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All Textbook Solutions for Understandable Statistics: Concepts and Methods

Basic Computation: Find Probabilities In Problems 514, assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probabilities. 11. P(x 30); = 20; = 3.4 Basic Computation: Find Probabilities In Problems 514, assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probabilities. 12. P(x 120); = 100; = 15 Basic Computation: Find Probabilities In Problems 514, assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probabilities. 13. P(x 90); = 100; = 15 Basic Computation: Find Probabilities In Problems 514, assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probabilities. 14. P(x 2); = 3; = 0.25 Basic Computation: Find z Values In Problems 1524, find the z value described and sketch the area described. 15. Find z such that 6% of the standard normal curve lies to the left of z.Basic Computation: Find z Values In Problems 1524, find the z value described and sketch the area described. 16. Find z such that 5.2% of the standard normal curve lies to the left of z.Basic Computation: Find z Values In Problems 1524, find the z value described and sketch the area described. 17. Find z such that 55% of the standard normal curve lies to the left of z.Basic Computation: Find z Values In Problems 1524, find the z value described and sketch the area described. 18. Find z such that 97.5% of the standard normal curve lies to the left of z.Basic Computation: Find z Values In Problems 1524, find the z value described and sketch the area described. 19. Find z such that 8% of the standard normal curve lies to the right of z.20P 21P Basic Computation: Find z Values In Problems 1524, find the z value described and sketch the area described. 22. Find z such that 95% of the standard normal curve lies to the right of z.23P 24P 25P 26P Archaeology: Hopi Village Thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally distributed, with a mean of 5.1 millimeters (mm) and a standard deviation of 0.9 mm (Source: Homolovi II: Archaeology of an Ancestral Hopi Village, Arizona, edited by E. C. Adams and K. A. Hays, University of Arizona Press). For a randomly found shard, what is the probability that the thickness is (a) less than 3.0 mm? (b) more than 7.0 mm? (c) between 3.0 mm and 7.0 mm?Law Enforcement: Police Response Time Police response time to an emergency call is the difference between the time the call is first received by the dispatcher and the time a patrol car radios that it has arrived at the scene (based on information from The Denver Post). Over a long period of time, it has been determined that the police response time has a normal distribution with a mean of 8.4 minutes and a standard deviation of 1.7 minutes. For a randomly received emergency call, what is the probability that the response time will be (a) between 5 and 10 minutes? (b) less than 5 minutes? (c) more than 10 minutes?29P Guarantee: Watches Accrotime is a manufacturer of quartz crystal watches. Accrotime researchers have shown that the watches have an average life of 28 months before certain electronic components deteriorate, causing the watch to become unreliable. The standard deviation of watch lifetimes is 5 months, and the distribution of lifetimes is normal. (a) If Accrotime guarantees a full refund on any defective watch for 2 years after purchase, what percentage of total production should the company expect to replace? (b) Inverse Normal Distribution If Accrotime does not want to make refunds on more than 12% of the watches it makes, how long should the guarantee period be (to the nearest month)?Expand Your Knowledge: Estimating the Standard Deviation Consumer Reports gave information about the ages at which various household products are replaced. For example, color TVs are replaced at an average age of = 8 years after purchase, and the (95% of data) range was from 5 to 11 years. Thus, the range was 11 5 = 6 years. Let x be the age (in years) at which a color TV is replaced. Assume that x has a distribution that is approximately normal. (a) The empirical rule (see Section 6.1) indicates that for a symmetric and bell-shaped distribution, approximately 95% of the data lies within two standard deviations of the mean. Therefore, a 95% range of data values extending from 2 to + 2 is often used for commonly occurring data values. Note that the interval from 2 to + 2 is 4 in length. This leads to a rule of thumb for estimating the standard deviation from a 95% range of data values. ESTIMATING THE STANDARD DEVIATION For a symmetric, bell-shaped distribution, standarddeviationrange4highvaluelowvalue4 where it is estimated that about 95% of the commonly occurring data values fall into this range. Use this rule of thumb to approximate the standard deviation of x values, where x is the age (in years) at which a color TV is replaced. (b) What is the probability that someone will keep a color TV more than 5 years before replacement? (c) What is the probability that someone will keep a color TV fewer than 10 years before replacement? (d) Inverse Normal Distribution Assume that the average life of a color TV is 8 years with a standard deviation of 1.5 years before it breaks. Suppose that a company guarantees color TVs and will replace a TV that breaks while under guarantee with a new one. However, the company does not want to replace more than 10% of the TVs under guarantee. For how long should the guarantee be made (rounded to the nearest tenth of a year)?Estimating the Standard Deviation: Refrigerator Replacement Consumer Reports indicated that the average life of a refrigerator before replacement is , = 14 years with a (95% of data) range from 9 to 19 years. Let x = age at which a refrigerator is replaced. Assume that x has a distribution that is approximately normal. (a) Find a good approximation for the standard deviation of x values. Hint: See Problem 31. (b) What is the probability that someone will keep a refrigerator fewer than 11 years before replacement? (c) What is the probability that someone will keep a refrigerator more than 18 years before replacement? (d) Inverse Normal Distribution Assume that the average life of a refrigerator is 14 years, with the standard deviation given in part (a) before it breaks. Suppose that a company guarantees refrigerators and will replace a refrigerator that breaks while under guarantee with a new one. However, the company does not want to replace more than 5% of the refrigerators under guarantee. For how long should the guarantee be made (rounded to the nearest tenth of a year)? 