cost will be less than 250? c. What is the probability that the cost will be between 250 and 450? d. If the cost for your car repair is in the lower 5% of automobile repair charges, what isyour cost?The average price for a gallon of gasoline in the United States is 3.73 and in Russia itis 3.40 (Bloomberg Businessweek, March 5March 11, 2012). Assume these averagesare the population means in the two countries and that the probability distributions arenormally distributed with a standard deviation of .25 in the United States and a standarddeviation of .20 in Russia. a. What is the probability that a randomly selected gas station in the United Statescharges less than 3.50 per gallon? b. What percentage of the gas stations in Russia charge less than 3.50 per gallon? c. What is the probability that a randomly selected gas station in Russia charged morethan the mean price in the United States?A person must score in the upper 2% of the population on an IQ test to qualify for membership in Mensa, the international high-IQ society. If IQ scores are normally distributedwith a mean of 100 and a standard deviation of 15, what score must a person have toqualify for Mensa?Television viewing reached a new high when the Nielsen Company reported a mean dailyviewing time of 8.35 hours per household (USA Today, November 11, 2009). Use a normal probability distribution with a standard deviation of 2.5 hours to answer the followingquestions about daily television viewing per household. a. what is the probability that a household views television between 5 and 10 hoursa day? b. How many hours of television viewing must a household have in order to be in the top3% of all television viewing households? c. what is the probability that a household views television more than 3 hours a day?The time needed to complete a final examination in a particular college course is normallydistributed with a mean of 80 minutes and a standard deviation of 10 minutes. Answer thefollowing questions. a. What is the probability of completing the exam in one hour or less? b. What is the probability that a student will complete the exam in more than 60 minutesbut less than 75 minutes? c. Assume that the class has 60 students and that the examination period is 90 minutesin length. How many students do you expect will be unable to complete the exam inthe allotted time?The American Automobile Association (AAA) reported that families planning to travelover the Labor Day weekend would spend an average of 749 (The Associated Press,August 12, 2012). Assume that the amount spent is normally distributed with a standarddeviation of 225. a. What is the probability of family expenses for the weekend being less that 400? b. What is the probability of family expenses for the weekend being 800 or more? c. What is the probability that family expenses for the weekend will be between 500and 1000? d. What would the Labor Day weekend expenses have to be for the 5% of the familieswith the most expensive travel plans?New York City is the most expensive city in the United States for lodging. The mean hotelroom rate is 204 per night (USA Today, April 30, 2012). Assume that room rates arenormally distributed with a standard deviation of 55. a. What is the probability that a hotel room costs 225 or more per night? b. What is the probability that a hotel room costs less than 140 per night? c. What is the probability that a hotel room costs between 200 and 300 per night? d. What is the cost of the 20% most expensive hotel rooms in New York City?A binomial probability distribution has p = .20 and n = 100. a. What are the mean and standard deviation? b. Is this situation one in which binomial probabilities can be approximated by the normal probability distribution? Explain. c. What is the probability of exactly 24 successes? d. What is the probability of 18 to 22 successes? e. What is the probability of 15 or fewer successes?Assume a binomial probability distribution has p = .60 and n = 200. a. What are the mean and standard deviation? b. Is this situation one in which binomial probabilities can be approximated by thenormal probability distribution? Explain. c. What is the probability of 100 to 110 successes? d. What is the probability of 130 or more successes? e. What is the advantage of using the normal probability distribution to approximate thebinomial probabilities? Use part (d) to explain the advantage.Although studies continue to show smoking leads to significant health problems, 20% ofadults in the United States smoke. Consider a group of 250 adults. a. What is the expected number of adults who smoke? b. What is the probability that fewer than 40 smoke? c. What is the probability that from 55 to 60 smoke? d. What is the probability that 70 or more smoke?A CBS News/New York Times survey found that 97% of Americans believe that textingwhile driving should be outlawed (CBS News website, January 5, 2015). a. For a sample of 10 Americans, what is the probability that at least 8 say that theybelieve texting while driving should be outlawed? Use the binomial distribution probability function discussed in Section 5.5 to answer this question. b. For a sample of 100 Americans, what is the probability that at least 95 say that theybelieve texting while driving should be outlawed? Use the normal approximation ofthe binomial distribution to answer this question. c. As the number of trials in a binomial distribution application becomes large, whatis the advantage of using the normal approximation of the binomial distribution tocompute probabilities? d. When the number of trials for a binominal distribution application becomes large,would developers of statistical software packages prefer to use the binomial distribution probability function shown in Section 5.5 or the normal approximation of thebinomial distribution discussed in Section 6.3? Explain.Playing video and computer games is very popular. Over 70% of households play suchgames. Of those individuals who play video and computer games, 18% are under 18 yearsold, 53% are 1859 years old, and 29% are over 59 years old (The Wall Street Journal, March6, 2012). a. For a sample of 800 people who play these games, how many would you expect to beunder 18 years of age? b. For a sample of 600 people who play these games, what is the probability that fewerthan 100 will be under 18 years of age? c. For a sample of 800 people who play these games, what is the probability that 200 ormore will be over 59 years of age?The Bank of America trends in Consumer Mobility Report indicates that in a typical day,51% of users of mobile phones use their phone at least once per hour, 26% use their phonea few times per day, 8% use their phone morning and evening, and 13% hardly ever usetheir phones. The remaining 2% indicated that they did not know how often they usedtheir mobile phone (USA today, July 7, 2014). Consider a sample of 150 mobile phoneusers. a. What is the probability that at least 70 use their phone at least once per hour? b. What is the probability that at least 75 but less than 80 use their phone at least onceper hour? c. What is the probability that less than 5 of the 150 phone users do not know how oftenthey use their phone?Consider the following exponential probability density function. f(x)=18ex/8forx0 a. Find P(x 6). b. Find P(x 4). c. Find P(x 6). d. Find P(4 x 6).Consider the following exponential probability density function. f(x)=13ex/3forx0 a. Write the formula for P(x x0). b. Find P(x 2). c. Find P(x 3). d. Find P(x 5). e. Find P(2 x 5).Battery life between charges for the Motorola Droid Razr Maxx is 20 hours when theprimary use is talk time (The Wall Street Journal, March 7, 2012). The battery life drops to7 hours when the phone is primarily used for Internet applications over cellular. Assumethat the battery life in both cases follows an exponential distribution. a. Show the probability density function for battery life for the Droid Razr Maxx phonewhen its primary use is talk time. b. what is the probability that the battery charge for a randomly selected Droid RazrMaxx phone will last no more than 15 hours when its primary use is talk time? c. what is the probability that the battery charge for a randomly selected Droid RazrMaxx phone will last more than 20 hours when its primary use is talk time? d. What is the probability that the battery charge for a randomly selected Droid Razr Maxxphone will last no more than 5 hours when its primary use is Internet applications?The time between arrivals of vehicles at a particular intersection follows an exponentialprobability distribution with a mean of 12 seconds. a. Sketch this exponential probability distribution. b. what is the probability that the arrival time between vehicles is 12 seconds or less? c. What is the probability that the arrival time between vehicles is 6 seconds or less? d. What is the probability of 30 or more seconds between vehicle arrivals?Comcast Corporation is the largest cable television company, the second largest Internet service provider, and the fourth largest telephone service provider in the United States. Generally known for quality and reliable service, the company periodically experiences unexpectedservice interruptions. On January 14, 2014, such an interruption occurred for the Comcastcustomers living in southwest Florida. When customers called the Comcast office, a recordedmessage told them that the company was aware of the service outage and that it was anticipated that service would be restored in two hours. Assume that two hours is the mean time todo the repair and that the repair time has an exponential probability distribution. a. What is the probability that the cable service will be repaired in one hour or less? b. What is the probability that the repair will take between one hour and two hours? c. For a customer who calls the Comcast office at 1:00 p.m., what is the probability thatthe cable service will not be repaired by 5:00 p.m.?Wendys restaurant has been recognized for having the fastest average service time amongfast food restaurants. In a benchmark study, Wendys average service time of 2.2 minuteswas less than those of Burger King, Chick-fil-A, Krystal, McDonalds, Taco Bell, andTaco Johns (QSR Magazine website, December 2014). Assume that the service time forWendys has an exponential distribution. a. What is the probability that a service time is less than or equal to one minute? b. What is the probability that a service time is between 30 seconds and one minute? c. Suppose a manager of a Wendys is considering instituting a policy such that if thetime it takes to serve you exceeds five minutes, your food is free. What is the probability that you will get your food for free? Comment.The Boston Fire Department receives 911 calls at a mean rate of 1.6 calls per hour(Mass.gov website, November 2012). Suppose the number of calls per hour follows aPoisson probability distribution. a. What is the mean time between 911 calls to the Boston Fire Department in minutes? b. Using the mean in part (a), show the probability density function for the time between911 calls in minutes. c. What is the probability that there will be less than one hour between 911 calls? d. What is the probability that there will be 30 minutes or more between 911 calls? e. What is the probability that there will be more than 5 minutes, but less than 20 minutesbetween 911 calls?A business executive, transferred from Chicago to Atlanta, needs to sell her house inChicago quickly. The executives employer has offered to buy the house for 210,000, butthe offer expires at the end of the week. The executive does not currently have a better offerbut can afford to leave the house on the market for another month. From conversations withher realtor, the executive believes the price she will get by leaving the house on the marketfor another month is uniformly distributed between 200,000 and 225,000. a. If she leaves the house on the market for another month, what is the mathematicalexpression for the probability density function of the sales price? b. If she leaves it on the market for another month, what is the probability she will get atleast 215,000 for the house? c. If she leaves it on the market for another month, what is the probability she will getless than 210,000? d. Should the executive leave the house on the market for another month? Why or why not?The NCAA estimates that the yearly value of a full athletic scholarship at in-state publicuniversities is 19,000 (The Wall Street Journal, March 12, 2012). Assume the scholarshipvalue is normally distributed with a standard deviation of 2100. a. For the 10% of athletic scholarships of least value, how much are they worth? b. what percentage of athletic scholarships are valued at 22,000 or more? c. For the 3% of athletic scholarships that are most valuable, how much are they worth?Motorola used the normal distribution to determine the probability of defects and thenumber of defects expected in a production process. Assume a production processproduces items with a mean weight of 10 ounces. Calculate the probability of a defectand the expected number of defects for a 1000-unit production run in the followingsituations. a. The process standard deviation is .15, and the process control is set at plus or minusone standard deviation. Units with weights less than 9.85 or greater than 10.15 ounceswill be classified as defects. b. Through process design improvements, the process standard deviation can be reducedto .05. Assume the process control remains the same, with weights less than 9.85 orgreater than 10.15 ounces being classified as defects. c. what is the advantage of reducing process variation, thereby causing process controllimits to be at a greater number of standard deviations from the mean?During early 2012, economic hardship was stretching the limits of Frances welfaresystem. One indicator of the level of hardship was the increase in the number of people bringing items to a Paris pawnbroker; the number of people bringing items to thepawnbroker had increased to 658 per day (Bloomberg Businessweek, March 5March 11,2012). Assume the number of people bringing items to the pawnshop per day in 2012 isnormally distributed with a mean of 658. a. Suppose you learn that on 3% of the days, 610 or fewer people brought items to thepawnshop. what is the standard deviation of the number of people bringing items tothe pawnshop per day? b. On any given day, what is the probability that between 600 and 700 people bring itemsto the pawnshop? c. How many people bring items to the pawnshop on the busiest 3% of days?The port of South Louisiana, located along 54 miles of the Mississippi River between NewOrleans and Baton Rouge, is the largest bulk cargo port in the world. The U.S. Army Corpsof Engineers reports that the port handles a mean of 4.5 million tons of cargo per week (USAtoday, September 25, 2012). Assume that the number of tons of cargo handled per week isnormally distributed with a standard deviation of .82 million tons. a. What is the probability that the port handles less than 5 million tons of cargo per week? b. what is the probability that the port handles 3 or more million tons of cargo per week? c. What is the probability that the port handles between 3 million and 4 million tons ofcargo per week? d. Assume that 85% of the time the port can handle the weekly cargo volume withoutextending operating hours. What is the number of tons of cargo per week that willrequire the port to extend its operating hours?Ward Doering Auto Sales is considering offering a special service contract that will coverthe total cost of any service work required on leased vehicles. From experience, the company manager estimates that yearly service costs are approximately normally distributed,with a mean of 150 and a standard deviation of 25. a. If the company offers the service contract to customers for a yearly charge of 200, whatis the probability that any one customers service costs will exceed the contract price of200? b. What is Wards expected profit per service contract?The XO Group Inc. conducted a survey of 13,000 brides and grooms married in the UnitedStates and found that the average cost of a wedding is 29,858 (XO Group website, January 5, 2015). Assume that the cost of a wedding is normally distributed with a mean of29,858 and a standard deviation of 5600. a. What is the probability that a wedding costs less than 20,000? b. What is the probability that a wedding costs between 20,000 and 30,000? c. For a wedding to be among the 5% most expensive, how much would it have to cost?Assume that the test scores from a college admissions test are normally distributed, with amean of 450 and a standard deviation of 100. a. What percentage of the people taking the test score between 400 and 500? b. Suppose someone receives a score of 630. What percentage of the people taking thetest score better? What percentage score worse? c. If a particular university will not admit anyone scoring below 480, what percentage ofthe persons taking the test would be acceptable to the university?According to the National Association of Colleges and Employers, the average startingsalary for new college graduates in health sciences is 51,541. The average starting salary for new college graduates in business is 53,901 (National Association of Collegesand Employers website, January 5, 2015). Assume that starting salaries are normally distributed and that the standard deviation for starting salaries for new college graduates inhealth sciences is 11,000. Assume that the standard deviation for starting salaries for newcollege graduates in business is 15,000. a. What is the probability that a new college graduate in business will earn a startingsalary of at least 65,000? b. What is the probability that a new college graduate in health sciences will earn a starting salary of at least 65,000? c. What is the probability that a new college graduate in health sciences will earn a starting salary less than 40,000? d. How much would a new college graduate in business have to earn in order to have astarting salary higher than 99% of all starting salaries of new college graduates in thehealth sciences?A machine fills containers with a particular product. The standard deviation of fillingweights is known from past data to be .6 ounce. If only 2% of the containers hold less than18 ounces, what is the mean filling weight for the machine? That is, what must 03BC equal?Assume the filling weights have a normal distribution.Consider a multiple-choice examination with 50 questions. Each question has four possible answers. Assume that a student who has done the homework and attended lectureshas a 75% probability of answering any question correctly. a. A student must answer 43 or more questions correctly to obtain a grade of A. Whatpercentage of the students who have done their homework and attended lectures willobtain a grade of A on this multiple-choice examination? b. A student who answers 35 to 39 questions correctly will receive a grade of C. Whatpercentage of students who have done their homework and attended lectures willobtain a grade of C on this multiple-choice examination? c. A student must answer 30 or more questions correctly to pass the examination. Whatpercentage of the students who have done their homework and attended lectures willpass the examination? d. Assume that a student has not attended class and has not done the homework for thecourse. Furthermore, assume that the student will simply guess at the answer to eachquestion. What is the probability that this student will answer 30 or more questionscorrectly and pass the examination?A blackjack player at a Las Vegas casino learned that the house will provide a freeroom if play is for four hours at an average bet of 50. The players strategy provides aprobability of .49 of winning on any one hand, and the player knows that there are60 hands per hour. Suppose the player plays for four hours at a bet of 50 per hand. a. What is the players expected payoff? b. What is the probability the player loses 1000 or more? c. What is the probability the player wins? d. Suppose the player starts with 1500. What is the probability of going broke?The Information Systems Audit and Control Association surveyed office workers tolearn about the anticipated usage of office computers for personal holiday shopping (USAToday, November 11, 2009). Assume that the number of hours a worker spends doingholiday shopping on an office computer follows an exponential distribution. a. The study reported that there is a .53 probability that a worker uses an office computerfor holiday shopping 5 hours or less. Is the mean time spent using an office computerfor holiday shopping closest to 5.8, 6.2, 6.6, or 7 hours? b. Using the mean time from part (a), what is the probability that a worker uses an officecomputer for holiday shopping more than 10 hours? c. What is the probability that a worker uses an office computer for holiday shoppingbetween 4 and 8 hours?The website for the Bed and Breakfast Inns of North America gets approximately sevenvisitors per minute (Time, September 2001). Suppose the number of website visitors perminute follows a Poisson probability distribution. a. What is the mean time between visits to the website? b. Show the exponential probability density function for the time between website visits. c. What is the probability no one will access the website in a 1-minute period? d. What is the probability no one will access the website in a 12-second period?Do you dislike waiting in line? Supermarket chain Kroger has used computer simulationand information technology to reduce the average waiting time for customers at 2300stores. Using a new system called QueVision, which allows Kroger to better predict whenshoppers will be checking out, the company was able to decrease average customer waiting time to just 26 seconds (InformationWeek website and The Wall Street Journal website, January 5, 2015). a. Assume that Kroger waiting times are exponentially distributed. Show the probabilitydensity function of waiting time at Kroger. b. What is the probability that a customer will have to wait between 15 and 30 seconds? c. What is the probability that a customer will have to wait more than 2 minutes?The time (in minutes) between telephone calls at an insurance claims office has the following exponential probability distribution. f(x)=.50e.50xforx0 a. What is the mean time between telephone calls? b. What is the probability of having 30 seconds or less between telephone calls? c. what is the probability of having 1 minute or less between telephone calls? d. what is the probability of having 5 or more minutes without a telephone call?Consider a finite population with five elements labeled A, B, C, D, and E. Ten possible simple random samples of size 2 can be selected. a. List the 10 samples beginning with AB, AC, and so on. b. Using simple random sampling, what is the probability that each sample of size 2 is selected? c. Assume random number 1 corresponds to A, random number 2 corresponds to B, and so on. List the simple random sample of size 2 that will be selected by using the random digits 8 0 5 7 5 3 2.Assume a finite population has 350 elements. Using the last three digits of each of the following five-digit random numbers (e.g., 601, 022, 448, ), determine the first four elements that will be selected for the simple random sample.Fortune publishes data on sales, profits, assets, stockholders equity, market value, and earnings per share for the 500 largest U.S. industrial corporations (Fortune 500, 2012). Assume that you want to select a simple random sample of 10 corporations from the Fortune 500 list. Use the last three digits in column 9 of Table 7. 1, beginning with 554. Read down the column and identify the numbers of the 10 corporations that would be selected.The 10 most active stocks on the New York Stock Exchange for a given week, are shown here. Exchange authorities decided to investigate trading practices using a sample of three of these stocks. a. Beginning with the first random digit in column 6 of Table 7.1, read down the column to select a simple random sample of three stocks for the exchange authorities. b. Using the information in the third Note and Comment, determine how many different simple random samples of size 3 can be selected from the list of 10 stocks.A student government organization is interested in estimating the proportion of students who favor a mandatory pass-fail grading policy for elective courses. A list of names and addresses of the 645 students enrolled during the current quarter is available from the registrars office. Using three-digit random numbers in row 10 of Table 7.1 and moving across the row from left to right, identify the first 10 students who would be selected using simple random sampling. The three-digit random numbers begin with 816, 283, and 610.The County and City Data Book, published by the Census Bureau, lists information on 3139 counties throughout the United States. Assume that a national study will collect data from 30 randomly selected counties. Use four-digit random numbers from the last column of Table 7.1 to identify the numbers corresponding to the first five counties selected for the sample. Ignore the first digits and begin with the four-digit random numbers 9945, 8364, 5702, and so on.Assume that we want to identify a simple random sample of 12 of the 372 doctors practicing in a particular city. The doctors names are available from a local medical organization. Use the eighth column of five-digit random numbers in Table 7.1 to identify the 12 doctors for the sample. Ignore the first two random digits in each five-digit grouping of the random numbers. This process begins with random number 108 and proceeds down the column of random numbers.The following stocks make up the Dow Jones Industrial Average (Barrons, July 30, 2012). 1. 1.3M 2. ATT 3. Alcoa 4. American Express 5. Bank of America 6. Boeing 7. Caterpillar 8. Chevron 9. Cisco Systems 10. Coca-Cola 11. Disney 12. DuPont 13. ExxonMobil 14. General Electric 15. Hewlett-Packard 16. Home Depot 17. IBM 18. Intel 19. Johnson Johnson 20. Kraft Foods 21. McDonalds 22. Merck 23. Microsoft 24. J.P. Morgan 25. Pfizer 26. Procter Gamble 27. Travelers 28. United Technologies 29. Verizon 30. Wal-Mart Suppose you would like to select a sample of six of these companies to conduct an in-depth study of management practices. Use the first two digits in each row of the ninth column of Table 7.1 to select a simple random sample of six companies.The Wall Street Journal provides the net asset value, the year-to-date percent return, and the three-year percent return for 748 mutual funds (The Wall Street Journal, December 15, 2014). Assume that a simple random sample of 12 of the 748 mutual funds will be selected for a follow-up study on the size and performance of mutual funds. Use the fourth column of the random numbers in Table 7.1, beginning with 51102, to select the simple random sample of 12 mutual funds. Begin with mutual fund 102 and use the last three digits in each row of the fourth column for your selection process. What are the numbers of the 12 mutual funds in the simple random sample?Indicate which of the following situations involve sampling from a finite population and which involve sampling from an infinite population. In cases where the sampled population is finite, describe how you would construct a frame. a. Obtain a sample of licensed drivers in the state of New York. b. Obtain a sample of boxes of cereal produced by the Breakfast Choice company. c. Obtain a sample of cars crossing the Golden Gate Bridge on a typical weekday. d. Obtain a sample of students in a statistics course at Indiana University. e. Obtain a sample of the orders that are processed by a mail-order firm.The following data are from a simple random sample. a. What is the point estimate of the population mean? b. What is the point estimate of the population standard deviation?A survey question for a sample of 150 individuals yielded 75 Yes responses, 55 No responses, and 20 No Opinions. a. What is the point estimate of the proportion in the population who respond Yes? b. What is the point estimate of the proportion in the population who respond No?A sample of 5 months of sales data provided the following information: a. Develop a point estimate of the population mean number of units sold per month. b. Develop a point estimate of the population standard deviation.Morningstar publishes ratings data on 1208 company stocks (Morningstar website, October 24, 2012). A sample of 40 of these stocks is contained in the DATAfile named Morningstar. Use the Morningstar data set to answer the following questions. a. Develop a point estimate of the proportion of the stocks that receive Morningstars highest rating of 5 Stars. b. Develop a point estimate of the proportion of the Morningstar stocks that are rated Above Average with respect to business risk. c. Develop a point estimate of the proportion of the Morningstar stocks that are rated 2 Stars or less.The National Football League (NFL) polls fans to develop a rating for each football game (NFL website, October 24, 2012). Each game is rated on a scale from 0 (forgettable) to 100 (memorable). The fan ratings for a random sample of 12 games follow. a. Develop a point estimate of mean fan rating for the population of NFL games. b. Develop a point estimate of the standard deviation for the population of NFL games.A sample of 426 U.S. adults age 50 and older were asked how important a variety of issues were in choosing whom to vote for in the 2012 presidential election (AARP Bulletin, March 2012). a. What is the sampled population for this study? b. Social Security and Medicare was cited as very important by 350 respondents. Estimate the proportion of the population of U.S. adults age 50 and over who believe this issue is very important. c. Education was cited as very important by 74% of the respondents. Estimate the number of respondents who believe this issue is very important. d. Job Growth was cited as very important by 354 respondents. Estimate the proportion of U.S. adults age 50 and over who believe job growth is very important. e. What is the target population for the inferences being made in parts (b) and (d)? Is it the same as the sampled population you identified in part (a)? Suppose you later learn that the sample was restricted to members of the American Association of Retired People (AARP). Would you still feel the inferences being made in parts (b) and (d) are valid? Why or why not?One of the questions in the Pew Internet American Life Project asked adults if they used the Internet at least occasionally (Pew website, October 23, 2012). The results showed that 454 out of 478 adults aged 1829 answered Yes; 741 out of 833 adults aged 3049 answered Yes; 1058 out of 1644 adults aged 50 and over answered Yes. a. Develop a point estimate of the proportion of adults aged 1829 who use the Internet. b. Develop a point estimate of the proportion of adults aged 3049 who use the Internet. c. Develop a point estimate of the proportion of adults aged 50 and over who use the Internet. d. Comment on any relationship between age and Internet use that seems apparent. e. Suppose your target population of interest is that of all adults (18 years of age and over). Develop an estimate of the proportion of that population who use the Internet.A population has a mean of 200 and a standard deviation of 50. A sample of size 100 will be taken and the sample mean x will be used to estimate the population mean. a. What is the expected value of x? b. What is the standard deviation of x? c. Show the sampling distribution of x. d. What does the sampling distribution of x show?A population has a mean of 200 and a standard deviation of 50. Suppose a sample of size 100 is selected and x is used to estimate . a. What is the probability that the sample mean will be within 5 of the population mean? b. What is the probability that the sample mean will be within 10 of the population mean?Assume the population standard deviation is = 25. Compute the standard error of the mean, x, for sample sizes of 50, 100, 150, and 200. What can you say about the size of the standard error of the mean as the sample size is increased?Suppose a random sample of size 50 is selected from a population with = 10. Find the value of the standard error of the mean in each of the following cases (use the finite population correction factor if appropriate). a. The population size is infinite. b. The population size is N = 50,000. c. The population size is N = 5000. d. The population size is N = 500.Refer to the EAI sampling problem. Suppose a simple random sample of 60 managers is used. a. Sketch the sampling distribution of x when simple random samples of size 60 are used. b. What happens to the sampling distribution of x if simple random samples of size 120 are used? c. What general statement can you make about what happens to the sampling distribution of x as the sample size is increased? Does this generalization seem logical? Explain.In the EAI sampling problem (see Figure 7.5), we showed that for n = 30, there was .5034 probability of obtaining a sample mean within 500 of the population mean. a. What is the probability that x is within 500 of the population mean if a sample of size 60 is used? b. Answer part (a) for a sample of size 120.Barrons reported that the average number of weeks an individual is unemployed is 17.5 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 17.5 weeks and that the population standard deviation is 4 weeks. Suppose you would like to select a sample of 50 unemployed individuals for a follow-up study. a. Show the sampling distribution of x, the sample mean average for a sample of 50 unemployed individuals. b. What is the probability that a simple random sample of 50 unemployed individuals will provide a sample mean within 1 week of the population mean? c. What is the probability that a simple random sample of 50 unemployed individuals will provide a sample mean within 1/2 week of the population mean?The College Board reported the following mean scores for the three parts of the Scholastic Aptitude Test (SAT) (The World Almanac, 2009): Critical Reading 502 Mathematics 515 Writing 494 Assume that the population standard deviation on each part of the test is = 100. a. What is the probability a sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 502 on the Critical Reading part of the test? b. What is the probability a sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 515 on the Mathematics part of the test? Compare this probability to the value computed in part (a). c. What is the probability a sample of 100 test takers will provide a sample mean test score within 10 of the population mean of 494 on the writing part of the test? Comment on the differences between this probability and the values computed in parts (a) and (b).For the year 2010, 33% of taxpayers with adjusted gross incomes between 30,000 and 60,000 itemized deductions on their federal income tax return (The Wall Street Journal, October 25, 2012). The mean amount of deductions for this population of taxpayers was 16,642. Assume the standard deviation is = 2400. a. What is the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within 200 of the population mean for each of the following sample sizes: 30, 50, 100, and 400? b. What is the advantage of a larger sample size when attempting to estimate the population mean?The Economic Policy Institute periodically issues reports on wages of entry level workers. The institute reported that entry level wages for male college graduates were 21.68 per hour and for female college graduates were 18.80 per hour in 2011 (Economic Policy Institute website, March 30, 2012). Assume the standard deviation for male graduates is 2.30, and for female graduates it is 2.05. a. What is the probability that a sample of 50 male graduates will provide a sample mean within .50 of the population mean, 21.68? b. What is the probability that a sample of 50 female graduates will provide a sample mean within .50 of the population mean, 18.80? c. In which of the preceding two cases, part (a) or part (b), do we have a higher probability of obtaining a sample estimate within .50 of the population mean? Why? d. What is the probability that a sample of 120 female graduates will provide a sample mean more than .30 below the population mean?The state of California has a mean annual rainfall of 22 inches, whereas the state of New York has a mean annual rainfall of 42 inches (Current Results website, October 27, 2012). Assume that the standard deviation for both states is 4 inches. A sample of 30 years of rainfall for California and a sample of 45 years of rainfall for New York has been taken. a. Show the probability distribution of the sample mean annual rainfall for California. b. What is the probability that the sample mean is within 1 inch of the population mean for California? c. What is the probability that the sample mean is within 1 inch of the population mean for New York? d. In which case, part (b) or part (c), is the probability of obtaining a sample mean within 1 inch of the population mean greater? Why?The mean preparation fee HR Block charged retail customers last year was 183 (The Wall Street Journal, March 7, 2012). Use this price as the population mean and assume the population standard deviation of preparation fees is 50. a. What is the probability that the mean price for a sample of 30 HR Block retail customers is within 8 of the population mean? b. What is the probability that the mean price for a sample of 50 HR Block retail customers is within 8 of the population mean? c. What is the probability that the mean price for a sample of 100 HR Block retail customers is within 8 of the population mean? d. Which, if any, of the sample sizes in parts (a), (b), and (c) would you recommend to have at least a .95 probability that the sample mean is within 8 of the population mean?To estimate the mean age for a population of 4000 employees, a simple random sample of 40 employees is selected. a. Would you use the finite population correction factor in calculating the standard error of the mean? Explain. b. If the population standard deviation is = 8.2 years, compute the standard error both with and without the finite population correction factor. What is the rationale for ignoring the finite population correction factor whenever n/N .05? c. What is the probability that the sample mean age of the employees will be within 2 years of the population mean age?A sample of size 100 is selected from a population with p = .40. a. What is the expected value of p? b. What is the standard error of p? c. Show the sampling distribution of p. d. What does the sampling distribution of p show?A population proportion is .40. A sample of size 200 will be taken and the sample proportion p will be used to estimate the population proportion. a. What is the probability that the sample proportion will be within .03 of the population proportion? b. What is the probability that the sample proportion will be within .05 of the population proportion?Assume that the population proportion is .55. Compute the standard error of the proportion, p, for sample sizes of 100, 200, 500, and 1000. What can you say about the size of the standard error of the proportion as the sample size is increased?The population proportion is .30. What is the probability that a sample proportion will be within .04 of the population proportion for each of the following sample sizes? a. n = 100 b. n = 200 c. n = 500 d. n = 1000 e. What is the advantage of a larger sample size?The president of Doerman Distributors, Inc., believes that 30% of the firms orders come from first-time customers. A random sample of 100 orders will be used to estimate the proportion of first-time customers. a. Assume that the president is correct and p = .30. What is the sampling distribution of p for this study? b. What is the probability that the sample proportion p will be between .20 and .40? c. What is the probability that the sample proportion will be between .25 and .35?The Wall Street Journal reported that the age at first startup for 55% of entrepreneurs was 29 years of age or less and the age at first startup for 45% of entrepreneurs was 30 years of age or more (The Wall Street Journal, March 19, 2012). a. Suppose a sample of 200 entrepreneurs will be taken to learn about the most important qualities of entrepreneurs. Show the sampling distribution of p where p is the sample proportion of entrepreneurs whose first startup was at 29 years of age or less. b. What is the probability that the sample proportion in part (a) will be within .05 of its population proportion? c. Suppose a sample of 200 entrepreneurs will be taken to learn about the most important qualities of entrepreneurs. Show the sampling distribution of p where p is now the sample proportion of entrepreneurs whose first startup was at 30 years of age or more. d. What is the probability that the sample proportion in part (c) will be within .05 of its population proportion? e. Is the probability different in parts (b) and (d)? Why? f. Answer part (b) for a sample of size 400. Is the probability smaller? Why?People end up tossing 12% of what they buy at the grocery store (Readers Digest, March 2009). Assume this is the true population proportion and that you plan to take a sample survey of 540 grocery shoppers to further investigate their behavior. a. Show the sampling distribution of p, the proportion of groceries thrown out by your sample respondents. b. What is the probability that your survey will provide a sample proportion within .03 of the population proportion? c. What is the probability that your survey will provide a sample proportion within .015 of the population proportion?Forty-two percent of primary care doctors think their patients receive unnecessary medical care (Readers Digest, December 2011/January 2012). a. Suppose a sample of 300 primary care doctors were taken. Show the sampling distribution of the proportion of the doctors who think their patients receive unnecessary medical care. b. What is the probability that the sample proportion will be within .03 of the population proportion? c. What is the probability that the sample proportion will be within .05 of the population proportion? d. What would be the effect of taking a larger sample on the probabilities in parts (b) and (c)? Why?In 2008 the Better Business Bureau settled 75% of complaints they received (USA Today, March 2, 2009). Suppose you have been hired by the Better Business Bureau to investigate the complaints they received this year involving new car dealers. You plan to select a sample of new car dealer complaints to estimate the proportion of complaints the Better Business Bureau is able to settle. Assume the population proportion of complaints settled for new car dealers is .75, the same as the overall proportion of complaints settled in 2008. a. Suppose you select a sample of 450 complaints involving new car dealers. Show the sampling distribution of p. b. Based upon a sample of 450 complaints, what is the probability that the sample proportion will be within .04 of the population proportion? c. Suppose you select a sample of 200 complaints involving new car dealers. Show the sampling distribution of p. d. Based upon the smaller sample of only 200 complaints, what is the probability that the sample proportion will be within .04 of the population proportion? e. As measured by the increase in probability, how much do you gain in precision by taking the larger sample in part (b)?The Grocery Manufacturers of America reported that 76% of consumers read the ingredients listed on a products label. Assume the population proportion is p = .76 and a sample of 400 consumers is selected from the population. a. Show the sampling distribution of the sample proportion p where p is the proportion of the sampled consumers who read the ingredients listed on a products label. b. What is the probability that the sample proportion will be within .03 of the population proportion? c. Answer part (b) for a sample of 750 consumers.The Food Marketing Institute shows that 17% of households spend more than 100 per week on groceries. Assume the population proportion is p = .17 and a sample of 800 households will be selected from the population. a. Show the sampling distribution of p, the sample proportion of households spending more than 100 per week on groceries. b. What is the probability that the sample proportion will be within .02 of the population proportion? c. Answer part (b) for a sample of 1600 households.Jack Lawler, a financial analyst, wants to prepare an article on the Shadow Stock portfolio developed by the American Association of Individual Investors (AAII). A list of the 30 companies in the Shadow Stock portfolio as of March 2014 is contained in the DATAfile named ShadowStocks (AAII website March 27, 2014). Jack would like to select a simple random sample of 5 of these companies for an interview concerning management practices. a. In the DATAfile the Shadow Stock companies are listed in column A of an Excel worksheet. In column B we have generated a random number for each of the companies. Use these random numbers to select a simple random sample of 5 of these companies for Jack. b. Generate a new set of random numbers and use them to select a new simple random sample. Did you select the same companies?The latest available data showed health expenditures were 8086 per person in the United States or 17.6% of gross domestic product (Centers for Medicare Medicaid Services website, April 1, 