31. Expand Your Knowledge: Estimating the Standard Deviation Consumer Reports gave information about the ages at which various household products are replaced. For example, color TVs are replaced at an average age of = 8 years after purchase, and the (95% of data) range was from 5 to 11 years. Thus, the range was 11 5 = 6 years. Let x be the age (in years) at which a color TV is replaced. Assume that x has a distribution that is approximately normal. (a) The empirical rule (see Section 6.1) indicates that for a symmetric and bell-shaped distribution, approximately 95% of the data lies within two standard deviations of the mean. Therefore, a 95% range of data values extending from 2 to + 2 is often used for commonly occurring data values. Note that the interval from 2 to + 2 is 4 in length. This leads to a rule of thumb for estimating the standard deviation from a 95% range of data values. (a) ESTIMATING THE STANDARD DEVIATION (b) For a symmetric, bell-shaped distribution, (c) standarddeviationrange4highvaluelowvalue4 (d) where it is estimated that about 95% of the commonly occurring data values fall into this range. (e) Use this rule of thumb to approximate the standard deviation of x values, where x is the age (in years) at which a color TV is replaced. (b) What is the probability that someone will keep a color TV more than 5 years before replacement? (c) What is the probability that someone will keep a color TV fewer than 10 years before replacement? (d) Inverse Normal Distribution Assume that the average life of a color TV is 8 years with a standard deviation of 1.5 years before it breaks. Suppose that a company guarantees color TVs and will replace a, TV that breaks while under guarantee with a new one. However, the company does not want to replace more than 10% of the TVs under guarantee. For how long should the guarantee be made (rounded to the nearest tenth of a year)?33P 34P Insurance: Satellites A relay microchip in a telecommunications satellite has a life expectancy that follows a normal distribution with a mean of 90 months and a standard deviation of 3.7 months. When this computer-relay microchip malfunctions, the entire satellite is useless. A large London insurance company is going to insure the satellite for 50 million. Assume that the only part of the satellite in question is the microchip. All other components will work indefinitely. (a) Inverse Normal Distribution For how many months should the satellite be insured to be 99% confident that it will last beyond the insurance date? (b) If the satellite is insured for 84 months, what is the probability that it will malfunction before the insurance coverage ends? (c) If the satellite is insured for 84 months, what is the expected loss to the insurance company? (d) If the insurance company charges 3 million for 84 months of insurance, how much profit does the company expect to make?Convertion Center: Exhibition Show Attendance Attendance at large exhibition shows in Denver averages about 8000 people per day, with standard deviation of about 500. Assume that the daily attendance figures follow a normal distribution. (a) What is the probability that the daily attendance will be fewer than 7200 people? (b) What is the probability that the daily attendance will be more than 8900 people? (c) What is the probability that the daily attendance will be between 7200 and 8900 people?Exhibition Shows: Inverse Normal Distribution Most exhibition shows open in the morning and close in the late evening. A study of Saturday arrival times showed that the average arrival time was 3 hours and 48 minutes after the doors opened, and the standard deviation was estimated at about 52 minutes. Assume that the arrival times follow a normal distribution. (a) At what time after the doors open will 90% of the people who are coming to the Saturday show have arrived? (b) At what time after the doors open will only 15% of the people who are coming to the Saturday show have arrived? (c) Do you think the probability distribution of arrival times for Friday might be different from the distribution of arrival times for Saturday? Explain.Budget: Maintenance The amount of money spent weekly on cleaning, maintenance, and repairs at a large restaurant was observed over a long period of time to be approximately normally distributed, with mean = 615 and standard deviation = 42. (a) If 646 is budgeted for next week, what is the probability that the actual costs will exceed the budgeted amount? (b) Inverse Normal Distribution How much should be budgeted for weekly repairs, cleaning, and maintenance so that the probability that the budgeted amount will be exceeded in a given week is only 0.10?39P 40P 1P 2P 3P 4P 5P 6P 7P 8P 9P Statistical Literacy What is the standard error of a sampling distribution?2P 3P 4P Basic Computation: Central Limit Theorem Suppose x has a distribution with a mean of 8 and a standard deviation of 16. Random samples of size n = 64 are drawn. (a) Describe the x distribution and compute the mean and standard deviation of the distribution. (b) Find the z value corresponding to x=9. (c) Find P(x19). (d) Interpretation Would it be unusual for a random sample of size 64 from the x distribution to have a sample mean greater than 9? Explain.Basic Computation: Central Limit Theorem Suppose x has a distribution with a mean of 20 and a standard deviation of 3. Random samples of size n = 36 are drawn. (a) Describe the x distribution and compute the mean and standard deviation of the distribution. (b) Find the z value corresponding to x=19. (c) Find P(x19). (d) Interpretation Would it be unusual for a random sample of size 36 from the x distribution to have a sample mean less than 19? Explain.7P 8P 9P 10P 11P Critical Thinking Suppose an x distribution has mean = 5. Consider two corresponding x distributions, the first based on samples of size n = 49 and the second