2012). Use 8086 as the population mean and suppose a survey research firm will take a sample of 100 people to investigate the nature of their health expenditures. Assume the population standard deviation is 2500. a. Show the sampling distribution of the mean amount of health care expenditures for a sample of 100 people. b. What is the probability the sample mean will be within 200 of the population mean? c. What is the probability the sample mean will be greater than 9000? If the survey research firm reports a sample mean greater than 9000, would you question whether the firm followed correct sampling procedures? Why or why not?Foot Locker uses sales per square foot as a measure of store productivity. Sales are currently running at an annual rate of 406 per square foot (The Wall Street Journal, March 7, 2012). You have been asked by management to conduct a study of a sample of 64 Foot Locker stores. Assume the standard deviation in annual sales per square foot for the population of all 3400 Foot Locker stores is 80. a. Show the sampling distribution of x, the sample mean annual sales per square foot for a sample of 64 Foot Locker stores. b. What is the probability that the sample mean will be within 15 of the population mean? c. Suppose you find a sample mean of 380. What is the probability of finding a sample mean of 380 or less? Would you consider such a sample to be an unusually low-performing group of stores?Allegiant Airlines charges a mean base fare of 89. In addition, the airline charges for making a reservation on its website, checking bags, and inflight beverages. These additional charges average 39 per passenger (Bloomberg Businessweek, October 814, 2012). Suppose a random sample of 60 passengers is taken to determine the total cost of their flight on Allegiant Airlines. The population standard deviation of total flight cost is known to be 40. a. What is the population mean cost per flight? b. What is the probability the sample mean will be within 10 of the population mean cost per flight? c. What is the probability the sample mean will be within 5 of the population mean cost per flight?After deducting grants based on need, the average cost to attend the University of Southern California (USC) is 27,175 (U.S. News World Report, Americas Best Colleges, 2009 ed.). Assume the population standard deviation is 7400. Suppose that a random sample of 60 USC students will be taken from this population. a. What is the value of the standard error of the mean? b. What is the probability that the sample mean will be more than 27,175? c. What is the probability that the sample mean will be within 1000 of the population mean? d. How would the probability in part (c) change if the sample size were increased to 100?Three firms carry inventories that differ in size. Firm As inventory contains 2000 items, firm Bs inventory contains 5000 items, and firm Cs inventory contains 10,000 items. The population standard deviation for the cost of the items in each firms inventory is = 144. A statistical consultant recommends that each firm take a sample of 50 items from its inventory to provide statistically valid estimates of the average cost per item. Managers of the small firm state that because it has the smallest population, it should be able to make the estimate from a much smaller sample than that required by the larger firms. However, the consultant states that to obtain the same standard error and thus the same precision in the sample results, all firms should use the same sample size regardless of population size. a. Using the finite population correction factor, compute the standard error for each of the three firms given a sample of size 50. b. What is the probability that for each firm the sample mean x will be within 25 of the population mean ?A researcher reports survey results by stating that the standard error of the mean is 20. The population standard deviation is 500. a. How large was the sample used in this survey? b. What is the probability that the point estimate was within 25 of the population mean?A production process is checked periodically by a quality control inspector. The inspector selects simple random samples of 30 finished products and computes the sample mean product weights x. If test results over a long period of time show that 5% of the x values are over 2.1 pounds and 5% are under 1.9 pounds, what are the mean and the standard deviation for the population of products produced with this process?Fifteen percent of Australians smoke. By introducing tough laws banning brand labels on cigarette packages, Australia hopes to reduce the percentage of people smoking to 10% by 2018 (Reuters website, October 23, 2012). Answer the following questions based on a sample of 240 Australians. a. Show the sampling distribution of p, the proportion of Australians who are smokers. b. What is the probability the sample proportion will be within .04 of the population proportion? c. What is the probability the sample proportion will be within .02 of the population proportion?A market research firm conducts telephone surveys with a 40% historical response rate. What is the probability that in a new sample of 400 telephone numbers, at least 150 individuals will cooperate and respond to the questions? In other words, what is the probability that the sample proportion will be at least 150/400 = .375?Advertisers contract with Internet service providers and search engines to place ads on websites. They pay a fee based on the number of potential customers who click on their ad. Unfortunately, click fraudthe practice of someone clicking on an ad solely for the purpose of driving up advertising revenuehas become a problem. Forty percent of advertisers claim they have been a victim of click fraud (Businessweek, March 13, 2006). Suppose a simple random sample of 380 advertisers will be taken to learn more about how they are affected by this practice. a. What is the probability that the sample proportion will be within .04 of the population proportion experiencing click fraud? b. What is the probability that the sample proportion will be greater than .45?The proportion of individuals insured by the All-Driver Automobile Insurance Company who received at least one traffic ticket during a five-year period is .15. a. Show the sampling distribution of p if a random sample of 150 insured individuals is used to estimate the proportion having received at least one ticket. b. What is the probability that the sample proportion will be within .03 of the population proportion?Lori Jeffrey is a successful sales representative for a major publisher of college textbooks. Historically, Lori obtains a book adoption on 25% of her sales calls. Viewing her sales calls for one month as a sample of all possible sales calls, assume that a statistical analysis of the data yields a standard error of the proportion of .0625. a. How large was the sample used in this analysis? That is, how many sales calls did Lori make during the month? b. Let p indicate the sample proportion of book adoptions obtained during the month. Show the sampling distribution of p. c. Using the sampling distribution of p, compute the probability that Lori will obtain book adoptions on 30% or more of her sales calls during a one-month period.Managerial Report Prepare a managerial report that addresses the following issues. 1. Calculate the probability that the mean score of Blugert given by the simple random sample of Marion Dairies customers will be 75 or less. 2. If the Marketing Department increases the sample size to 150, what is the probability that the mean score of Blugert given by the simple random sample of Marion Dairies customers will be 75 or less? 3. Explain to Marion Dairies senior management why the probability that the mean score of Blugert given by the simple random sample of Marion Dairies customers will be 75 or less differs for these two sample sizes.A simple random sample of 40 items resulted in a sample mean of 25. The population standard deviation is = 5. a. What is the standard error of the mean, x? b. At 95% confidence, what is the margin of error?A simple random sample of 50 items from a population with = 6 resulted in a sample mean of 32. a. Provide a 90% confidence interval for the population mean. b. Provide a 95% confidence interval for the population mean. c. Provide a 99% confidence interval for the population mean.A simple random sample of 60 items resulted in a sample mean of 80. The population standard deviation is = 15. a. Compute the 95% confidence interval for the population mean. b. Assume that the same sample mean was obtained from a sample of 120 items. Provide a 95% confidence interval for the population mean. c. What is the effect of a larger sample size on the interval estimate?A 95% confidence interval for a population mean was reported to be 152 to 160. If = 15, what sample size was used in this study?Data were collected on the amount spent by 64 customers for lunch at a major Houston restaurant. These data are contained in the file named Houston. Based upon past studies the population standard deviation is known with = 6. a. At 99% confidence, what is the margin of error? b. Develop a 99% confidence interval estimate of the mean amount spent for lunch.In an attempt to assess total daily travel taxes in various cities, the Global Business Travel Association conducted a study of daily travel taxes on lodging, rental car, and meals (GBTA Foundation website, October 30, 2012). The data contained in the DATAfile named TravelTax are consistent with the findings of that study for business travel to Chicago. Assume the population standard deviation is known to be 8.50 and develop a 95% confidence interval of the population mean total daily travel taxes for Chicago.The average annual cost of the first year of owning and caring for a large dog is 1843 (US News and World Report, September 9, 2013). The Irish Red and White Setter Association of America has requested a study to estimate the annual first-year cost for owners of this breed. A sample of 50 will be used. Based on past studies, the population standard deviation is assumed known with = 255. a. What is the margin of error for a 95% confidence interval of the mean cost of the first year of owning and caring for this breed? b. The DATAfile Setters contains data collected from fifty owners of Irish Setters on the cost of the first year of owning and caring for their dogs. Use these data data set to compute the sample mean. Using this sample, what is the 95% confidence interval for the mean cost of the first year of owning and caring for an Irish Red and White Setter ?Studies show that massage therapy has a variety of health benefits and it is not too expensive (The Wall Street Journal, March 13, 2012). A sample of 10 typical one-hour massage therapy sessions showed an average charge of 59. The population standard deviation for a one-hour session is = 5.50. a. What assumptions about the population should we be willing to make if a margin of error is desired? b. Using 95% confidence, what is the margin of error? c. Using 99% confidence, what is the margin of error?The mean cost to repair the smoke and fire damage that results from home fires of all causes is 11,389 (HomeAdvisor website, December 2014). How does the damage that results from home fires caused by careless use of tobacco compare? The DATAfile named TobaccoFires provides the cost to repair smoke and fire damage associated with a sample of 55 fires caused by careless use of tobacco products. Using past years data, the population standard deviation can be assumed known with = 3027. What is the 95% confidence interval estimate of the mean cost to repair smoke and fire damage that results from home fires caused by careless use of tobacco? How does this compare with the mean cost to repair the smoke and fire damage that results from home fires of all causes?Costs are rising for all kinds of medical care. The mean monthly rent at assisted-living facilities was reported to have increased 17% over the last five years to 3486 (The Wall Street Journal, October 27, 2012). Assume this cost estimate is based on a sample of 120 facilities and, from past studies, it can be assumed that the population standard deviation is = 650. a. Develop a 90% confidence interval estimate of the population mean monthly rent. b. Develop a 95% confidence interval estimate of the population mean monthly rent. c. Develop a 99% confidence interval estimate of the population mean monthly rent. d. What happens to the width of the confidence interval as the confidence level is increased? Does this seem reasonable? Explain.For a t distribution with 16 degrees of freedom, find the area, or probability, in each region. a. To the right of 2.120 b. To the left of 1.337 c. To the left of 1.746 d. To the right of 2.583 e. Between 2.120 and 2.120 f. Between 1.746 and 1.746Find the t value(s) for each of the following cases. a. Upper tail area of .025 with 12 degrees of freedom b. Lower tail area of .05 with 50 degrees of freedom c. Upper tail area of .01 with 30 degrees of freedom d. Where 90% of the area falls between these two t values with 25 degrees of freedom e. Where 95% of the area falls between these two t values with 45 degrees of freedomThe following sample data are from a normal population: 10, 8, 12, 15, 13, 11, 6, 5. a. What is the point estimate of the population mean? b. What is the point estimate of the population standard deviation? c. With 95% confidence, what is the margin of error for the estimation of the population mean? d. What is the 95% confidence interval for the population mean?A simple random sample with n = 54 provided a sample mean of 22.5 and a sample standard deviation of 4.4. a. Develop a 90% confidence interval for the population mean. b. Develop a 95% confidence interval for the population mean. c. Develop a 99% confidence interval for the population mean. d. What happens to the margin of error and the confidence interval as the confidence level is increased?Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 65 weekly reports showed a sample mean of 19.5 customer contacts per week. The sample standard deviation was 5.2. Provide 90% and 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel.A sample containing years to maturity and yield for 40 corporate bonds are contained in the data file named CorporateBonds (Barrons, April 2, 2012). a. What is the sample mean years to maturity for corporate bonds and what is the sample standard deviation? b. Develop a 95% confidence interval for the population mean years to maturity. c. What is the sample mean yield on corporate bonds and what is the sample standard deviation? d. Develop a 95% confidence interval for the population mean yield on corporate bonds.The International Air Transport Association surveys business travelers to develop quality ratings for transatlantic gateway airports. The maximum possible rating is 10. Suppose a simple random sample of 50 business travelers is selected and each traveler is asked to provide a rating for the Miami International Airport. The ratings obtained from the sample of 50 business travelers follow. Develop a 95% confidence interval estimate of the population mean rating for Miami.Older people often have a hard time finding work. AARP reported on the number of weeks it takes a worker aged 55 plus to find a job. The data on number of weeks spent searching for a job contained in the file JobSearch are consistent with the AARP findings. a. Provide a point estimate of the population mean number of weeks it takes a worker aged 55 plus to find a job. b. At 95% confidence, what is the margin of error? c. What is the 95% confidence interval estimate of the mean? d. Discuss the degree of skewness found in the sample data. What suggestion would you make for a repeat of this study?The mean cost of a meal for two in a mid-range restaurant in Tokyo is 40 (Numbeo.com website, December 14, 2014). How do prices for comparable meals in Hong Kong compare? The DATAfile HongKongMeals contains the costs for a sample of 42 recent meals for two in Hong Kong mid-range restaurants. a. With 95% confidence, what is the margin of error? b. What is the 95% confidence interval estimate of the population mean? c. How do prices for meals for two in mid-range restaurants in Hong Kong compare to prices for comparable meals in Tokyo restaurants?The average annual premium for automobile insurance in the United States is 1503 (Insure.com website, March 6, 2014). The following annual premiums () are representative of the websites findings for the state of Michigan. 1905 3112 2312 2725 2545 2981 2677 2525 2627 2600 2370 2857 2962 2545 2675 2184 2529 2115 2332 2442 Assume the population is approximately normal. a. Provide a point estimate of the mean annual automobile insurance premium in Michigan. b. Develop a 95% confidence interval for the mean annual automobile insurance premium in Michigan. c. Does the 95% confidence interval for the annual automobile insurance premium in Michigan include the national average for the United States? What is your interpretation of the relationship between auto insurance premiums in Michigan and the national average?Health insurers are beginning to offer telemedicine services online that replace the common office visit. Wellpoint provides a video service that allows subscribers to connect with a physician online and receive prescribed treatments (Bloomberg Businessweek, March 49, 2014). Wellpoint claims that users of its LiveHealth Online service saved a significant amount of money on a typical visit. The data shown below (), for a sample of 20 online doctor visits, are consistent with the savings per visit reported by Wellpoint. 92 34 40 105 83 55 56 49 40 76 48 96 93 74 73 78 93 100 53 82 Assuming the population is roughly symmetric, construct a 95% confidence interval for the mean savings for a televisit to the doctor as opposed to an office visit.Marvel Studios motion picture Guardians of the Galaxy opened over the first two days of the 2014 Labor Day weekend to a record-breaking 94.3 million in ticket sales revenue in North America (The Hollywood Reporter, August 3, 2014). The ticket sales revenue in dollars for a sample of 30 theaters is as follows. a. What is the 95% confidence interval estimate for the mean ticket sales revenue per theater? Interpret this result. b. Using the movie ticket price of 8.11 per ticket, what is the estimate of the mean number of customers per theater? c. The movie was shown in 4080 theaters. Estimate the total number of customers who saw Guardians of the Galaxy and the total box office ticket sales for the weekend.How large a sample should be selected to provide a 95% confidence interval with a margin of error of 10? Assume that the population standard deviation is 40.The range for a set of data is estimated to be 36. a. What is the planning value for the population standard deviation? b. At 95% confidence, how large a sample would provide a margin of error of 3? c. At 95% confidence, how large a sample would provide a margin of error of 2?Refer to the Scheer Industries example in Section 8.2. Use 6.84 days as a planning value for the population standard deviation. a. Assuming 95% confidence, what sample size would be required to obtain a margin of error of 1.5 days? b. If the precision statement was made with 90% confidence, what sample size would be required to obtain a margin of error of 2 days?The U.S. Energy Information Administration (US EIA) reported that the average price for a gallon of regular gasoline is 3.94 (US EIA website, April 6, 2012). The US EIA updates its estimates of average gas prices on a weekly basis. Assume the standard deviation is .25 for the price of a gallon of regular gasoline and recommend the appropriate sample size for the US EIA to use if they wish to report each of the following margins of error at 95% confidence. a. The desired margin of error is .10. b. The desired margin of error is .07. c. The desired margin of error is .05.Annual starting salaries for college graduates with degrees in business administration are generally expected to be between 30,000 and 45,000. Assume that a 95% confidence interval estimate of the population mean annual starting salary is desired. What is the planning value for the population standard deviation? How large a sample should be taken if the desired margin of error is a. 500? b. 200? c. 100? d. Would you recommend trying to obtain the 100 margin of error? Explain.Many medical professionals believe that eating too much red meat increases the risk of heart disease and cancer (WebMD website, March 12, 2014). Suppose you would like to conduct a survey to determine the yearly consumption of beef by a typical American and want to use 3 pounds as the desired margin of error for a confidence interval estimate of the population mean amount of beef consumed annually. Use 25 pounds as a planning value for the population standard deviation and recommend a sample size for each of the following situations. a. A 90% confidence interval is desired for the mean amount of beef consumed. b. A 95% confidence interval is desired for the mean amount of beef consumed. c. A 99% confidence interval is desired for the mean amount of beef consumed. d. When the desired margin of error is set, what happens to the sample size as the confidence level is increased? Would you recommend using a 99% confidence interval in this case? Discuss.Customers arrive at a movie theater at the advertised movie time only to find that they have to sit through several previews and prepreview ads before the movie starts. Many complain that the time devoted to previews is too long (The Wall Street Journal, October 12, 2012). A preliminary sample conducted by the Wall Street Journal showed that the standard deviation of the amount of time devoted to previews was 4 minutes. Use that as a planning value for the standard deviation in answering the following questions. a. If we want to estimate the population mean time for previews at movie theaters with a margin of error of 75 seconds, what sample size should be used? Assume 95% confidence. b. If we want to estimate the population mean time for previews at movie theaters with a margin of error of 1 minute, what sample size should be used? Assume 95% confidence.There has been a trend toward less driving in the last few years, especially by young people. From 2001 to 2009 the annual vehicle miles traveled by people from 16 to 34 years of age decreased from 10,300 to 7900 miles per person (U.S. PIRG and Education Fund website, April 6, 2012). Assume the standard deviation was 2000 miles in 2009. Suppose you would like to conduct a survey to develop a 95% confidence interval estimate of the annual vehicle-miles per person for people 16 to 34 years of age at the current time. A margin of error of 100 miles is desired. How large a sample should be used for the current survey?A simple random sample of 400 individuals provides 100 Yes responses. a. What is the point estimate of the proportion of the population that would provide Yes responses? b. What is your estimate of the standard error of the proportion,p? c. Compute the 95% confidence interval for the population proportion.A simple random sample of 800 elements generates a sample proportion p=.70. a. Provide a 90% confidence interval for the population proportion. b. Provide a 95% confidence interval for the population proportion.In a survey, the planning value for the population proportion is p = .35. How large a sample should be taken to provide a 95% confidence interval with a margin of error of .05?At 95% confidence, how large a sample should be taken to obtain a margin of error of .03 for the estimation of a population proportion? Assume that past data are not available for developing a planning value for p.The Consumer Reports National Research Center conducted a telephone survey of 2000 adults to learn about the major economic concerns for the future (Consumer Reports, January 2009). The survey results showed that 1760 of the respondents think the future health of Social Security is a major economic concern. a. What is the point estimate of the population proportion of adults who think the future 1. health of Social Security is a major economic concern. b. At 90% confidence, what is the margin of error? c. Develop a 90% confidence interval for the population proportion of adults who think 2. the future health of Social Security is a major economic concern. d. Develop a 95% confidence interval for this population proportion.According to statistics reported on CNBC, a surprising number of motor vehicles are not covered by insurance. Sample results, consistent with the CNBC report, showed 46 of 200 vehicles were not covered by insurance. a. What is the point estimate of the proportion of vehicles not covered by insurance? b. Develop a 95% confidence interval for the population proportion.One of the questions on a survey of 1000 adults asked if todays children will be better off than their parents (Rasmussen Reports website, October 26, 2012). Representative data are shown in the DATAfile named ChildOutlook. A response of Yes indicates that the adult surveyed did think todays children will be better off than their parents. A response of No indicates that the adult surveyed did not think todays children will be better off than their parents. A response of Not Sure was given by 23% of the adults surveyed. a. What is the point estimate of the proportion of the population of adults who do think that todays children will be better off than their parents? b. At 95% confidence, what is the margin of error? c. What is the 95% confidence interval for the proportion of adults who do think that todays children will be better off than their parents? d. What is the 95% confidence interval for the proportion of adults who do not think that todays children will be better off than their parents? e. Which of the confidence intervals in parts (c) and (d) has the smaller margin of error? Why?According to Franchise business review, over 50% of all food franchises earn a profit of less than 50,000 a year. In a sample of 142 casual dining restaurants, 81 earned a profit of less than 50,000 last year. a. What is the point estimate of the proportion of casual dining restaurants that earned a profit of less than 50,000 last year? b. Determine the margin of error and provide a 95% confidence interval for the proportion of casual dining restaurants that earned a profit of less than 50,000 last year. c. How large a sample is needed if the desired margin of error is .03?In 16% of all homes with a stay-at-home parent, the father is the stay-at-home parent (Pew Research, June 5, 2014). An independent research firm has been charged with conducting a sample survey to obtain more current information. a. What sample size is needed if the research firms goal is to estimate the current proportion of homes with a stay-at-home parent in which the father is the stay-at-home parent with a margin of error of .03? Use a 95% confidence level. b. Repeat part (a) using a 99% confidence level.For many years businesses have struggled with the rising cost of health care. But recently, the increases have slowed due to less inflation in health care prices and employees paying for a larger portion of health care benefits. A recent Mercer survey showed that 52% of U.S. employers were likely to require higher employee contributions for health care coverage in 2009. Suppose the survey was based on a sample of 800 companies. Compute the margin of error and a 95% confidence interval for the proportion of companies likely to require higher employee contributions for health care coverage in 2009.Fewer young people are driving. In 1983, 87% of 19-year-olds had a drivers license. Twenty-five years later that percentage had dropped to 75% (University of Michigan Transportation Research Institute website, April 7, 2012). Suppose these results are based on a random sample of 1200 19-year-olds in 1983 and again in 2008. a. At 95% confidence, what is the margin of error and the interval estimate of the number of 19-year-old drivers in 1983? b. At 95% confidence, what is the margin of error and the interval estimate of the number of 19-year-old drivers in 2008? c. Is the margin of error the same in parts (a) and (b)? Why or why not?A poll for the presidential campaign sampled 491 potential voters in June. A primary purpose of the poll was to obtain an estimate of the proportion of potential voters who favored each candidate. Assume a planning value of p = .50 and a 95% confidence level. a. For p = .50, what was the planned margin of error for the June poll? b. Closer to the November election, better precision and smaller margins of error are desired. Assume the following margins of error are requested for surveys to be conducted during the presidential campaign. Compute the recommended sample size for each survey. Survey Margin of Error September .04 October .03 Early November .02 Pre-Election Day .01The Pew Research Center Internet Project, conducted on the 25th anniversary of the Internet, involved a survey of 857 Internet users (Pew Research Center website, April 1, 2014). It provided a variety of statistics on Internet users. For instance, in 2014, 87% of American adults were Internet users. In 1995 only 14% of American adults used the Internet. a. The sample survey showed that 90% of respondents said the Internet has been a good thing for them personally. Develop a 95% confidence interval for the proportion of respondents who say the Internet has been a good thing for them personally. b. The sample survey showed that 67% of Internet users said the Internet has generally strengthened their relationship with family and friends. Develop a 95% confidence interval for the proportion of respondents who say the Internet has strengthened their relationship with family and friends. c. Fifty-six percent of Internet users have seen an online group come together to help a person or community solve a problem, whereas only 25% have left an online group because of unpleasant interaction. Develop a 95% confidence interval for the proportion of Internet users who say online groups have helped solve a problem. d. Compare the margin of error for the interval estimates in parts (a), (b), and (c). How is the margin of error related to the sample proportion?A sample survey of 54 discount brokers showed that the mean price charged for a trade of 100 shares at 50 per share was 33.77. The survey is conducted annually. With the historical data available, assume a known population standard deviation of 15. a. Using the sample data, what is the margin of error associated with a 95% confidence interval? b. Develop a 95% confidence interval for the mean price charged by discount brokers for a trade of 100 shares at 50 per share.A survey conducted by the American Automobile Association showed that a family of four spends an average of 215.60 per day while on vacation. Suppose a sample of 64 families of four vacationing at Niagara Falls resulted in a sample mean of 252.45 per day and a sample standard deviation of 74.50. a. Develop a 95% confidence interval estimate of the mean amount spent per day by a family of four visiting Niagara Falls. b. based on the confidence interval from part (a), does it appear that the population mean amount spent per day by families visiting Niagara Falls differs from the mean reported by the American Automobile Association? Explain.The 92 million Americans of age 50 and over control 50 percent of all discretionary income. AARP estimates that the average annual expenditure on restaurants and carryout food was 1873 for individuals in this age group. Suppose this estimate is based on a sample of 80 persons and that the sample standard deviation is 550. a. At 95% confidence, what is the margin of error? b. What is the 95% confidence interval for the population mean amount spent on restaurants and carryout food? c. What is your estimate of the total amount spent by Americans of age 50 and over on restaurants and carryout food? d. If the amount spent on restaurants and carryout food is skewed to the right, would you expect the median amount spent to be greater or less than 1873?Russia has recently started a push for stronger smoking regulations much like those in Western countries concerning cigarette advertising, smoking in public places, and so on. The DATAfile named Russia contains sample data on smoking habits of Russians that are consistent with those reported by the Wall Street Journal (The Wall Street Journal, October 16, 2012). Analyze the data using Excel and answer the following questions. a. Develop a point estimate and a 95% confidence interval for the proportion of Russians who smoke. b. Develop a point estimate and a 95% confidence interval for the mean annual per capita consumption (number of cigarettes) of a Russian. c. For those Russians who do smoke, estimate the number of cigarettes smoked per day.The Health Care Cost Institute tracks health care expenditures for beneficiaries under the age of 65 who are covered by employer-sponsored private health insurance (Health Care Cost Institute website, November 4, 2012). The data contained in the DATAfile named DrugCost are consistent with the institutes findings concerning annual prescription costs per employee. Analyze the data using Excel and answer the following questions. a. Develop a 90% confidence interval for the annual cost of prescription drugs. b. Develop a 90% confidence interval for the amount of out-of-pocket expense per employee. c. What is your point estimate of the proportion of employees who incurred no prescription drug costs? d. Which, if either, of the confidence intervals in parts (a) and (b) has a larger margin of error. Why?A recent article reported that there are approximately 11 minutes of actual playing time in a typical National Football League (NFL) game (The Wall Street Journal, January 15, 2010). The article included information about the amount of time devoted to replays, the amount of time devoted to commercials, and the amount of time the players spend standing around between plays. Data consistent with the findings published in The Wall Street Journal are in the DATAfile named Standing. These data provide the amount of time players spend standing around between plays for a sample of 60 NFL games. a. Use the Standing data set to develop a point estimate of the number of minutes during an NFL game that players are standing around between plays. Compare this to the actual playing time reported in the article. Are you surprised? b. What is the sample standard deviation? c. Develop a 95% confidence interval for the number of minutes players spend standing around between plays.Mileage tests are conducted for a particular model of automobile. If a 98% confidence interval with a margin of error of 1 mile per gallon is desired, how many automobiles should be used in the test? Assume that preliminary mileage tests indicate the standard deviation is 2.6 miles per gallon.In developing patient appointment schedules, a medical center wants to estimate the mean time that a staff member spends with each patient. How large a sample should be taken if the desired margin of error is two minutes at a 95% level of confidence? How large a sample should be taken for a 99% level of confidence? Use a planning value for the population standard deviation of eight minutes.Annual salary plus bonus data for chief executive officers are presented in the business-Week Annual Pay Survey. A preliminary sample showed that the standard deviation is 675 with data provided in thousands of dollars. How many chief executive officers should be in a sample if we want to estimate the population mean annual salary plus bonus with a margin of error of 100,000? (Note: The desired margin of error would be E = 100 if the data are in thousands of dollars.) Use 95% confidence.The National Center for Education Statistics reported that 47% of college students work to pay for tuition and living expenses. Assume that a sample of 450 college students was used in the study. a. Provide a 95% confidence interval for the population proportion of college students who work to pay for tuition and living expenses. b. Provide a 99% confidence interval for the population proportion of college students who work to pay for tuition and living expenses. c. What happens to the margin of error as the confidence is increased from 95% to 99%?A USA Today/CNN/Gallup survey of 369 working parents found 200 who said they spend too little time with their children because of work commitments. a. What is the point estimate of the proportion of the population of working parents who feel they spend too little time with their children because of work commitments? b. At 95% confidence, what is the margin of error? c. What is the 95% confidence interval estimate of the population proportion of working parents who feel they spend too little time with their children because of work commitments?The Pew Research Center has conducted extensive research on the young adult population (Pew Research website, November 6, 2012). One finding was that 93% of adults aged 18 to 29 use the Internet. Another finding was that 21% of those aged 18 to 28 are married. Assume the sample size associated with both findings is 500. a. Develop a 95% confidence interval for the proportion of adults aged 18 to 29 who use the Internet. b. Develop a 99% confidence interval for the proportion of adults aged 18 to 28 who are married. c. In which case, part (a) or part (b), is the margin of error larger? Explain why.A survey of 750 likely voters in Ohio was conducted by the Rasmussen Poll just prior to the general election (Rasmussen Reports website, November 4, 2012). The state of the economy was thought to be an important determinant of how people would vote. Among other things, the survey found that 165 of the respondents rated the economy as good or excellent and 315 rated the economy as poor. a. Develop a point estimate of the proportion of likely voters in Ohio who rated the economy as good or excellent. b. Construct a 95% confidence interval for the proportion of likely voters in Ohio who rated the economy as good or excellent. c. Construct a 95% confidence interval for the proportion of likely voters in Ohio who rated the economy as poor. d. Which of the confidence intervals in parts (b) and (c) is wider? Why?The Centers for Disease Control reported the percentage of people 18 years of age and older who smoke (CDC website, December 14, 2014). Suppose that a study designed to collect new data on smokers and nonsmokers uses a preliminary estimate of the proportion who smoke of .30. a. How large a sample should be taken to estimate the proportion of smokers in the population with a margin of error of .02? Use 95% confidence. b. Assume that the study uses your sample size recommendation in part (a) and finds 520 smokers. What is the point estimate of the proportion of smokers in the population? c. What is the 95% confidence interval for the proportion of smokers in the population?A well-known bank credit card firm wishes to estimate the proportion of credit card holders who carry a nonzero balance at the end of the month and incur an interest charge. Assume that the desired margin of error is .03 at 98% confidence. a. How large a sample should be selected if it is anticipated that roughly 70% of the firms card holders carry a nonzero balance at the end of the month? b. How large a sample should be selected if no planning value for the proportion could be specified?Workers in several industries were surveyed to determine the proportion of workers who feel their industry is understaffed. In the government sector, 37% of the respondents said they were understaffed, in the health care sector 33% said they were understaffed, and in the education sector 28% said they were understaffed (USA Today, January 11, 2010). Suppose that 200 workers were surveyed in each industry. a. Construct a 95% confidence interval for the proportion of workers in each of these industries who feel their industry is understaffed. b. Assuming the same sample size will be used in each industry, how large would the sample need to be to ensure that the margin of error is .05 or less for each of the three confidence intervals?Although airline schedules and cost are important factors for business travelers when choosing an airline carrier, a USA Today survey found that business travelers list an airlines frequent flyer program as the most important factor. From a sample of n = 1993 business travelers who responded to the survey, 618 listed a frequent flyer program as the most important factor. a. What is the point estimate of the proportion of the population of business travellers who believe a frequent flyer program is the most important factor when choosing an airline carrier? b. Develop a 95% confidence interval estimate of the population proportion. c. How large a sample would be required to report the margin of error of .01 at 95% confidence? Would you recommend that USA Today attempt to provide this degree of precision? Why or why not?Young Professional Magazine Young Professional magazine was developed for a target audience of recent college graduates who are in their first 10 years in a business/professional career. In its two years of publication, the magazine has been fairly successful. Now the publisher is interested in expanding the magazines advertising base. Potential advertisers continually ask about the demographics and interests of subscribers to Young Professional. To collect this information. the magazine commissioned a survey to develop a profile of its subscribers. The survey results will be used to help the magazine choose articles of interest and provide advertisers with a profile of subscribers. As a new employee of the magazine, you have been asked to help analyze the survey results. Some of the survey questions follow: Professional 1. What is your age? 2. Are you: Male_______ Female_______ 3. Do you plan to make any real estate purchases in the next two years? Yes____ No____ 4. What is the approximate total value of financial investments, exclusive of your home, owned by you or members of your household? 5. How many stock/bond/mutual fund transactions have you made in the past year? 6. Do you have broadband access to the Internet at home? Yes____ No____ 7. Please indicate your total household income last year._______ 8. Do you have children? Yes____ No____ The file entitled Professional contains the responses to these questions. Table 8.6 shows the portion of the file pertaining to the first five survey respondents. Managerial Report Prepare a managerial report summarizing the results of the survey. In addition to statistical summaries, discuss how the magazine might use these results to attract advertisers. You might also comment on how the survey results could be used by the magazines editors to identify topics that would be of interest to readers. Your report should address the following issues, but do not limit your analysis to just these areas. 1. Develop appropriate descriptive statistics to summarize the data. 2. Develop 95% confidence intervals for the mean age and household income of subscribers. 3. Develop 95% confidence intervals for the proportion of subscribers who have broadband access at home and the proportion of subscribers who have children. 4. Would Young Professional be a good advertising outlet for online brokers? Justify your conclusion with statistical data. 5. Would this magazine be a good place to advertise for companies selling educational software and computer games for young children? 6. Comment on the types of articles you believe would be of interest to readers of Young Professional.Gulf Real Estate Properties Gulf Real Estate Properties, Inc., is a real estate firm located in southwest Florida. The company, which advertises itself as expert in the real estate market, monitors condominium sales by collecting data on location, list price, sale price, and number of days it takes to sell each unit. Each condominium is classified as Gulf View if it is located directly on the Gulf of Mexico or No Gulf View if it is located on the bay or a golf course, near but not on the Gulf. Sample data from the Multiple Listing Service in Naples, Florida, provided recent sales data for 40 Gulf view condominiums and 18 No Gulf view condominiums. Prices are in thousands of dollars. The data are shown in Table 8.7. Managerial Report 1. Use appropriate descriptive statistics to summarize each of the three variables for the 40 Gulf View condominiums. 2. Use appropriate descriptive statistics to summarize each of the three variables for the 18 No Gulf View condominiums. 3. Compare your summary results. Discuss any specific statistical results that would help a real estate agent understand the condominium market. 4. Develop a 95% confidence interval estimate of the population mean sales price and population mean number of days to sell for Gulf View condominiums. Interpret your results. 5. Develop a 95% confidence interval estimate of the population mean sales price and population mean number of days to sell for No Gulf View condominiums. Interpret your results. 