based on samples of size n = 81. (a) What is the value of the mean of each of the two x distributions? (b) For which x distribution is P(x 6) smaller? Explain. (c) For which x distribution is P(4 x 6) greater? Explain.13P Vital Statistics: Heights of Men The heights of 18-year-old men are approximately normally distributed, with mean 68 inches and standard deviation 3 inches (based on information from Statistical Abstract of the United States, 112th edition). (a) What is the probability that an 18-year-old man selected at random is between 67 and 69 inches tall? (b) If a random sample of nine 18-year-old men is selected, what is the probability that the mean height x is between 67 and 69 inches? (c) Interpretation Compare your answers to parts (a) and (b). Is the probability in part (b) much higher? Why would you expect this?15P Medical: White Blood Cells Let x be a random variable that represents white blood cell count per cubic milliliter of whole blood. Assume that x has a distribution that is approximately normal, with mean = 7500 and estimated standard deviation = 1750 (see reference in Problem 15). A test result of x 3500 is an indication of leukopenia. This indicates bone marrow depression that may be the result of a viral infection. (a) What is the probability that, on a single test, x is less than 3500? (b) Suppose a doctor uses the average x for two tests taken about a week apart. What can we say about the probability distribution of x? What is the probability of x3500? (c) Repeat part (b) for n = 3 tests taken a week apart. (d) Interpretation Compare your answers to parts (a), (b), and (c). How did the probabilities change as n increased? If a person had x3500 based on three tests, what conclusion would you draw as a doctor or a nurse? 15. Medical: Blood Glucose Let x be a random variable that represents the level of glucose in the blood (milligrams per deciliter of blood) after a 12-hour fast. Assume that for people under 50 years old, x has a distribution that is approximately normal, with mean = 85 and estimated standard deviation = 25 (based on information from Diagnostic Tests with Nursing Applications, edited by S. Loeb, Springhouse). A test result x 40 is an indication of severe excess insulin, and medication is usually prescribed. (a) What is the probability that, on a single test, x 40? (b) Suppose a doctor uses the average x for two tests taken about a week apart. What can we say about the probability distribution of x? Hint: See Theorem 6.1. What is the probability that x40? (c) Repeat part (b) for n = 3 tests taken a week apart. (d) Repeat part (b) for n = 5 tests taken a week apart. (e) Interpretation Compare your answers to parts (a), (b), (c), and (d). Did the probabilities decrease as n increased? Explain what this might imply if you were a doctor or a nurse. If a patient had a test result of x40 based on five tests, explain why either you are looking at an extremely rare event or (more likely) the person has a case of excess insulin. THEOREM 6.1 For a Normal Probability Distribution Let x be a random variable with a normal distribution whose mean is and whose standard deviation is . Let x be the sample mean corresponding to random samples of size n taken from the x distribution. Then the following are true: (a) The x distribution is a normal distribution. (b) The mean of the x distribution is . (c) The standard deviation of the x distribution is /n. We conclude from Theorem 6.1 that when x has a normal distribution, the x distribution will be normal for any sample size n. Furthermore, we can convert the x distribution to the standard normal z distribution using the following formulas. x=x=nz=xxx=x/n where n is the sample size, is the mean of the x distribution, and is the standard deviation of the x distribution. Theorem 6.1 is a wonderful theorem! It states that the x distribution will be normal provided the x distribution is normal. The sample size n could be 2, 3, 4, or any other (fixed) sample size we wish. Furthermore, the mean of the x distribution is (same as for the x distribution), but the standard deviation is /n (which is, of course, smaller than ). The next example illustrates Theorem 6.1.Wildlife: Deer Let x be a random variable that represents the weights in kilograms (kg) of healthy adult female deer (does) in December in Mesa Verde National Park. Then x has a distribution that is approximately normal, with mean = 63.0 kg and standard deviation = 7.1 kg (Source: The Mule Deer of Mesa Verde National Park, by G. W. Mierau and J. L. Schmidt, Mesa Verde Museum Association). Suppose a doe that weighs less than 54 kg is considered undernourished. (a) What is the probability that a single doe captured (weighed and released) at random in December is undernourished? (b) If the park has about 2200 does, what number do you expect to be undernourished in December? (c) Interpretation To estimate the health of the December doe population, park rangers use the rule that the average weight of n = 50 does should be more than 60 kg. If the average weight is less than 60 kg, it is thought that the entire population of does might be undernourished. What is the probability that the average weight x for a random sample of 50 does is less than 60 kg (assume a healthy population)? (d) Interpretation Compute the probability that x64.2 kg for 50 does (assume a healthy population). Suppose park rangers captured, weighed, and released 50 does in December, and the average weight was x=64.2 kg. Do you think the doe population is undernourished or not? Explain.Focus Problem: Impulse Buying Let x represent the dollar amount spent on supermarket impulse buying in a 10-minute (unplanned) shopping interval. Based on a Denver Post article, the mean of the x distribution is about 20 and the estimated standard deviation is about 7. (a) Consider a random sample of n = 100 customers, each of whom has 10 minutes of unplanned shopping time in a supermarket. From the central limit theorem, what can you say about the probability distribution of x, the average amount spent by these customers due to impulse buying? What are the mean and standard deviation of the x distribution? Is it necessary to make any assumption about the x distribution? Explain. (b) What