6. Assume the branch manager requested estimates of the mean selling price of Gulf View condominiums with a margin of error of 40,000 and the mean selling price of No Gulf View condominiums with a margin of error of 15,000. Using 95% confidence, how large should the sample sizes be? 7. Gulf Real Estate Properties just signed contracts for two new listings: a Gulf View condominium with a list price of 589,000 and a No Gulf View condominium with a list price of 285,000. What is your estimate of the final selling price and number of days required to sell each of these units?Metropolitan Research, Inc. Metropolitan Research, Inc., a consumer research organization, conducts surveys designed to evaluate a wide variety of products and services available to consumers. In one particular study, Metropolitan looked at consumer satisfaction with the performance of automobiles produced by a major Detroit manufacturer. A questionnaire sent to owners of one of the manufacturers full-sized cars revealed several complaints about early transmission problems. To learn more about the transmission failures, Metropolitan used a sample of actual transmission repairs provided by a transmission repair firm in the Detroit area. The following data show the actual number of miles driven for 50 vehicles at the time of transmission failure. TABLE 8.7 SALES DATA FOR GULF REAL ESTATE PROPERTIES Managerial Report 1. Use appropriate descriptive statistics to summarize the transmission failure data. 2. Develop a 95% confidence interval for the mean number of miles driven until transmission failure for the population of automobiles with transmission failure. Provide a managerial interpretation of the interval estimate. 3. Discuss the implication of your statistical findings in terms of the belief that some owners of the automobiles experienced early transmission failures. 4. How many repair records should be sampled if the research firm wants the population mean number of miles driven until transmission failure to be estimated with a margin of error of 5000 miles? Use 95% confidence. 5. What other information would you like to gather to evaluate the transmission failure problem more fully?The manager of the Danvers-Hilton Resort Hotel stated that the mean guest bill for aweekend is 600 or less. A member of the hotels accounting staff noticed that the totalcharges for guest bills have been increasing in recent months. The accountant will use asample of future weekend guest bills to test the managers claim. a. Which form of the hypotheses should be used to test the managers claim? Explain. H0: 600 H0: 600 H0: = 600 Ha: 600 Ha: 600 Ha: 600 b. What conclusion is appropriate when H0 cannot be rejected? c. What conclusion is appropriate when H0 can be rejected?The manager of an automobile dealership is considering a new bonus plan designed toincrease sales volume. Currently, the mean sales volume is 14 automobiles per month. Themanager wants to conduct a research study to see whether the new bonus plan increasessales volume. To collect data on the plan, a sample of sales personnel will be allowed tosell under the new bonus plan for a one-month period. a. Develop the null and alternative hypotheses most appropriate for this situation. b. Comment on the conclusion when H0 cannot be rejected. c. comment on the conclusion when H0 can be rejected.A production line operation is designed to fill cartons with laundry detergent to a meanweight of 32 ounces. A sample of cartons is periodically selected and weighed to determine whether underfilling or overfilling is occurring. If the sample data lead to a conclusion of underfilling or overfilling, the production line will be shut down and adjusted toobtain proper filling. a. Formulate the null and alternative hypotheses that will help in deciding whether to shutdown and adjust the production line. b. Comment on the conclusion and the decision when H0 cannot be rejected. c. Comment on the conclusion and the decision when H0 can be rejected.Because of high production-changeover time and costs, a director of manufacturing mustconvince management that a proposed manufacturing method reduces costs before thenew method can be implemented. The current production method operates with a meancost of 220 per hour. A research study will measure the cost of the new method over asample production period. a. Develop the null and alternative hypotheses most appropriate for this study. b. Comment on the conclusion when H0 cannot be rejected. c. Comment on the conclusion when H0 can be rejected.Duke Energy reported that the cost of electricity for an efficient home in a particularneighborhood of Cincinnati, Ohio, was 104 per month (Home Energy Report, dukeenergy, March 2012). A researcher believes that the cost of electricity for a comparableneighborhood in Chicago, Illinois, is higher. a sample of homes in this Chicago neighborhood will be taken and the sample mean monthly cost of electricity will be used to test thefollowing null and alternative hypotheses. H0: 104 Ha: 104 a. Assume the sample data led to rejection of the null hypothesis. What would be yourconclusion about the cost of electricity in the Chicago neighborhood? b. What is the Type I error in this situation? what are the consequences of making thiserror? c. What is the type II error in this situation? what are the consequences of making thiserror?The label on a 3-quart container of orange juice states that the orange juice contains anaverage of 1 gram of fat or less. Answer the following questions for a hypothesis test thatcould be used to test the claim on the label. a. Develop the appropriate null and alternative hypotheses. b. What is the Type I error in this situation? What are the consequences of making this error? c. What is the Type II error in this situation? What are the consequences of making this error?Carpetland salespersons average 8000 per week in sales. Steve Contois, the firms vicepresident, proposes a compensation plan with new selling incentives. Steve hopes that theresults of a trial selling period will enable him to conclude that the compensation planincreases the average sales per salesperson. a. Develop the appropriate null and alternative hypotheses. b. What is the Type I error in this situation? What are the consequences of making this error? c. What is the Type II error in this situation? What are the consequences of making this error?Suppose a new production method will be implemented if a hypothesis test supports theconclusion that the new method reduces the mean operating cost per hour. a. State the appropriate null and alternative hypotheses if the mean cost for the currentproduction method is 220 per hour. b. What is the Type I error in this situation? What are the consequences of making this error? c. What is the Type II error in this situation? What are the consequences of making this error?consider the following hypothesis test: H0: 20 Ha: 20 A sample of 50 provided a sample mean of 19.4. The population standard deviation is 2. a. Compute the value of the test statistic. b. What is the p-value? c. Using = .05, what is your conclusion? d. What is the rejection rule using the critical value? What is your conclusion?Consider the following hypothesis test: H0: 25 Ha: 25 A sample of 40 provided a sample mean of 26.4. The population standard deviation is 6. a. Compute the value of the test statistic. b. What is the p-value? c. At = .01, what is your conclusion? d. What is the rejection rule using the critical value? What is your conclusion?Consider the following hypothesis test: H0: = 15 Ha: 15 A sample of 50 provided a sample mean of 14.15. The population standard deviation is 3. a. Compute the value of the test statistic. b. What is the p-value? c. At = .05, what is your conclusion? d. What is the rejection rule using the critical value? What is your conclusion?Consider the following hypothesis test: H0: 80 Ha: 80 A sample of 100 is used and the population standard deviation is 12. Compute the p-value and state your conclusion for each of the following sample results. Use = .01. a. x = 78.5 b. x = 77 c. x= 75.5 d. x = 81Consider the following hypothesis test: H0: 50 Ha: 50 A sample of 60 is used and the population standard deviation is 8. Use the critical value approach to state your conclusion for each of the following sample results. Use = .05. a. x = 52.5 b. x = 51 c. x = 51.8Consider the following hypothesis test: H0: = 22 Ha: 22 A sample of 75 is used and the population standard deviation is 10. Compute the p-value and state your conclusion for each of the following sample results. Use = .01. a. x = 23 b. x = 25.1 c. x = 20Individuals filing federal income tax returns prior to March 31 received an averagerefund of 1056. Consider the population of last-minute filers who mail their taxreturn during the last five days of the income tax period (typically April 10 to April 15). a. A researcher suggests that a reason individuals wait until the last five days is that onaverage these individuals receive lower refunds than do early filers. Develop appropriate hypotheses such that rejection of H0 will support the researchers contention. b. For a sample of 400 individuals who filed a tax return between April 10 and 15, thesample mean refund was 910. Based on prior experience a population standard deviation of = 1600 may be assumed. What is the p-value? c. At = .05, what is your conclusion? d. Repeat the preceding hypothesis test using the critical value approach.In a study entitled How Undergraduate Students Use Credit Cards, it was reported thatundergraduate students have a mean credit card balance of 3173. This figure was anall-time high and had increased 44% over the previous five years. Assume that a currentstudy is being conducted to determine if it can be concluded that the mean credit card balance for undergraduate students has continued to increase compared to the original report.Based on previous studies, use a population standard deviation = 1000. a. State the null and alternative hypotheses. b. What is the p-value for a sample of 180 undergraduate students with a sample meancredit card balance of 3325? c. Using a .05 level of significance, what is your conclusion?The mean hourly wage for employees in goods-producing industries is currently 24.57(Bureau of Labor Statistics website, April, 12, 2012). Suppose we take a sample ofemployees from the manufacturing industry to see if the mean hourly wage differs fromthe reported mean of 24.57 for the goods-producing industries. a. State the null and alternative hypotheses we should use to test whether the populationmean hourly wage in the manufacturing industry differs from the population meanhourly wage in the goods-producing industries. b. Suppose a sample of 30 employees from the manufacturing industry showed a samplemean of 23.89 per hour. Assume a population standard deviation of 2.40 per hourand compute the p-value. c. With = .05 as the level of significance, what is your conclusion? d. Repeat the preceding hypothesis test using the critical value approach.Young millennials, adults aged 18 to 34, are viewed as the future of the restaurant industry. During 2011, this group consumed a mean of 192 restaurant meals per person (NPDGroup website, November 7, 2012). Conduct a hypothesis test to determine if the pooreconomy caused a change in the frequency of consuming restaurant meals by young millennials in 2012. a. Formulate hypotheses that can be used to determine whether the annual mean numberof restaurant meals per person has changed for young millennials in 2012. b. Based on a sample, the NPD Group stated that the mean number of restaurant mealsconsumed by young millennials in 2012 was 182. Assume the sample size was 150and that, based on past studies, the population standard deviation can be assumed tobe = 55. Use the sample results to compute the test statistic and p-value for yourhypothesis test. c. At = .05, what is your conclusion?The Internal Revenue Service (IRS) provides a toll-free help line for taxpayers to call inand get answers to questions as they prepare their tax returns. In recent years, the IRShas been inundated with taxpayer calls and has redesigned its phone service as well asposted answers to frequently asked questions on its website (The Cincinnati Enquirer,January 7, 2010). According to a report by a taxpayer advocate, callers using the newsystem can expect to wait on hold for an unreasonably long time of 12 minutes beforebeing able to talk to an IRS employee. Suppose you select a sample of 50 callers afterthe new phone service has been implemented; the sample results show a mean waitingtime of 10 minutes before an IRS employee comes on the line. Based upon data from pastyears, you decide it is reasonable to assume that the standard deviation of waiting timeis 8 minutes. using your sample results, can you conclude that the actual mean waitingtime turned out to be significantly less than the 12-minute claim made by the taxpayeradvocate? Use = .05.Annual expenditure for prescription drugs was 838 per person in the Northeast of thecountry (Hospital Care Cost Institute website, November 7, 2012). A sample of 60 individuals in the Midwest showed a per person annual expenditure for prescription drugsof 745. Use a population standard deviation of 300 to answer the following questions. a. Formulate hypotheses for a test to determine whether the sample data support theconclusion that the population annual expenditure for prescription drugs per person islower in the Midwest than in the Northeast. b. What is the value of the test statistic? c. What is the p-value? d. At = .01, what is your conclusion?Fowle Marketing Research, Inc., bases charges to a client on the assumption that telephone surveys can be completed in a mean time of 15 minutes or less. If a longer meansurvey time is necessary, a premium rate is charged. A sample of 35 surveys providedthe survey times shown in the file named Fowle. Based upon past studies, the populationstandard deviation is assumed known with = 4 minutes. Is the premium rate justified? a. Formulate the null and alternative hypotheses for this application. b. Compute the value of the test statistic. c. What is the p-value? d. At = .01, what is your conclusion?CCN and ActMedia provided a television channel targeted to individuals waiting in supermarket checkout lines. The channel showed news, short features, and advertisements. Thelength of the program was based on the assumption that the population mean time a shopper stands in a supermarket checkout line is 8 minutes. A sample of actual waiting timeswill be used to test this assumption and determine whether actual mean waiting time differs from this standard. a. Formulate the hypotheses for this application. b. A sample of 120 shoppers showed a sample mean waiting time of 8.4 minutes. Assumea population standard deviation of = 3.2 minutes. What is the p-value? c. At = .05, what is your conclusion? d. Compute a 95% confidence interval for the population mean. Does it support yourconclusion?Consider the following hypothesis test: H0: 12 Ha: 12 A sample of 25 provided a sample mean x= 14 and a sample standard deviation s = 4.32. a. Compute the value of the test statistic. b. Use the t distribution table (Table 2 in Appendix B) to compute a range for the p-value. c. At = .05, what is your conclusion? d. What is the rejection rule using the critical value? What is your conclusion?Consider the following hypothesis test: H0: = 18 Ha: 18 A sample of 48 provided a sample mean x= 17 and a sample standard deviation s = 4.5. a. Compute the value of the test statistic. b. Use the t distribution table (Table 2 in Appendix B) to compute a range for the p-value. c. At = .05, what is your conclusion? d. What is the rejection rule using the critical value? What is your conclusion?Consider the following hypothesis test: H0: 45 Ha: 45 A sample of 36 is used. Identify the p-value and state your conclusion for each of the following sample results. Use = .01. a. x = 44 and s = 5.2 b. x = 43 and s = 4.6 c. x = 46 and s = 5.0Consider the following hypothesis test: H0: = 100 Ha: 100 A sample of 65 is used. Identify the p-value and state your conclusion for each of the following sample results. Use = .05. a. x = 103 and s = 11.5 b. x = 96.5 and s = 11.0 c. x = 102 and s = 10.5Which is cheaper: eating out or dining in? The mean cost of a flank steak, broccoli, andrice bought at the grocery store is 13.04 (Money.msn website, November 7, 2012). Asample of 100 neighborhood restaurants showed a mean price of 12.75 and a standarddeviation of 2 for a comparable restaurant meal. a. Develop appropriate hypotheses for a test to determine whether the sample data support the conclusion that the mean cost of a restaurant meal is less than fixing a comparable meal at home. b. Using the sample from the 100 restaurants, what is the p-value? c. At = .05, what is your conclusion? d. Repeat the preceding hypothesis test using the critical value approach.A shareholders group, in lodging a protest, claimed that the mean tenure for a chief exective office (CEO) was at least nine years. A survey of companies reported in The WallStreet Journal found a sample mean tenure of x= 7.27 years for CEOs with a standarddeviation of s = 6.38 years. a. Formulate hypotheses that can be used to challenge the validity of the claim made bythe shareholders group. b. Assume 85 companies were included in the sample. what is the p-value for yourhypothesis test? c. At = .01, what is your conclusion?The national mean annual salary for a school administrator is 90,000 a year (The Cincinnati Enquirer, April 7, 2012). A school official took a sample of 25 school administratorsin the state of Ohio to learn about salaries in that state to see if they differed from thenational average. a. Formulate hypotheses that can be used to determine whether the population meanannual administrator salary in Ohio differs from the national mean of 90,000. b. The sample data for 25 Ohio administrators is contained in the file named Administrator. What is the p-value for your hypothesis test in part (a)? c. At = .05, can your null hypothesis be rejected? What is your conclusion? d. Repeat the preceding hypothesis test using the critical value approach.The time married men with children spend on child care averages 6.4 hours perweek (Time, March 12, 2012). You belong to a professional group on family practicesthat would like to do its own study to determine if the time married men in your area spendon child care per week differs from the reported mean of 6.4 hours per week. A sampleof 40 married couples will be used with the data collected showing the hours per week thehusband spends on child care. The sample data are contained in the file ChildCare. a. What are the hypotheses if your group would like to determine if the populationmean number of hours married men are spending in child care differs from the meanreported by Time in your area? b. What is the sample mean and the p-value? c. Select your own level of significance. What is your conclusion?The Coca-Cola Company reported that the mean per capita annual sales of its beverages inthe United States was 423 eight-ounce servings (Coca-Cola Company website, February3, 2009). Suppose you are curious whether the consumption of Coca-Cola beverages ishigher in Atlanta, Georgia, the location of Coca-Colas corporate headquarters. A sampleof 36 individuals from the Atlanta area showed a sample mean annual consumption of460.4 eight-ounce servings with a standard deviation of s = 101.9 ounces. Using = .05,do the sample results support the conclusion that mean annual consumption of Coca-Colabeverage products is higher in Atlanta?According to the National Automobile Dealers Association, the mean price for used carsis 10,192. A manager of a Kansas City used car dealership reviewed a sample of 50recent used car sales at the dealership in an attempt to determine whether the populationmean price for used cars at this particular dealership differed from the national mean. Theprices for the sample of 50 cars are shown in the file named UsedCars. a. Formulate the hypotheses that can be used to determine whether a difference exists inthe mean price for used cars at the dealership. b. What is the p-value? c. At = .05, what is your conclusion?The mean annual premium for automobile insurance in the United States is 1503(Insure.com website, March 6, 2014). Being from Pennsylvania, you believe automobileinsurance is cheaper there and wish to develop statistical support for your opinion. Asample of 25 automobile insurance policies from the state of Pennsylvania showed amean annual premium of 1440 with a standard deviation of s = 165. a. Develop a hypothesis test that can be used to determine whether the mean annualpremium in Pennsylvania is lower than the national mean annual premium. b. What is a point estimate of the difference between the mean annual premium in Pennsylvania and the national mean? c. At = .05, test for a significant difference. What is your conclusion?Joans Nursery specializes in custom-designed landscaping for residential areas. Theestimated labor cost associated with a particular landscaping proposal is based onthe number of plantings of trees, shrubs, and so on to be used for the project. Forcost-estimating purposes, managers use two hours of labor time for the planting of a medium-sized tree. Actual times from a sample of 10 plantings during the past month follow (times in hours). With a .05 level of significance, test to see whether the mean tree-planting time differs from two hours. a. State the null and alternative hypotheses. b. Compute the sample mean. c. Compute the sample standard deviation. d. What is the p-value? e. What is your conclusion?Consider the following hypothesis test: H0: p = .20 Ha: p .20 A sample of 400 provided a sample proportion p= .175. a. Compute the value of the test statistic. b. What is the p-value? c. At = .05, what is your conclusion? d. What is the rejection rule using the critical value? What is your conclusion?Consider the following hypothesis test: H0: p .75 Ha: p .75 A sample of 300 items was selected. Compute the p-value and state your conclusion for each of the following sample results. Use = .05. a. p = .68 b. p = .72 c. p = .70 d. p = .77The U.S. Bureau of Labor Statistics reports that 11.3% of U.S. workers belong to unions(BLS website, January 2014). Suppose a sample of 400 U.S. workers is collected in 2014to determine whether union efforts to organize have increased union membership. a. Formulate the hypotheses that can be used to determine whether union membershipincreased in 2014. b. If the sample results show that 52 of the workers belonged to unions, what is thep-value for your hypothesis test? c. At = .05, what is your conclusion?A study by Consumer Reports showed that 64% of supermarket shoppers believe supermarket brands to be as good as national name brands. To investigate whether this resultapplies to its own product, the manufacturer of a national name-brand ketchup asked asample of shoppers whether they believed that supermarket ketchup was as good as thenational brand ketchup. a. Formulate the hypotheses that could be used to determine whether the percentage ofsupermarket shoppers who believe that the supermarket ketchup was as good as thenational brand ketchup differed from 64%. b. If a sample of 100 shoppers showed 52 stating that the supermarket brand was as goodas the national brand, what is the p-value? c. At = .05, what is your conclusion? d. Should the national brand ketchup manufacturer be pleased with this conclusion?Explain.What percentage of the population live in their state of birth? According to the U.S.Census Bureaus American Community Survey, the figure ranges from 25% in Nevadato 78.7% in Louisiana (AARP Bulletin, March 2014). The average percentage across allstates and the District of Columbia is 57.7%. The data in the DATAfile Homestate areconsistent with the findings in the American Community Survey. The data are for a random sample of 120 Arkansas residents and for a random sample of 180 Virginia residents. a. Formulate hypotheses that can be used to determine whether the percentage of stay-at-home residents in the two states differs from the overall average of 57.7%. b. Estimate the proportion of stay-at-home residents in Arkansas. Does this proportiondiffer significantly from the mean proportion for all states? Use = .05. c. Estimate the proportion of stay-at-home residents in Virginia. Does this proportiondiffer significantly from the mean proportion for all states? use = .05. d. Would you expect the proportion of stay-at-home residents to be higher in Virginia thanin Arkansas? Support your conclusion with the results obtained in parts (b) and (c).Last year, 46% of business owners gave a holiday gift to their employees. A survey ofbusiness owners conducted this year indicates that 35% plan to provide a holiday giftto their employees. Suppose the survey results are based on a sample of 60 businessowners. a. How many business owners in the survey plan to provide a holiday gift to theiremployees this year? b. Suppose the business owners in the sample did as they plan. Compute the p-value fora hypothesis test that can be used to determine if the proportion of business ownersproviding holiday gifts had decreased from last year. c. Using a .05 level of significance, would you conclude that the proportion of businessowners providing gifts decreased? what is the smallest level of significance for whichyou could draw such a conclusion?Ten years ago 53% of American families owned stocks or stock funds. Sample data collected by the Investment Company Institute indicate that the percentage is now 46% (TheWall Street Journal, October 5, 2012). a. Develop appropriate hypotheses such that rejection of H0 will support the conclusionthat a smaller proportion of American families own stocks or stock funds in 2012 than10 years ago. b. Assume the Investment Company Institute sampled 300 American families to estimatethat the percent owning stocks or stock funds was 46% in 2012. What is the p-valuefor your hypothesis test? c. At = .01, what is your conclusion?According to the University of Nevada Center for Logistics Management, 6% of all merchandise sold in the United States gets returned. A Houston department store sampled 80items sold in January and found that 12 of the items were returned. a. Construct a point estimate of the proportion of items returned for the population ofsales transactions at the Houston store. b. Construct a 95% confidence interval for the proportion of returns at the Houston store. c. Is the proportion of returns at the Houston store significantly different from the returnsfor the nation as a whole? Provide statistical support for your answer.Eagle Outfitters is a chain of stores specializing in outdoor apparel and camping gear.They are considering a promotion that involves mailing discount coupons to all their creditcard customers. This promotion will be considered a success if more than 10% of thosereceiving the coupons use them. Before going national with the promotion, coupons weresent to a sample of 100 credit card customers. a. Develop hypotheses that can be used to test whether the population proportion of thosewho will use the coupons is sufficient to go national. b. The file Eagle contains the sample data. Develop a point estimate of the populationproportion. c. Use = .05 to conduct your hypothesis test. Should Eagle go national with thepromotion?One of the reasons health care costs have been rising rapidly in recent years is the increasing cost of malpractice insurance for physicians. Also, fear of being sued causes doctors torun more precautionary tests (possibly unnecessary) just to make sure they are not guilty of missing something (Readers Digest, October 2012). These precautionary tests also add to health care costs. Data in the DATAfile named LawSuit are consistent with findings inthe Readers Digest article and can be used to estimate the proportion of physicians overthe age of 55 who have been sued at least once. a. Formulate hypotheses that can be used to see if these data can support a finding thatmore than half of physicians over the age of 55 have been sued at least once. b. Use Excel and the DATAfile named LawSuit to compute the sample proportion ofphysicians over the age of 55 who have been sued at least once. What is the p-valuefor your hypothesis test? c. At = .01, what is your conclusion?The American Association of Individual Investors conducts a weekly survey of its members to measure the percent who are bullish, bearish, and neutral on the stock market for thenext six months. For the week ending November 7, 2012, the survey results showed 38.5%bullish, 21.6% neutral, and 39.9% bearish (AAII website, November 12, 2012). Assumethese results are based on a sample of 300 AAII members. a. Over the long term, the proportion of bullish AAII members is .39. Conduct a hypothesis test at the 5% level of significance to see if the current sample results show thatbullish sentiment differs from its long term average of .39. What are your findings? b. Over the long term, the proportion of bearish AAII members is .30. Conduct a hypothesis test at the 1% level of significance to see if the current sample results show thatbearish sentiment is above its long term average of .30. What are your findings? c. Would you feel comfortable extending these results to all investors? Why or why not?consider the following hypothesis test. H0: 10 Ha: 10 The sample size is 120 and the population standard deviation is assumed known with = 5. Use = .05. a. If the population mean is 9, what is the probability that the sample mean leads to theconclusion do not reject H0? b. What type of error would be made if the actual population mean is 9 and we concludethat H0: 10 is true? c. What is the probability of making a Type II error if the actual population mean is 8?Consider the following hypothesis test. H0: = 20 Ha: 20 A sample of 200 items will be taken and the population standard deviation is = 10.Use = .05. Compute the probability of making a Type II error if the population mean is: a. = 18.0 b. = 22.5 c. = 21.0Fowle Marketing Research, Inc., bases charges to a client on the assumption that telephone surveys can be completed within 15 minutes or less. If more time is required, apremium rate is charged. With a sample of 35 surveys, a population standard deviation of4 minutes, and a level of significance of .01, the sample mean will be used to test the nullhypothesis H0: 15. a. What is your interpretation of the Type II error for this problem? What is its impacton the firm? b. What is the probability of making a Type II error when the actual mean time is = 17 minutes? c. What is the probability of making a Type II error when the actual mean time is = 18 minutes? d. Sketch the general shape of the power curve for this test.A consumer research group is interested in testing an automobile manufacturers claim that a new economy model will travel at least 25 miles per gallon of gasoline (H0: 25). a. With a .02 level of significance and a sample of 30 cars, what is the rejection rule basedon the value of x for the test to determine whether the manufacturers claim should berejected? Assume that is 3 miles per gallon. b. What is the probability of committing a Type II error if the actual mileage is 23 milesper gallon? c. What is the probability of committing a Type II error if the actual mileage is 24 milesper gallon? d. What is the probability of committing a Type II error if the actual mileage is 25.5 milesper gallon?Young Adult magazine states the following hypotheses about the mean age of its subscribers. H0: = 28 Ha: 28 a. What would it mean to make a Type II error in this situation? b. The population standard deviation is assumed known at = 6 years and the samplesize is 100. With = .05, what is the probability of accepting H0 for equal to 26,27, 29, and 30? c. What is the power at = 26? What does this result tell you?Refer to exercise 48. Assume the firm selects a sample of 50 surveys and repeat parts (b)and (c). What observation can you make about how increasing the sample size affects theprobability of making a Type II error? 48. Fowle Marketing Research, Inc., bases charges to a client on the assumption that telephone surveys can be completed within 15 minutes or less. If more time is required, apremium rate is charged. With a sample of 35 surveys, a population standard deviation of4 minutes, and a level of significance of .01, the sample mean will be used to test the nullhypothesis H0: 15. a. What is your interpretation of the Type II error for this problem? What is its impacton the firm? b. What is the probability of making a Type II error when the actual mean time is = 17 minutes? c. What is the probability of making a Type II error when the actual mean time is = 18 minutes? d. Sketch the general shape of the power curve for this test.Sparr Investments, Inc., specializes in tax-deferred investment opportunities for its clients.Recently Sparr offered a payroll deduction investment program for the employees of aparticular company. Sparr estimates that the employees are currently averaging 100 orless per month in tax-deferred investments. A sample of 40 employees will be used to testSparrs hypothesis about the current level of investment activity among the population ofemployees. Assume the employee monthly tax-deferred investment amounts have a standard deviation of 75 and that a .05 level of significance will be used in the hypothesis test. a. What is the Type II error in this situation? b. What is the probability of the Type II error if the actual mean employee monthlyinvestment is 120? c. What is the probability of the Type II error if the actual mean employee monthlyinvestment is 130? d. Assume a sample size of 80 employees is used and repeat parts (b) and (c).Consider the following hypothesis test. H0: 10 Ha: 10 The sample size is 120 and the population standard deviation is 5. Use = .05. If the actual population mean is 9, the probability of a Type II error is .2912. Suppose the researcherwants to reduce the probability of a Type II error to .10 when the actual population meanis 9. What sample size is recommended?Consider the following hypothesis test. H0: = 20 Ha: 20 The population standard deviation is 10. Use = .05. How large a sample should be taken if the researcher is willing to accept a .05 probability of making a Type II error when theactual population mean is 22?Suppose the project director for the Hilltop Coffee study (see Section 9.3) asked for a .10probability of claiming that Hilltop was not in violation when it really was underfilling by1 ounce (a = 2.9375 pounds). What sample size would have been recommended?A special industrial battery must have a life of at least 400 hours. A hypothesis test is tobe conducted with a .02 level of significance. If the batteries from a particular productionrun have an actual mean use life of 385 hours, the production manager wants a samplingprocedure that only 10% of the time would show erroneously that the batch is acceptable.What sample size is recommended for the hypothesis test? Use 30 hours as an estimate ofthe population standard deviation.Young Adult magazine states the following hypotheses about the mean age of its subscribers. H0: = 28 Ha: 28 If the manager conducting the test will permit a .15 probability of making a Type II error when the true mean age is 29, what sample size should be selected? Assume = 6 and a.05 level of significance.An automobile mileage study tested the following hypotheses. Hypothesis Conclusion H0: 25 mpg Manufacturers claim supported Ha: 25 mpg Manufacturers claim rejected; average mileage per gallon less than stated For = 3 and a .02 level of significance, what sample size would be recommended if the researcher wants an 80% chance of detecting that is less than 25 miles per gallon when it is actually 24?A production line operates with a mean filling weight of 16 ounces per container. Overfilling or underfilling presents a serious problem and when detected requires the operator to shut down the production line to readjust the filling mechanism. From past data, a population standard deviation = .8 ounces is assumed. A quality control inspector selects a sample of 30 items every hour and at that time makes the decision of whether to shut down the line for readjustment. The level of significance is = .05. a. State the hypothesis test for this quality control application. b. If a sample mean of x=16.32 ounces were found, what is the p-value? What action would you recommend? c. If a sample mean of x=15.82 ounces were found, what is the p-value? What action would you recommend? d. Use the critical value approach. What is the rejection rule for the preceding hypothesis testing procedure? Repeat parts (b) and (c). Do you reach the same conclusion?At Western University the historical mean of scholarship examination scores for freshman applications is 900. A historical population standard deviation = 180 is assumed known. Each year, the assistant dean uses a sample of applications to determine whether the mean examination score for the new freshman applications has changed. a. State the hypotheses. b. What is the 95% confidence interval estimate of the population mean examination score if a sample of 200 applications provided a sample mean x=935? c. Use the confidence interval to conduct a hypothesis test. Using = .05, what is your conclusion? d. What is the p-value?Young children in the United States are exposed to an average of 4 hours of background television per day (CNN website, November 13, 2012). Having the television on in the background while children are doing other activities may have adverse consequences on a childs well-being. You have a research hypothesis that children from low-income families are exposed to more than 4 hours of daily background television. In order to test this hypothesis, you have collected a random sample of 60 children from low-income families and found that these children were exposed to a sample mean of 4.5 hours of daily background television. a. Develop hypotheses that can be used to test your research hypothesis. b. Based on a previous study, you are willing to assume that the population standard deviation is = 0.5 hours. What is the p-value based on your sample of 60 children from low-income families? c. Use = .01 as the level of significance. What is your conclusion?The Wall Street Journal reported that bachelors degree recipients with majors in business received average starting salaries of 53,900 in 2012 (The Wall Street Journal, March 17, 2014). The results for a sample of 100 business majors receiving a bachelors degree in 2013 showed a mean starting salary of 55,144 with a sample standard deviation of 5200. Conduct a hypothesis test to determine whether the mean starting salary for business majors in 2013 is greater than the mean starting salary in 2012. Use = .01 as the level of significance.Data from the Office for National Statistics show that the mean age at which men in Great Britain get married is 30.8 years (The Guardian, February 15, 2013). A news reporter noted that this represents a continuation of the trend of waiting until a later age to wed. A new sample of 47 recently wed British men provided their age at the time of marriage. These data are contained in the DATAfile BritainMarriages. Do these data indicate that the mean age of British men at the time of marriage exceeds the mean age in 2013? Test this hypothesis at = .05. What is your conclusion?.A recent issue of the AARP Bulletin reported that the average weekly pay for a woman with a high school school degree is 520 (AARP Bulletin, January-February, 2010). Suppose you would like to determine if the average weekly pay for all working women is significantly greater than that for women with a high school degree. Data providing the weekly pay for a sample of 50 working women are available in the file named WeeklyPay. These data are consistent with the findings reported in the AARP article. a. State the hypotheses that should be used to test whether the mean weekly pay for all women is significantly greater than the mean weekly pay for women with a high school degree. b. Use the data in the file named WeeklyPay to compute the sample mean, the test statistic, and the p-value. c. Use = .05. What is your conclusion? d. Repeat the hypothesis test using the critical value approach.The chamber of commerce of a Florida Gulf coast community advertises that area residential property is available at a mean cost of 125,000 or less per lot. Suppose a sample of 32 properties provided a sample mean of 130,000 per lot and a sample standard deviation of 12,500. Use a .05 level of significance to test the validity of the advertising claim.In Hamilton County, Ohio, the mean number of days needed to sell a house is 86 days (Cincinnati Multiple Listing Service, April, 2012). Data for the sale of 40 houses in a nearby county showed a sample mean of 80 days with a sample standard deviation of 20 days. Conduct a hypothesis test to determine whether the mean number of days until a house is sold is different than the Hamilton County mean of 86 days in the nearby county. Use = .05 for the level of significance, and state your conclusion.On December 25, 2009, an airline passenger was subdued while attempting to blow up a Northwest Airlines flight headed for Detroit, Michigan. The passenger had smuggled explosives hidden in his underwear past a metal detector at an airport screening facility. As a result, the Transportation Security Administration (TSA) proposed installing full-body scanners to replace the metal detectors at the nations largest airports. This proposal resulted in strong objections from privacy advocates who considered the scanners an invasion of privacy. On January 5-6, 2010, USA Today conducted a poll of 542 adults to learn what proportion of airline travelers approved of using full-body scanners (USA Today, January 11, 2010). The poll results showed that 455 of the respondents felt that full-body scanners would improve airline security and 423 indicated that they approved of using the devices. a. conduct a hypothesis test to determine if the results of the poll justify concluding that over 80% of airline travelers feel that the use of full-body scanners will improve airline security. Use = .05. b. Suppose the TSA will go forward with the installation and mandatory use of full-body scanners if over 75% of airline travelers approve of using the devices. You have been told to conduct a statistical analysis using the poll results to determine if the TSA should require mandatory use of the full-body scanners. Because this is viewed as a very sensitive decision, use = .01. What is your recommendation?An airline promotion to business travelers is based on the assumption that two-thirds of business travelers use a laptop computer on overnight business trips. a. State the hypotheses that can be used to test the assumption. b. What is the sample proportion from an American Express sponsored survey that found 355 of 546 business travelers use a laptop computer on overnight business trips? c. What is the p-value? d. Use = .05. What is your conclusion?Members of the millennial generation are continuing to be dependent on their parents (either living with or otherwise receiving support from parents) into early adulthood (The Enquirer, March 16, 2014). A family research organization has claimed that, in past generations, no more than 30% of individuals aged 18 to 32 continued to be dependent on their parents. Suppose that a sample of 400 individuals aged 18 to 32 showed that 136 of them continue to be dependent on their parents. a. Develop hypotheses for a test to determine whether the proportion of millennials continuing to be dependent on their parents is higher than for past generations. b. What is your point estimate of the proportion of millennials that are continuing to be dependent on their parents? c. What is the p-value provided by the sample data? d. What is your hypothesis testing conclusion? Use = .05 as the level of significance.The unemployment rate for 18- to 34-year-olds was reported to be 10.8% (The Cincinnati Enquirer, November 6, 2012). Assume that this report was based on a random sample of four hundred 18- to 34-year-olds. a. A political campaign manager wants to know if the sample results can be used to conclude that the unemployment rate for 18- to 34-years-olds is significantly higher than the unemployment rate for all adults. According to the Bureau of Labor Statistics, the unemployment rate for all adults was 7.9%. Develop a hypothesis test that can be used to see if the conclusion that the unemployment rate is higher for 18- to 34-year-olds can be supported. b. Use the sample data collected for the 18- to 34-year-olds to compute the p-value for the hypothesis test in part (a). Using = .05, what is your conclusion? c. Explain to the campaign manager what can be said about the observed level of significance for the hypothesis testing results using the p-value.A radio station in Myrtle Beach announced that at least 90% of the hotels and motels would be full for the Memorial Day weekend. The station advised listeners to make reservations in advance if they planned to be in the resort over the weekend. On Saturday night a sample of 58 hotels and motels showed 49 with a no-vacancy sign and 9 with vacancies. What is your reaction to the radio stations claim after seeing the sample evidence? Use = .05 in making the statistical test. What is the p-value?In recent years more people have been working past the age of 65. In 2005, 27% of people aged 65-69 worked. A recent report from the Organization for Economic Cooperation and Development (OECD) claimed that the percentage working had increased (USA Today, November 16, 2012). The findings reported by the OECD were consistent with taking a sample of 600 people aged 65-69 and finding that 180 of them were working. a. Develop a point estimate of the proportion of people aged 65-69 who are