is the probability that x is between 18 and 22? (c) Let us assume that x has a distribution that is approximately normal. What is the probability that x is between 18 and 22? (d) Interpretation: In part (b), we used x, the average amount spent, computed for 100 customers. In part (c), we used x, the amount spent by only one customer. The answers to parts (b) and (c) are very different. Why would this happen? In this example, x is a much more predictable or reliable statistic than x. Consider that almost all marketing strategies and sales pitches are designed for the average customer and not the individual customer. How does the central limit theorem tell us that the average customer is much more predictable than the individual customer?Finance: Templeton Funds Templeton World is a mutual fund that invests in both U.S. and foreign markets. Let x be a random variable that represents the monthly percentage return for the Templeton World fund. Based on information from the Morningstar Guide to Mutual Funds (available in most libraries), x has mean = 1.6% and standard deviation = 0.9%. (a) Templeton World fund has over 250 stocks that combine together to give the overall monthly percentage return x. We can consider the monthly return of the stocks in the fund to be a sample from the population of monthly returns of all world stocks. Then we see that the overall monthly return x for Templeton World fund is itself an average return computed using all 250 stocks in the fund. Why would this indicate that x has an approximately normal distribution? Explain. Hint: See the discussion after Theorem 6.2. (b) After 6 months, what is the probability that the average monthly percentage return x will be between 1% and 2%? Hint: See Theorem 6.1, and assume that x has a normal distribution as based on part (a). (c) After 2 years, what is the probability that x will be between 1% and 2%? (d) Compare your answers to parts (b) and (c). Did the probability increase as n (number of months) increased? Why would this happen? (e) Interpretation: If after 2 years the average monthly percentage return x was less than 1%, would that tend to shake your confidence in the statement that = 1.6%? Might you suspect that has slipped below 1.6%? Explain. THEOREM 6.1 For a Normal Probability Distribution Let x be a random variable with a normal distribution whose mean is and whose standard deviation is . Let x be the sample mean corresponding to random samples of size n taken from the x distribution. Then the following are true: (a) The x distribution is a normal distribution. (b) The mean of the x distribution is . (c) The standard deviation of the x distribution is /n. We conclude from Theorem 6.1 that when x has a normal distribution, the x distribution will be normal for any sample size n. Furthermore, we can convert the x distribution to the standard normal z distribution using the following formulas. x=x=nz=xxx=x/n where n is the sample size, is the mean of the x distribution, and is the standard deviation of the x distribution. Theorem 6.1 is a wonderful theorem! It states that the x distribution will be normal provided the x distribution is normal. The sample size n could be 2, 3, 4, or any other (fixed) sample size we wish. Furthermore, the mean of the x distribution is (same as for the x distribution), but the standard deviation is /n (which is, of course, smaller than ). The next example illustrates Theorem 6.1. THEOREM 6.2 The Central Limit Theorem for Any Probability Distribution If x possesses any distribution with mean and standard deviation , then the sample mean x based on a random sample of size n will have a distribution that approaches the distribution of a normal random variable with mean and standard deviation /n as n increases without limit. The central limit theorem is indeed surprising! It says that x can have any distribution whatsoever, but that as the sample size gets larger and larger, the distribution of x will approach a normal distribution. From this relation, we begin to appreciate the scope and significance of the normal distribution.Finance: European Growth Fund A European growth mutual fund specializes in stocks from the British Isles, continental Europe, and Scandinavia. The fund has over 100 stocks. Let x be a random variable that represents the monthly percentage return for this fund. Based on information from Morningstar (see Problem 19), x has mean = 1.4% and standard deviation = 0.8%. (a) Lets consider the monthly return of the stocks in the European growth fund to be a sample from the population of monthly returns of all European stocks. Is it reasonable to assume that x (the average monthly return on the 100 stocks in the European growth fund) has a distribution that is approximately normal? Explain. Hint: See Problem 19, part (a). (b) After 9 months, what is the probability that the average monthly percentage return x will be between 1% and 2%? Hint: See Theorem 6.1 and the results of part (a). (c) After 18 months, what is the probability that the average monthly percentage return x will be between 1% and 2%? (d) Compare your answers to parts (b) and (c). Did the probability increase as n (number of months) increased? Why would this happen? (e) Interpretation: If after 18 months the average monthly percentage return x is more than 2%, would that tend to shake your confidence in the statement that = 1.4%? If this happened, do you think the European stock market might be heating up? Explain. 19. Finance: Templeton Funds Templeton World is a mutual fund that invests in both U.S. and foreign markets. Let x be a random variable that represents the monthly percentage return for the Templeton World fund. Based on information from the Morningstar Guide to Mutual Funds (available in most libraries), x has mean = 1.6% and standard deviation = 0.9%. (a) Templeton World fund has over 250 stocks that combine together to give the overall monthly percentage return x. We can consider the monthly return of the stocks in the fund to be a sample from the population of monthly returns of all world stocks. Then we see that the overall monthly return x for Templeton World fund is itself an average return computed using all 250 stocks in the fund. Why would this indicate that x has an approximately normal distribution? Explain. Hint: See the discussion after Theorem 6.2. (b) After 6 months, what is the probability that the average monthly percentage return x will be between 1% and 2%? Hint: See Theorem 6.1, and assume that x has a normal distribution as based on part (a). (c) After 2 years, what is the probability that x will be between 1% and 2%? (d) Compare your answers to parts (b) and (c). Did the probability increase as n (number of months) increased? Why would this happen? (e) Interpretation: If after 2 years the average monthly percentage return x was less than 1%, would that tend to shake your confidence in the statement that = 1.6%? Might you suspect that has slipped below 1.6%? Explain. THEOREM 6.1 For a Normal Probability Distribution Let x be a random variable with a normal distribution whose mean is and whose standard deviation is . Let x be the sample mean corresponding to random samples of size n taken from the x distribution. Then the following are true: (a) The x distribution is a normal distribution. (b) The mean of the x distribution is . (c) The standard deviation of the x distribution is /n. We conclude from Theorem 6.1 that when x has a normal distribution, the x distribution will be normal for any sample size n. Furthermore, we can convert the x distribution to the standard normal z distribution using the following formulas. x=x=nz=xxx=x/n where n is the sample size, is the mean of the x distribution, and is the standard deviation of the x distribution. Theorem 6.1 is a wonderful theorem! It states that the x distribution will be normal provided the x distribution is normal. The sample size n could be 2, 3, 4, or any other (fixed) sample size we wish. Furthermore, the mean of the x distribution is (same as for the x distribution), but the standard deviation is /n (which is, of course, smaller than ). The next example illustrates Theorem 6.1. THEOREM 6.2 The Central Limit Theorem for Any Probability Distribution If x possesses any distribution with mean and standard deviation , then the sample mean x based on a random sample of size n will have a distribution that approaches the distribution of a normal random variable with mean and standard deviation /n as n increases without limit. The central limit theorem is indeed surprising! It says that x can have any distribution whatsoever, but that as the sample size gets larger and larger, the distribution of x will approach a normal distribution. From this relation, we begin to appreciate the scope and significance of the normal distribution.Expand Your Knowledge: Totals Instead of Averages Let x be a random variable that represents checkout time (time spent in the actual checkout process) in minutes in the express lane of a large grocery. Based on a consumer survey, the mean of the x distribution is about = 2.7 minutes, with standard deviation = 0.6 minute. Assume that the express lane always has customers waiting to be checked out and that the distribution of x values is more or less symmetric and mound-shaped. What is the probability that the total checkout time for the next 30 customers is less than 90 minutes? Let us solve this problem in steps. (a) Let xi (for i = 1, 2, 3, , 30) represent the checkout time for each customer. For example, x1 is the checkout time for the first customer, x2 is the checkout time for the second customer, and so forth. Each xi has mean = 2.7 minutes and standard deviation = 0.6 minute. Let w = x1 + x2 + + x30. Explain why the problem is asking us to compute the probability that w is less than 90. (b) Use a little algebra and explain why w 90 is mathematically equivalent to w/30 3. Since w is the total of the 30 x values, then w/30=x. Therefore, the statement x3 is equivalent to the statement w 90. From this we conclude that the probabilities P(x3) and P(w 90) are equal. (c) What does the central limit theorem say about the probability distribution of x? Is it approximately normal? What are the mean and standard deviation of the x distribution? (d) Use the result of part (c) to compute P(x3). What does this result tell you about P(w 90)?22P 23P 1P 2P Basic Computation: Normal Approximation to a Binomial Distribution Suppose we have a binomial experiment with n = 40 trials and a probability of success p = 0.50. (a) Is it appropriate to use a normal approximation to this binomial distribution? Why? (b) Compute and of the approximating normal distribution. (c) Use a continuity correction factor to convert the statement r 23 successes to a statement about the corresponding normal variable x. (d) Estimate P(r 23). (e) Interpretation Is it unusual for a binomial experiment with 40 trials and probability of success 0.50 to have 23 or more successes? Explain.Basic Computation: Normal Approximation to a Binomial Distribution Suppose we have a binomial experiment with n = 40 trials and probability of success p = 0.85. (a) Is it appropriate to use a normal approximation to this binomial distribution? Why? (b) Compute and of the approximating normal distribution. (c) Use a continuity correction factor to convert the statement r 30 successes to a statement about the corresponding normal variable x. (d) Estimate P(r 30). (e) Interpretation Is it unusual for a binomial experiment with 40 trials and probability of success 0.85 to have fewer than 30 successes? Explain.Critical Thinking You need to compute the probability of 5 or fewer successes for a binomial experiment with 10 trials. The probability of success on a single trial is 0.43. Since this probability of success is not in the table, you decide to use the normal approximation to the binomial. Is this an appropriate strategy? Explain.Critical Thinking Consider a binomial experiment with 20 trials and probability 0.45 of success on a single trial. (a) Use the binomial distribution to find the probability of exactly 10 successes. (b) Use the normal distribution to approximate the probability of exactly 10 successes. (c) Compare the results of parts (a) and (b).In the following problems, check that it is appropriate to use the normal approximation to the binomial. Then use the normal distribution to estimate the requested probabilities. Health: Lead Contamination