working. b. Set up a hypothesis test so that the rejection of H0 will allow you to conclude that the proportion of people aged 65-69 working has increased from 2005. c. Conduct your hypothesis test using = .05. What is your conclusion?Shorney Construction Company bids on projects assuming that the mean idle time per worker is 72 or fewer minutes per day. A sample of 30 construction workers will be used to test this assumption. Assume that the population standard deviation is 20 minutes. a. State the hypotheses to be tested. b. What is the probability of making a Type II error when the population mean idle time is 80 minutes? c. What is the probability of making a Type II error when the population mean idle time is 75 minutes? d. What is the probability of making a Type II error when the population mean idle time is 70 minutes? e. Sketch the power curve for this problem.A federal funding program is available to low-income neighborhoods. To qualify for the funding, a neighborhood must have a mean household income of less than 15,000 per year. Neighborhoods with mean annual household income of 15,000 or more do not qualify. Funding decisions are based on a sample of residents in the neighborhood. A hypothesis test with a .02 level of significance is conducted. If the funding guidelines call for a maximum probability of .05 of not funding a neighborhood with a mean annual household income of 14,000, what sample size should be used in the funding decision study? Use = 4000 as a planning value.H0: = 120 and Ha: 120 are used to test whether a bath soap production process is meeting the standard output of 120 bars per batch. Use a .05 level of significance for the test and a planning value of 5 for the standard deviation. a. If the mean output drops to 117 bars per batch, the firm wants to have a 98% chance of concluding that the standard production output is not being met. How large a sample should be selected? b. With your sample size from part (a), what is the probability of concluding that the process is operating satisfactorily for each of the following actual mean outputs: 117, 118, 119, 121, 122, and 123 bars per batch? That is, what is the probability of a Type II error in each case?Quality Associates, Inc. Quality Associates, Inc., a consulting firm, advises its clients about sampling and statistical procedures that can be used to control their manufacturing processes. In one particular application, a client gave Quality Associates a sample of 800 observations taken during a time in which that clients process was operating satisfactorily. The sample standard deviation for these data was .21; hence, with so much data, the population standard deviation was assumed to be .21. Quality Associates then suggested that random samples of size 30 be taken periodically to monitor the process on an ongoing basis. By analyzing the new samples, the client could quickly learn whether the process was operating satisfactorily. When the process was not operating satisfactorily, corrective action could be taken to eliminate the problem. The design specification indicated the mean for the process should be 12. The hypothesis test suggested by Quality Associates follows. H0: = 12 Ha: 12 Corrective action will be taken any time H0 is rejected. The following samples were collected at hourly intervals during the first day of operation of the new statistical process control procedure. These data are available in the data set Quality. Sample 1 Sample 2 Sample 3 Sample 4 11.55 11.62 11.91 12.02 11.62 11.69 11.36 12.02 11.52 11.59 11.75 12.05 11.75 11.82 11.95 12.18 11.90 11.97 12.14 12.11 11.64 11.71 11.72 12.07 11.80 11.87 11.61 12.05 12.03 12.10 11.85 11.64 11.94 12.01 12.16 12.39 11.92 11.99 11.91 11.65 12.13 12.20 12.12 12.11 12.09 12.16 11.61 11.90 11.93 12.00 12.21 12.22 12.21 12.28 11.56 11.88 12.32 12.39 11.95 12.03 11.93 12.00 12.01 12.35 11.85 11.92 12.06 12.09 11.76 11.83 11.76 11.77 12.16 12.23 11.82 12.20 11.77 11.84 12.12 11.79 12.00 12.07 11.60 12.30 12.04 12.11 11.95 12.27 11.98 12.05 11.96 12.29 12.30 12.37 12.22 12.47 12.18 12.25 11.75 12.03 11.97 12.04 11.96 12.17 12.17 12.24 11.95 11.94 11.85 11.92 11.89 11.97 12.30 12.37 11.88 12.23 12.15 12.22 11.93 12.25 Managerial Report 1. Conduct a hypothesis test for each sample at the .01 level of significance and determine what action, if any, should be taken. Provide the test statistic and p-value for each test. 2. Compute the standard deviation for each of the four samples. Does the assumption of .21 for the population standard deviation appear reasonable? 3. Compute limits for the sample mean x around = 12 such that, as long as a new sample mean is within those limits, the process will be considered to be operating satisfactorily. If x exceeds the upper limit or if x is below the lower limit, corrective action will be taken. These limits are referred to as upper and lower control limits for quality control purposes. 4. Discuss the implications of changing the level of significance to a larger value. What mistake or error could increase if the level of significance is increased?Ethical Behavior of Business Students at Bayview University During the global recession of 2008 and 2009, there were many accusations of unethical behavior by Wall Street executives, financial managers, and other corporate officers. At that time, an article appeared that suggested that part of the reason for such unethical business behavior may stem from the fact that cheating has become more prevalent among business students (Chronicle of Higher Education, February 10, 2009). The article reported that 56% of business students admitted to cheating at some time during their academic career as compared to 47% of nonbusiness students. Cheating has been a concern of the dean of the College of Business at Bayview University for several years. Some faculty members in the college believe that cheating is more widespread at Bayview than at other universities, while other faculty members think that cheating is not a major problem in the college. To resolve some of these issues, the dean commissioned a study to assess the current ethical behavior of business students at Bay-view. As part of this study, an anonymous exit survey was administered to a sample of 90 business students from this years graduating class. Responses to the following questions were used to obtain data regarding three types of cheating. During your time at Bayview, did you ever present work copied off the Internet as your own? Yes ______ No ______ During your time at Bayview, did you ever copy answers off another students exam? Yes ________ No _______ During your time at Bayview, did you ever collaborate with other students on projects that were supposed to be completed individually? Yes ________ No _______ Any student who answered Yes to one or more of these questions was considered to have been involved in some type of cheating. A portion of the data collected follows. The complete data set is in the file named Bayview. Student Copied from Internet Copied on Exam Collaborated on Individual project Gender 1 No No No Female 2 No No No Male 3 Yes No Yes Male 4 Yes Yes No Male 5 No No Yes Male 6 Yes No No Female . . . . . . . . . . 88 No No No Male 89 No Yes Yes Male 90 No No No Female Managerial Report Prepare a report for the dean of the college that summarizes your assessment of the nature of cheating by business students at Bayview University. Be sure to include the following items in your report. 1. Use descriptive statistics to summarize the data and comment on your findings. 2. Develop 95% confidence intervals for the proportion of all students, the proportion of male students, and the proportion of female students who were involved in some type of cheating. 3. Conduct a hypothesis test to determine if the proportion of business students at Bayview University who were involved in some type of cheating is less than that of business students at other institutions as reported by the Chronicle of Higher Education. 4. Conduct a hypothesis test to determine if the proportion of business students at Bay-view University who were involved in some form of cheating is less than that of non-business students at other institutions as reported by the Chronicle of higher Education. 5. What advice would you give to the dean based upon your analysis of the data?The following results come from two independent random samples taken of two populations. Sample 1 Sample 2 n1 = 50 n2 = 35 x1=13.6 x2=11.6 1 = 2.2 2 = 3.0 a. What is the point estimate of the difference between the two population means? b. Provide a 90% confidence interval for the difference between the two population means. c. Provide a 95% confidence interval for the difference between the two population means.Consider the following hypothesis test. H0:120Ha:120 The following results are for two independent samples taken from the two populations. Sample 1 Sample 2 n1 = 40 n2 = 50 x1=25.2 x2=22.8 1 = 5.2 2 = 6.0 a. What is the value of the test statistic? b. What is the p-value? c. With = .05, what is your hypothesis testing conclusion?Consider the following hypothesis test. H0:12=0Ha:120 The following results are for two independent samples taken from the two populations. Sample 1 Sample 2 n1 = 80 n2 = 70 x1=104 x2=106 1 = 8.4 2 = 7.6 a. What is the value of the test statistic? b. What is the p-value? c. With = .05, what is your hypothesis testing conclusion?Cond Nast Traveler conducts an annual survey in which readers rate their favorite cruise ship. All ships are rated on a 100-point scale, with higher values indicating better service. A sample of 37 ships that carry fewer than 500 passengers resulted in an average rating of 85.36, and a sample of 44 ships that carry 500 or more passengers provided an average rating of 81.40. Assume that the population standard deviation is 4.55 for ships that carry fewer than 500 passengers and 3.97 for ships that carry 500 or more passengers. a. What is the point estimate of the difference between the population mean rating for ships that carry fewer than 500 passengers and the population mean rating for ships that carry 500 or more passengers? b. At 95% confidence, what is the margin of error? c. What is a 95% confidence interval estimate of the difference between the population mean ratings for the two sizes of ships?The USA Today reports that the average expenditure on Valentines Day is 100.89. Do male and female consumers differ in the amounts they spend? The average expenditure in a sample survey of 40 male consumers was 135.67, and the average expenditure in a sample survey of 30 female consumers was 68.64. Based on past surveys, the standard deviation for male consumers is assumed to be 35, and the standard deviation for female consumers is assumed to be 20. a. What is the point estimate of the difference between the population mean expenditure for males and the population mean expenditure for females? b. At 99% confidence, what is the margin of error? c. Develop a 99% confidence interval for the difference between the two population means.Suppose that you are responsible for making arrangements for a business convention and that you have been charged with choosing a city for the convention that has the least expensive hotel rooms. You have narrowed your choices to Atlanta and Houston. The file named Hotel contains samples of prices for rooms in Atlanta and Houston that are consistent with the results reported by Smith Travel Research (SmartMoney, March 2009). Because considerable historical data on the prices of rooms in both cities are available, the population standard deviations for the prices can be assumed to be 20 in Atlanta and 25 in Houston. Based on the sample data, can you conclude that the mean price of a hotel room in Atlanta is lower than one in Houston?Consumer Reports uses a survey of readers to obtain customer satisfaction ratings for the nations largest retailers (Consumer Reports, March 2012). Each survey respondent is asked to rate a specified retailer in terms of six factors: quality of products, selection, value, checkout efficiency, service, and store layout. An overall satisfaction score summarizes the rating for each respondent with 100 meaning the respondent is completely satisfied in terms of all six factors. Sample data representative of independent samples of Target and Walmart customers are shown below. Target Wahmart n1 = 25 n2 = 30 x1=79 x2=71 a. Formulate the null and alternative hypotheses to test whether there is a difference between the population mean customer satisfaction scores for the two retailers. b. Assume that experience with the Consumer Reports satisfaction rating scale indicates that a population standard deviation of 12 is a reasonable assumption for both retailers. Conduct the hypothesis test and report the p-value. At a .05 level of significance what is your conclusion? c. Which retailer, if either, appears to have the greater customer satisfaction? Provide a 95% confidence interval for the difference between the population mean customer satisfaction scores for the two retailers.Will improving customer service result in higher stock prices for the companies providing the better service? When a companys satisfaction score has improved over the prior years results and is above the national average (75.7), studies show its shares have a good chance of outperforming the broad stock market in the long run. The following satisfaction scores of three companies for the 4th quarters of two previous years were obtained from the American Customer Satisfaction Index. Assume that the scores are based on a poll of 60 customers from each company. Because the polling has been done for several years, the standard deviation can be assumed to equal 6 points in each case. Company Year 1 Year 2 Rite Aid 73 76 Expedia 75 77 J.C. Penney 77 78 a. For Rite Aid, is the increase in the satisfaction score from year 1 to year 2 statistically significant? Use = .05. What can you conclude? b. Can you conclude that the year 2 score for Rite Aid is above the national average of 75.7? Use = .05. c. For Expedia, is the increase from year 1 to year 2 statistically significant? Use = .05. d. When conducting a hypothesis test with the values given for the standard deviation, sample size, and , how large must the increase from year 1 to year 2 be for it to be statistically significant? e. Use the result of part (d) to state whether the increase for J.C. Penney from year 1 to year 2 is statistically significant.The following results are for independent random samples taken from two populations. Sample 1 Sample 2 n1 = 20 n2 = 30 x1=22.5 x2=20.1 s1 = 2.5 s2 = 4.8 a. What is the point estimate of the difference between the two population means? b. What is the degrees of freedom for the t distribution? c. At 95% confidence, what is the margin of error? d. What is the 95% confidence interval for the difference between the two population means?Consider the following hypothesis test. H0:12=0Ha:120 The following results are from independent samples taken from two populations. Sample 1 Sample 2 n1 = 35 n2 = 40 x1=13.6 x2=10.1 s1 = 5.2 s2 = 8.5 a. What is the value of the test statistic? b. What is the degrees of freedom for the t distribution? c. What is the p-value? d. At = .05, what is your conclusion?Consider the following data for two independent random samples taken from two normal populations. a. Compute the two sample means. b. Compute the two sample standard deviations. c. What is the point estimate of the difference between the two population means? d. What is the 90% confidence interval estimate of the difference between the two population means?The U.S. Department of Transportation provides the number of miles that residents of the 75 largest metropolitan areas travel per day in a car. Suppose that for a simple random sample of 50 Buffalo residents the mean is 22.5 miles a day and the standard deviation is 8.4 miles a day, and for an independent simple random sample of 40 Boston residents the mean is 18.6 miles a day and the standard deviation is 7.4 miles a day. a. What is the point estimate of the difference between the mean number of miles that Buffalo residents travel per day and the mean number of miles that Boston residents travel per day? b. What is the 95% confidence interval for the difference between the two population means?The average annual cost (including tuition, room, board, books, and fees) to attend a public college takes nearly a third of the annual income of a typical family with college-age children (Money, April 2012). At private colleges, the average annual cost is equal to about 60% of the typical familys income. The following random samples show the annual cost of attending private and public colleges. Data are in thousands of dollars. a. Compute the sample mean and sample standard deviation for private and public colleges. b. What is the point estimate of the difference between the two population means? Interpret this value in terms of the annual cost of attending private and public colleges. c. Develop a 95% confidence interval of the difference between the mean annual cost of attending private and public colleges.Are nursing salaries in Tampa, Florida, lower than those in Dallas, Texas? As reported by the Tampa Tribune, salary data show staff nurses in Tampa earn less than staff nurses in Dallas. Suppose that in a follow-up study of 40 staff nurses in Tampa and 50 staff nurses in Dallas you obtain the following results. Tampa Dallas n1 = 40 n2 = 50 x1=56,100 x2=59,400 s1 = 6000 s2 = 7000 a. Formulate hypothesis so that, if the null hypothesis is rejected, we can conclude that salaries for staff nurses in Tampa are significantly lower than for those in Dallas. Use = .05. b. What is the value of the test statistic? c. What is the p-value? d. What is your conclusion?Commercial real estate prices and rental rates suffered substantial declines in 2008 and 2009 (Newsweek, July 27, 2009). These declines were particularly severe in Asia; annual lease rates in Tokyo, Hong Kong, and Singapore declined by 40% or more. Even with such large declines, annual lease rates in Asia were still higher than those in many cities in Europe. Annual lease rates for a sample of 30 commercial properties in Hong Kong showed a mean of 1,114 per square meter with a standard deviation of 230. Annual lease rates for a sample of 40 commercial properties in Paris showed a mean lease rate of 989 per square meter with a standard deviation of 195. a. On the basis of the sample results, can we conclude that the mean annual lease rate is higher in Hong Kong than in Paris? Develop appropriate null and alternative hypotheses. b. Use = .01. What is your conclusion?The College Board provided comparisons of Scholastic Aptitude Test (SAT) scores based on the highest level of education attained by the test takers parents. A research hypothesis was that students whose parents had attained a higher level of education would on aver age score higher on the SAT. The overall mean SAT math score was 514 (College Board website, January 8, 2012). SAT math scores for independent samples of students follow. The first sample shows the SAT math test scores for students whose parents are college graduates with a bachelors degree. The second sample shows the SAT math test scores for students whose parents are high school graduates but do not have a college degree. a. Formulate the hypotheses that can be used to determine whether the sample data support the hypothesis that students show a higher population mean math score on the SAT if their parents attained a higher level of education. b. What is the point estimate of the difference between the means for the two populations? c. Compute the p-value for the hypothesis test. d. At = .05, what is your conclusion?Periodically, Merrill Lynch customers are asked to evaluate Merrill Lynch financial consultants and services. Higher ratings on the client satisfaction survey indicate better service, with 7 the maximum service rating. Independent samples of service ratings for two financial consultants are summarized here. Consultant A has 10 years of experience, whereas consultant B has 1 year of experience. Use = .05 and test to see whether the consultant with more experience has the higher population mean service rating. Consultant A Consultant B n1 = 16 n2 = 10 x1=6.82 x2=6.25 s1 = .64 s2 = .75 a. State the null and alternative hypotheses. b. Compute the value of the test statistic. c. What is the p-value? d. What is your conclusion?Researchers at Purdue University and Wichita State University found that airlines are doing a better job of getting passengers to their destinations on time (Associated Press, April 2, 2012). AirTran Airways and Southwest Airlines were among the leaders in on-time arrivals with both having 88% of their flights arriving on time. But for the 12% of flights that were delayed, how many minutes were these flights late? Sample data showing the number of minutes that delayed flights were late are provided in the file named AirDelay. Data are shown for both airlines. a. Formulate the hypotheses that can be used to test for a difference between the population mean minutes late for delayed flights by these two airlines. b. What is the sample mean number of minutes late for delayed flights for each of these two airlines? c. Using a .05 level of significance, what is the p-value and what is your conclusion?Consider the following hypothesis test. H0:d0Ha:d0 The following data are from matched samples taken from two populations. Element Population 1 2 1 21 20 2 28 26 3 18 18 4 20 20 5 26 24 a. Compute the difference value for each element. b. Compute d. c. Compute the standard deviation sd. d. Conduct a hypothesis test using = .05. What is your conclusion?The following data are from matched samples taken from two populations. Element Population 1 2 1 11 8 2 7 8 3 9 6 4 12 7 5 13 10 6 15 15 7 15 14 a. Compute the difference value for each element. b. Compute d. c. Compute the standard deviation sd. d. What is the point estimate of the difference between the two population means? e. Provide a 95% confidence interval for the difference between the two population means.A market research firm used a sample of individuals to rate the purchase potential of a particular product before and after the individuals saw a new television commercial about the product. The purchase potential ratings were based on a 0 to 10 scale, with higher values indicating a higher purchase potential. The null hypothesis stated that the mean rating after would be less than or equal to the mean rating before. Rejection of this hypothesis would show that the commercial improved the mean purchase potential rating. Use = .05 and the following data to test the hypothesis and comment on the value of the commercial. Individual Purchase Rating After Before 1 6 5 2 6 4 3 7 7 4 4 3 5 3 5 6 9 8 7 7 5 8 6 6The price per share of stock for a sample of 25 companies was recorded at the beginning of 2012 and then again at the end of the 1st quarter of 2012 (The Wall Street Journal, April 2, 2012). How stocks perform during the 1st quarter is an indicator of what is ahead for the stock market and the economy. Use the sample data in the file entitled StockPrices to answer the following. a. Let di denote the change in price per share for company i where di = 1st quarter of 2012 price per share minus the beginning of 2012 price per share. Use the sample mean of these values to estimate the dollar amount a share of stock has changed during the 1st quarter. b. What is the 95% confidence interval estimate of the population mean change in the price per share of stock during the first quarter? Interpret this result.Bank of Americas Consumer Spending Survey collected data on annual credit card charges in seven different categories of expenditures: transportation, groceries, dining out, household expenses, home furnishings, apparel, and entertainment. Using data from a sample of 42 credit card accounts, assume that each account was used to identify the annual credit card charges for groceries (population 1) and the annual credit card charges for dining out (population 2). Using the difference data, the sample mean difference was d=850, and the sample standard deviation was sd = 1123. a. Formulate the null and alternative hypotheses to test for no difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out. b. Use a .05 level of significance. Can you conclude that the population means differ? What is the p-value? c. Which category, groceries or dining out, has a higher population mean annual credit card charge? What is the point estimate of the difference between the population means? What is the 95% confidence interval estimate of the difference between the population means?The Global Business Travel Association reported the domestic airfare for business travel for the current year and the previous year (INC. Magazine, February 2012). Below is a sample of 12 flights with their domestic airfares shown for both years. Current Year Previous Year 345 315 526 463 420 462 216 206 285 275 405 432 635 585 710 650 605 545 517 547 570 508 610 580 a. Formulate the hypotheses and test for a significant increase in the mean domestic airfare for business travel for the one-year period. What is the p-value? Using a .05 level of significance, what is your conclusion? b. What is the sample mean domestic airfare for business travel for each year? c. What is the percentage change in the airfare for the one-year period?The College Board SAT college entrance exam consists of three parts: math, writing, and critical reading (The World Almanac, 2012). Sample data showing the math and writing scores for a sample of 12 students who took the SAT follow. Student Math Writing 1 540 474 2 432 380 3 528 463 4 574 612 5 448 420 6 502 526 7 480 430 8 499 459 9 610 615 10 572 541 11 390 335 12 593 613 a. Use a .05 level of significance and test for a difference between the population mean for the math scores and the population mean for the writing scores. What is the p-value and what is your conclusion? b. What is the point estimate of the difference between the mean scores for the two tests? What are the estimates of the population mean scores for the two tests? Which test reports the higher mean score?Scores in the first and fourth (final) rounds for a sample of 20 golfers who competed in PGA tournaments are shown in the following table. Suppose you would like to determine if the mean score for the first round of a PGA Tour event is significantly different than the mean score for the fourth and final round. Does the pressure of playing in the final round cause scores to go up? Or does the increased player concentration cause scores to come down? Player First Round Final Round Michael Letzig 70 72 Scott Verplank 71 72 D. A. Points 70 75 Jerry Kelly 72 71 Soren Hansen 70 69 D. J. Trahan 67 67 Bubba Watson 71 67 Reteif Goosen 68 75 Jeff Klauk 67 73 Kenny Perry 70 69 Aron Price 72 72 Charles Howell 72 70 Jason Dufner 70 73 Mike Weir 70 77 Carl Pettersson 68 70 Bo Van Pelt 68 65 Ernie Els 71 70 Cameron Beckman 70 68 Nick Watney 69 68 Tommy Armour III 67 71 a. Use = .10 to test for a statistically significantly difference between the population means for first- and fourth-round scores. What is the p-value? What is your conclusion? b. What is the point estimate of the difference between the two population means? For which round is the population mean score lower? c. What is the margin of error for a 90% confidence interval estimate for the difference between the population means? Could this confidence interval have been used to test the hypothesis in part (a)? Explain.A manufacturer produces both a deluxe and a standard model of an automatic sander designed for home use. Selling prices obtained from a sample of retail outlets follow. Retail Outlet Model Price () Deluxe Standard 1 39 27 2 39 28 3 45 35 4 38 30 5 40 30 6 39 34 7 35 29 a. The manufacturers suggested retail prices for the two models show a 10 price differential. Use a .05 level of significance and test that the mean difference between the prices of the two models is 10. b. What is the 95% confidence interval for the difference between the mean prices of the two models?Consider the following results for independent samples taken from two populations. Sample 1 Sample 2 n1 = 400 n2 = 300 p1=.48 p2=.36 a. What is the point estimate of the difference between the two population proportions? b. Develop a 90% confidence interval for the difference between the two population proportions. c. Develop a 95% confidence interval for the difference between the two population proportions.Consider the hypothesis test H0:p1p20Ha:p1p20 The following results are for independent samples taken from the two populations. Sample 1 Sample 2 n1 = 200 n2 = 300 p1=.22 p2=.16 a. What is the p-value? b. With = .05, what is your hypothesis testing conclusion?A Businessweek/Harris survey asked senior executives at large corporations their opinions about the economic outlook for the future. One question was, Do you think that there will be an increase in the number of full-time employees at your company over the next 12 months? In the current survey, 220 of 400 executives answered Yes, while in a previous year survey, 192 of 400 executives had answered Yes. Provide a 95% confidence interval estimate for the difference between the proportions at the two points in time. What is your interpretation of the interval estimate?Forbes reports that women trust recommendations from Pinterest more than recommendations from any other social network platform (Forbes website, April 10, 2012). But does trust in Pinterest differ by gender? The following sample data show the number of women and men who stated in a recent sample that they trust recommendations made on Pinterest. Women Men Sample 150 170 Trust Recommendations Made on Pinterest 117 102 a. What is the point estimate of the proportion of women who trust recommendations made on Pinterest? b. What is the point estimate of the proportion of men who trust recommendations made on Pinterest? c. Provide a 95% confidence interval estimate of the difference between the proportion of women and men who trust recommendations made on Pinterest.Researchers with Oceana, a group dedicated to preserving the ocean ecosystem, reported finding that 33% of fish sold in retail outlets, grocery stores, and sushi bars throughout the United States had been mislabeled (San Francisco Chronicle website, February 21, 2013). Does this mislabeling differ for different species of fish? The following data show the number labeled incorrectly for samples of tuna and mahi mahi. Tuna Mahi Mahi Sample 220 160 Mislabeled 99 56 a. What is the point estimate of the proportion of tuna that is mislabeled? b. What is the point estimate of the proportion of mahi mahi that is mislabeled? c. Provide a 95% confidence interval estimate of the difference between the proportion of tuna and mahi mahi that is mislabeled.Minnesota had the highest turnout rate of any state for the 2012 presidential election (United States Election Project website, February 9, 2013). Political analysts wonder if turnout in rural Minnesota was higher than turnout in the urban areas of the state. A sample shows that 663 of 884 registered voters from rural Minnesota voted in the 2012 presidential election, while 414 out of 575 registered voters from urban Minnesota voted. a. Formulate the null and alternative hypotheses that can be used to test whether registered voters in rural Minnesota were more likely than registered voters in urban Minnesota to vote in the 2012 presidential election. b. What is the proportion of sampled registered voters in rural Minnesota that voted in the 2012 presidential election? c. What is the proportion of sampled registered voters in urban Minnesota that voted in the 2012 presidential election? d. At = .05, test the political analysts hypothesis. What is the p-value, and what conclusion do you draw from your results?Oil wells are expensive to drill, and dry wells are a great concern to oil exploration companies. The domestic oil and natural gas producer Aegis Oil, LLC describes on its web-site how improvements in technologies such as three-dimensional seismic imaging have dramatically reduced the number of dry (nonproducing) wells it and other oil exploration companies drill. The following sample data for wells drilled in 2005 and 2012 show the number of dry wells that were drilled in each year. 2005 2012 Wells Drilled 119 162 Dry Wells 24 18 a. Formulate the null and alternative hypotheses that can be used to test whether the wells drilled in 2005 were more likely to be dry than wells drilled in 2012. b. What is the point estimate of the proportion of wells drilled in 2005 that were dry? c. What is the point estimate of the proportion of wells drilled in 2012 that were dry? d. What is the p-value of your hypothesis test? At = .05, what conclusion do you draw from your results?The Adecco Workplace Insights Survey sampled men and women workers and asked if they expected to get a raise or promotion this year (USA Today, February 16, 2012). Suppose the survey sampled 200 men and 200 women. If 104 of the men replied Yes and 74 of the women replied Yes, are the results statistically significant so that you can conclude a greater proportion of men expect to get a raise or a promotion this year? a. State the hypothesis test in terms of the population proportion of men and the population proportion of women. b. What is the sample proportion for men? For women? c. Use a .01 level of significance. What is the p-value and what is your conclusion?Winter visitors are extremely important to the economy of Southwest Florida. Hotel occupancy is an often-reported measure of visitor volume and visitor activity (Naples Daily News, March 22, 2012). Hotel occupancy data for February in two consecutive years are as follows. Current Year Previous Year Occupied Rooms 1470 1458 Total Rooms 1750 1800 a. Formulate the hypothesis test that can be used to determine if there has been an increase in the proportion of rooms occupied over the one-year period. b. What is the estimated proportion of hotel rooms occupied each year? c. Using a .05 level of significance, what is your hypothesis test conclusion? What is the p-value? d. What is the 95% confidence interval estimate of the change in occupancy for the one-year period? Do you think area officials would be pleased with the results?The Adecco Workplace Insights Survey sampled men and women workers and asked if they expected to get a raise or promotion this year (USA Today, February 16, 2012). Suppose the survey sampled 200 men and 200 women. If 104 of the men replied Yes and 74 of the women replied Yes, are the results statistically significant in that you can conclude a greater proportion of men are expecting to get a raise or a promotion this year? a. State the hypothesis test in terms of the population proportion of men and the population proportion of women? b. What is the sample proportion for men? For women? c. Use a .01 level of significance. What is the p-value and what is your conclusion?Safegate Foods, Inc., is redesigning the checkout lanes in its supermarkets throughout the country and is considering two designs. Tests on customer checkout times conducted at two stores where the two new systems have been installed result in the following summary of the data. System A System B n1 = 120 n2 = 100 x1=4.1minutes x2=3.4minutes 1 = 2.2 minutes 2 = 1.5 minutes Test at the .05 level of significance to determine whether the population mean checkout times of the two systems differ. Which system is preferred?Home values tend to increase over time under normal conditions, but the recession of 2008 and 2009 has reportedly caused the sales price of existing homes to fall nationwide (Businessweek, March 9, 2009). You would like to see if the data support this conclusion. The file HomePrices contains data on 30 existing home sales in 2006 and 40 existing home sales in 2009. a. Provide a point estimate of the difference between the population mean prices for the two years. b. Develop a 99% confidence interval estimate of the difference between the resale prices of houses in 2006 and 2009. c. Would you feel justified in concluding that resale prices of existing homes have declined from 2006 to 2009? Why or why not?Mutual funds are classified as load or no-load funds. Load funds require an investor to pay an initial fee based on a percentage of the amount invested in the fund. The no-load funds do not require this initial fee. Some financial advisors argue that the load mutual funds may be worth the extra fee because these funds provide a higher mean rate of return than the no-load mutual funds. A sample of 30 load mutual funds and a sample of 30 no-load mutual funds were selected. Data were collected on the annual return for the funds over a five-year period. The data are contained in the data set Mutual. The data for the first five load and first five no-load mutual funds are as follows. Mutual FundsLoad Return American National Growth 15.51 Arch Small Cap Equity 14.57 Bartlett Cap Basic 17.73 Calvert World International 10.31 Colonial Fund A 16.23 Amana Income Fund 13.24 Berger One Hundred 12.13 Columbia International Stock 12.17 Dodge Cox Balanced 16.06 Evergreen Fund 17.61 a. Formulate H0 and Ha such that rejection of H0 leads to the conclusion that the load mutual funds have a higher mean annual return over the five-year period. b. Use the 60 mutual funds in the data set Mutual to conduct the hypothesis test. What is the p-value? At = .05, what is your conclusion?The National Association of Home Builders provided data on the cost of the most popular home remodeling projects. Sample data on cost in thousands of dollars for two types of remodeling projects are as follows. Kitchen Master Bedroom 25.2 18.0 17.4 22.9 22.8 26.4 21.9 24.8 19.7 26.9 23.0 17.8 19.7 24.6 16.9 21.0 21.8 23.6 a. Develop a point estimate of the difference between the population mean remodeling costs for the two types of projects. b. Develop a 90% confidence interval for the difference between the two population means.In Born TogetherReared Apart: The Landmark Minnesota Twin Study (2012), Nancy Segal discusses the efforts of research psychologists at the University of Minnesota to understand similarities and differences between twins by studying sets of twins who were raised separately. Below are critical reading SAT scores for several pairs of identical twins (twins who share all of their genes), one of whom was raised in a family with no other children (no siblings) and one of whom was raised in a family with other children (with siblings). a. What is the mean difference between the critical reading SAT scores for the twins raised with no siblings and the twins raised with siblings? b. Provide a 90% confidence interval estimate of the mean difference between the critical reading SAT scores for the twins raised with no siblings and the twins raised with siblings. c. Conduct a hypothesis test of equality of the critical reading SAT scores for the twins raised with no siblings and the twins raised with siblings at = .01. What is your conclusion?Country Financial, a financial services company, uses surveys of adults age 18 and older to determine if personal financial fitness is changing over time (USA Today, April 4, 2012). In February 2012, a sample of 1000 adults showed 410 indicating that their financial security was more than fair. In February 2010, a sample of 900 adults showed 315 indicating that their financial security was more than fair. a. State the hypotheses that can be used to test for a significant difference between the population proportions for the two years. b. What is the sample proportion indicating that their financial security was more than fair in 2012? In 2010? c. Conduct the hypothesis test and compute the p-value. At a .05 level of significance, what is your conclusion? d. What is the 95% confidence interval estimate of the difference between the two population proportions?A large automobile insurance company selected samples of single and married male policyholders and recorded the number who made an insurance claim over the preceding three-year period. Single Policyholders Married Policyholders n1 = 400 n2 = 900 Number making claims = 76 Number making claims = 90 a. Use = .05. Test to determine whether the claim rates differ between single and married male policyholders. b. Provide a 95% confidence interval for the difference between the proportions for the two populations.Medical tests were conducted to learn about drug-resistant tuberculosis. Of 142 cases tested in New Jersey, 9 were found to be drug-resistant. Of 268 cases tested in Texas, 5 were found to be drug-resistant. Do these data suggest a statistically significant difference between the proportions of drug-resistant cases in the two states? Use a .02 level of significance. What is the p-value, and what is your conclusion?Vacation occupancy rates were expected to be up during March 2008 in Myrtle Beach, South Carolina (The Sun News, February 29, 2008). Data in the file Occupancy will allow you to replicate the findings presented in the newspaper. The data show units rented and not rented for a random sample of vacation properties during the first week of March 2007 and March 2008. a. Estimate the proportion of units rented during the first week of March 2007 and the first week of March 2008. b. Provide a 95% confidence interval for the difference in proportions. c. On the basis of your findings, does it appear March rental rates for 2008 will be up from those a year earlier?Winter visitors are extremely important to the economy of Southwest Florida. Hotel occupancy is an often-reported measure of visitor volume and visitor activity (Naples Daily News, March 22, 2012). Hotel occupancy data for February in two consecutive years are as follows. Current Year Previous Year Occupied Rooms 1470 1458 Total Rooms 1750 1800 a. Formulate the hypothesis test that can be used to determine if there has been an increase in the proportion of rooms occupied over the one-year period. b. What is the estimated proportion of hotel rooms occupied each year? c. Using a .05 level of significance, what is your hypothesis test conclusion? What is the p-value? d. What is the 95% confidence interval estimate of the change in occupancy for the one-year period? Do you think area officials would be pleased with the results?One of the researchers voiced concern about the effect of the new coating on driving distances. Par would like the new cut-resistant ball to offer driving distances comparable to those of the current-model golf ball. To compare the driving distances for the two balls, 40 balls of both the new and current models were subjected to distance tests. The testing was performed with a mechanical hitting machine so that any difference between the mean distances for the two models could be attributed to a difference in the two models. The results of the tests, with distances measured to the nearest yard, follow. These data are available on the website that accompanies the text. Managerial Report 1. Formulate and present the rationale for a hypothesis test that Par could use to compare the driving distances of the current and new golf balls. 2. Analyze the data to provide the hypothesis testing conclusion. What is the p-value for your test? What is your recommendation for Par, Inc.? 3. Provide descriptive statistical summaries of the data for each model. 4. What is the 95% confidence interval for the population mean driving distance of each model, and what is the 95% confidence interval for the difference between the means of the two populations? 5. Do you see a need for larger sample sizes and more testing with the golf balls? Discuss.Find the following chi-square distribution values from Table 11.1 or Table 3 of Appendix B. a. 052with df = 5 b. 0252with df = 15 c. 9752with df = 20 d. 012with df = 10 e. 952with df = 18A sample of 20 items provides a sample standard deviation of 5. a. Compute the 90% confidence interval estimate of the population variance. b. Compute the 95% confidence interval estimate of the population variance. c. Compute the 95% confidence interval estimate of the population standard deviation.A sample of 16 items provides a sample standard deviation of 9.5. Test the following hypotheses using = .05. What is your conclusion? Use both the p-value approach and the critical value approach. H0:2 50 H4:2 50The variance in drug weights is critical in the pharmaceutical industry. For a specific drug, with weights measured in grams, a sample of 18 units provided a sample variance of s2 = .36. a. Construct a 90% confidence interval estimate of the population variance for the weight of this drug. b. Construct a 90% confidence interval estimate of the population standard deviation.John Calipari, head basketball coach for the 2012 national champion university of Kentucky Wildcats, is the highest paid coach in college basketball with an annual salary of 5.4 million (USA Today, March 29, 2012). The sample below shows the head basketball coachs salary for a sample of 10 schools playing NCAA Division I basketball. Salary data are in millions of dollars. a. Use the sample mean for the 10 schools to estimate the population mean annual salary for head basketball coaches at colleges and universities playing NCAA Division I basketball. b. Use the data to estimate the population standard deviation for the annual salary for head basketball coaches. c. What is the 95% confidence interval for the population variance? d. What is the 95% confidence interval for the population standard deviation?