More than a decade ago, high levels of lead in the blood put 88% of children at risk. A concerted effort was made to remove lead from the environment. Now, according to the Third National Health and Nutrition Examination Survey (NHANES III) conducted by the Centers for Disease Control and Prevention, only 9% of children in the United States are at risk of high blood-lead levels. (a) In a random sample of 200 children taken more than a decade ago, what is the probability that 50 or more had high blood-lead levels? (b) In a random sample of 200 children taken now, what is the probability that 50 or more have high blood-lead levels?In the following problems, check that it is appropriate to use the normal approximation to the binomial. Then use the normal distribution to estimate the requested probabilities. Insurance: Claims Do you try to pad an insurance claim to cover your deductible? About 40% of all U.S. adults will try to pad their insurance claims! (Source: Are You Normal?, by Bernice Kanner, St. Martins Press.) Suppose that you are the director of an insurance adjustment office. Your office has just received 128 insurance claims to be processed in the next few days. What is the probability that (a) half or more of the claims have been padded? (b) fewer than 45 of the claims have been padded? (c) from 40 to 64 of the claims have been padded? (d) more than 80 of the claims have not been padded?9P 10P 11P 12P 13P In the following problems, check that it is appropriate to use the normal approximation to the binomial. Then use the normal distribution to estimate the requested probabilities. Ice Cream: Flavors Whats your favorite ice cream flavor? For people who buy ice cream, the all-time favorite is still vanilla. About 25% of ice cream sales are vanilla. Chocolate accounts for only 9% of ice cream sales. (See reference in Problem 8.) Suppose that 175 customers go to a grocery store in Cheyenne, Wyoming today to buy ice cream. (a) What is the probability that 50 or more will buy vanilla? (b) What is the probability that 12 or more will buy chocolate? (c) A customer who buys ice cream is not limited to one container or one flavor. What is the probability that someone who is buying ice cream will buy chocolate or vanilla? Hint: Chocolate flavor and vanilla flavor are not mutually exclusive events. Assume that the choice to buy one flavor is independent of the choice to buy another flavor. Then use the multiplication rule for independent events, together with the addition rule for events that are not mutually exclusive, to compute the requested probability. (See Section 4.2.) (d) What is the probability that between 50 and 60 customers will buy chocolate or vanilla ice cream? Hint: Use the probability of success computed in part (c). 8. Insurance: Claims Do you try to pad an insurance claim to cover your deductible? About 40% of all U.S. adults will try to pad their insurance claims! (Source: Are You Normal?, by Bernice Kanner, St. Martins Press.) Suppose that you are the director of an insurance adjustment office. Your office has just received 128 insurance claims to be processed in the next few days. What is the probability that (a) half or more of the claims have been padded? (b) fewer than 45 of the claims have been padded? (c) from 40 to 64 of the claims have been padded? (d) more than 80 of the claims have not been padded?15P 16P 17P 18P 19P Basic Computation: p Distribution Suppose we have a binomial experiment in which success is defined to be a particular quality or attribute that interests us. (a) Suppose n = 33 and p = 0.21. Can we approximate the p distribution by a normal distribution? Why? What are the values of p and p? (b) Suppose n = 25 and p = 0.15. Can we safely approximate the p distribution by a normal distribution? Why or why not? (c) Suppose n = 48 and p = 0.15. Can we approximate the p distribution by a normal distribution? Why? What are the values of p and p?21P 1CRP 2CRP Statistical Literacy Is a process in control if the corresponding control chart for data having a normal distribution shows a value beyond 3 standard deviations of the mean?4CRP 5CRP 6CRP 7CRP 8CRP 9CRP 10CRP 11CRP Basic Computation: Probability Given that x is a normal variable with mean = 110 and standard deviation = 12, find (a) P(x 120) (b) P(x 80) (c) P(108 x 117)13CRP 14CRP 15CRP 16CRP 17CRP 18CRP 19CRP 20CRP 21CRP 22CRP 23CRP 24CRP 25CRP 26CRP Break into small groups and discuss the following topics. Organize a brief outline in which you summarize the main points of your group discussion. Iris setosa is a beautiful wildflower that is found in such diverse places as Alaska, the Gulf of St. Lawrence, much of North America, and even in English meadows and parks. R. A. Fisher, with his colleague Dr. Edgar Anderson, studied these flowers extensively. Dr. Anderson described how he collected information on irises: I have studied such irises as I could get to see, in as great detail as possible, measuring iris standard after iris standard and iris fall after iris fall, sitting squat-legged with record book and ruler in mountain meadows, in cypress swamps, on lake beaches, and in English parks. [E. Anderson, The Irises of the Gasp Peninsula, Bulletin, American Iris Society, Vol. 59 pp. 25, 1935.] The data in Table 6-11 were collected by Dr. Anderson and were published by his friend and colleague R. A. Fisher in a paper titled The Use of Multiple Measurements in Taxonomic Problems (Annals of Eugenics, part II, pp. 179188, 1936). To find these data, visit the Carnegie Mellon University Data and Story Library (DASL) web site. From the DASL site, look under Biology and Wild iris select Fisher's Irises Story. Let x be a random variable representing petal length. Using a TI-84Plus/ TI-83Plus/TI-nspire calculator, it was found that the sample mean is x = 1.46 centimeters (cm) and the sample standard deviation is s = 0.17 cm. Figure 6-39 shows a histogram for the given data generated on a TI-84Plus/TI-83Plus/TI-nspire calculator. (a) Examine the histogram for petal lengths. Would you say that the distribution is approximately mound-shaped and symmetric? Our sample has only 50 irises; if many thousands of irises had been used, do you think the distribution would look even more like a normal curve? Let x be the petal length of Iris setosa. Research has shown that x has an approximately normal distribution, with mean = 1.5 cm and standard deviation = 0.2 cm. (b) Use the empirical rule with = 1.5 and = 0.2 to get an interval into which approximately 68% of the petal lengths will fall. Repeat this for 95% and 99.7%. Examine the raw data and compute the percentage of the raw data that actually fall into each of these intervals (the 68% interval, the 95% interval, and the 99.7% interval). Compare your computed percentages with those given by the empirical rule. (c) Compute the probability that a petal length is between 1.3 and 1.6 cm. Compute the probability that a petal length is greater than 1.6 cm. (d) Suppose that a random sample of 30 irises is obtained. Compute the probability that the average petal length for this sample is between 1.3 and 1.6 cm. Compute the probability that the average petal length is greater than 1.6 cm. (e) Compare your answers to parts (c) and (d). Do you notice any differences? Why would these differences occur?1LC 2LC 3LC 4LC Discuss each of the following topics in class or review the topics on your own. Then write a brief but complete essay in which you summarize the main points. Please include formulas and graphs as appropriate. In a way, the central limit theorem can be thought of as a kind of grand central station. It is a connecting hub or center for a great deal of statistical work. We will use it extensively in Chapters 7, 8, and 9. Put in a very elementary way, the central limit theorem states that as the sample size n increases, the distribution of the sample mean x will always approach a normal distribution, no matter where the original x variable came from. For most people, it is the complete generality of the central limit theorem that is so awe inspiring: It applies to practically everything. List and discuss at least three variables from everyday life for which you expect the variable x itself not to follow a normal or bell-shaped distribution. Then discuss what would happen to the sampling distribution x if the sample size were increased. Sketch diagrams of the x distributions as the sample size n increases.1UT 2UT 1CURP 2CURP 3CURP 4CURP 5CURP 6CURP 7CURP 8CURP 9CURP 10CURP 11CURP 12CURP 13CURP 14CURP 15CURP 16CURP 17CURP 18CURP 19CURP 20CURP In Problems 18, answer true or false. Explain your answer. 1. Statistical Literacy The value zc is a value from the standard normal distribution such that P(zc x zc) c.In Problems 18, answer true or false. Explain your answer. 2. Statistical Literacy The point estimate for the population mean of an x distribution is x, computed from a random sample of the x distribution.In Problems 18, answer true or false. Explain your answer. 3. Statistical Literacy Consider a random sample of size n from an x distribution. For such a sample, the margin of error for estimating is the magnitude of the difference between x and .In Problems 18, answer true or false. Explain your answer. 4. Statistical Literacy Every random sample of the same size from a given population will produce exactly the same confidence interval for .In Problems 18, answer true or false. Explain your answer. 5. Statistical Literacy A larger sample size produces a longer confidence interval for .In Problems 18, answer true or false. Explain your answer. 6. Statistical Literacy If the original x distribution has a relatively small standard deviation, the confidence interval for will be relatively short.In Problems 18, answer true or false. Explain your answer. 7. Statistical Literacy If the sample mean x of a random sample from an x distribution is relatively small, then the confidence interval for will be relatively short.In Problems 18, answer true or false. Explain your answer. 8. Statistical Literacy For the same random sample, when the confidence level c is reduced, the confidence interval for becomes shorter.Critical Thinking Sam computed a 95% confidence interval for from a specific random sample. His confidence interval was 10.1 12.2. He claims that the probability that is in this interval is 0.95. What is wrong with his claim?Critical Thinking Sam computed a 90% confidence interval for from a specific random sample of size n. He claims that at the 90% confidence level, his confidence interval contains . Is his claim correct? Explain.Basic Computation: Confidence Interval Suppose x has a normal distribution with = 6. A random sample of size 16 has sample mean 50. (a) Check Requirements Is it appropriate to use a normal distribution to compute a confidence interval for the population mean ? Explain. (b) Find a 90% confidence interval for . (c) Interpretation Explain the meaning of the confidence interval you computed.Basic Computation: Confidence Interval Suppose x has a mound-shaped distribution with = 9. A random sample of size 36 has sample mean 20. (a) Check Requirements Is it appropriate to use a normal distribution to compute a confidence interval for the population mean ? Explain. (b) Find a 95% confidence interval for . (c) Interpretation Explain the meaning of the confidence interval you computed.Basic Computation: Sample Size Suppose x has a mound-shaped distribution with = 3. (a) Find the minimal sample size required so that for a 95% confidence interval, the maximal margin of error is E = 0.4. (b) Check Requirements Based on this sample size, can we assume that the x distribution is approximately normal? Explain.Basic Computation: Sample Size Suppose x has a normal distribution with = 1.2. (a) Find the minimal sample size required so that for a 90% confidence interval, the maximal margin of error is E = 0.5. (b) Check Requirements Based on this sample size and the x distribution, can we assume that the x distribution is approximately normal? Explain.Zoology: Hummingbirds Allens hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther (Reference: Hummingbirds by K. Long and W. Alther). A small group of 15 Allens hummingbirds has been under study in Arizona. The average weight for these birds is x = 3.15 grams. Based on previous studies, we can assume that the weights of Allens hummingbirds have a normal distribution, with = 0.33gram. (a) Find an 80% confidence interval for the average weights of Allens hummingbirds in the study region. What is the margin of error? (b) What conditions are necessary for your calculations? (c) Interpret your results in the context of this problem. (d) Sample Size Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.08 for the mean weights of the hummingbirds.Diagnostic Tests: Uric Acid Overproduction of uric acid in the body can be an indication of cell breakdown. This may be an advance indication of illness such as gout, leukemia, or lymphoma (Reference: Manual of Laboratory and Diagnostic Tests by F. Fischbach). Over a period of months, an adult male patient has taken eight blood tests for uric acid. The mean concentration was x = 5.35mg/dl. The distribution of uric acid in healthy adult males can be assumed to be normal, with = 1.85mg/dl. (a) Find a 95% confidence interval for the population mean concentration of uric acid in this patients blood. What is the margin of error? (b) What conditions are necessary for your calculations? (c) Interpret your results in the context of this problem. (d) Sample Size Find the sample size necessary for a 95% confidence level with maximal margin of error E = 1.10 for the mean concentration of uric acid in this patients blood.Diagnostic Tests: Plasma Volume Total plasma volume is important in determining the required plasma component in blood replacement therapy for a person undergoing surgery. Plasma volume is influenced by the overall health and physical activity of an individual. (Reference: See Problem 16.) Suppose that a random sample of 45 male firefighters are tested and that they have a plasma volume sample mean of x = 37.5ml/kg (milliliters plasma per kilogram body weight). Assume that = 7.50ml/kg for the distribution of blood plasma. (a) Find a 99% confidence interval for the population mean blood plasma volume in male firefighters. What is the margin of error? (b) What conditions are necessary for your calculations? (c) Interpret your results in the context of this problem. (d) Sample Size Find the sample size necessary for a 99% confidence level with maximal margin of error E 5 2.50 for the mean plasma volume in male firefighters.Agriculture: Watermelon What price do farmers get for their watermelon crops? In the third week of July, a random sample of 40 farming regions gave a sample mean of x = 6.88 per 100 pounds of watermelon. Assume that is known to be 1.92 per 100 pounds (Reference: Agricultural Statistics, U.S. Department of Agriculture). (a) Find a 90% confidence interval for the population mean price (per 100 pounds) that farmers in this region get for their watermelon crop. What is the margin of error? (b) Sample Size Find the sample size necessary for a 90% confidence level with maximal margin of error E 5 0.3 for the mean price per 100 pounds of watermelon. (c) A farm brings 15 tons of watermelon to market. Find a 90% confidence interval for the population mean cash value of this crop. What is the margin of error? Hint: 1 ton is 2000 pounds.19P Confidence Intervals: Values of A random sample of size 36 is drawn from an x distribution. The sample mean is 100. (a) Suppose the x distribution has = 30. Compute a 90% confidence interval for . What is the value of the margin of error? (b) Suppose the x distribution has = 20. Compute a 90% confidence interval for . What is the value of the margin of error? (c) Suppose the x distribution has = 10. Compute a 90% confidence interval for . What is the value of the margin of error? (d) Compare the margins of error for parts (a) through (c). As the standard deviation decreases, does the margin of error decrease? (e) Critical Thinking Compare the lengths of the confidence intervals for parts (a) through (c). As the standard deviation decreases, does the length of a 90% confidence interval decrease?Confidence Intervals: Sample Size A random sample is drawn from a population with = 12. The sample mean is 30. (a) Compute a 95% confidence interval for based on a sample of size 49. What is the value of the margin of error? (b) Compute a 95% confidence interval for based on a sample of size 100. What is the value of the margin of error? (c) Compute a 95% confidence interval for based on a sample of size 225. What is the value of the margin of error? (d) Compare the margins of error for parts (a) through (c). As the sample size increases, does the margin of error decrease? (e) Critical Thinking Compare the lengths of the confidence intervals for parts (a) through (c). As the sample size increases, does the length of a 90% confidence interval decrease?Ecology: Sand Dunes At wind speeds above 1000 centimeters per second (cm/sec), significant sand-moving events begin to occur. Wind speeds below 1000 cm/sec deposit sand, and wind speeds above 1000 cm/sec move sand to new locations. The cyclic nature of wind and moving sand determines the shape and location of large dunes (Reference: Hydraulic, Geologic, and Biologic Research at Great Sand Dunes National Monument and Vicinity, Colorado, Proceedings of the National Park Service Research Symposium). At a test site, the prevailing direction of the wind did not change noticeably. However, the velocity did change. Sixty wind speed readings gave an average velocity of x = 1075 cm/sec. Based on long-term experience, can be assumed to be 265 cm/sec. (a) Find a 95% confidence interval for the population mean wind speed at this site. (b) Interpretation Does the confidence interval indicate that the population mean wind speed is such that the sand is always moving at this site? Explain.23P 24P 25P Use Table 6 of Appendix II to find tc for a 0.95 confidence level when the sample size is 18.2P 3P 4P 5P 6P 7P 8P 9P 10P Basic Computation: Confidence Interval Suppose x has a mound-shaped symmetric distribution. A random sample of size 16 has sample mean 10 and sample standard deviation 2. (a) Check Requirements Is it appropriate to use a Students t distribution to compute a confidence interval for the population mean ? Explain. (b) Find a 90% confidence interval for . (c) Interpretation Explain the meaning of the confidence interval you computed.Basic Computation: Confidence Interval A random sample of size 81 has sample mean 20 and sample standard deviation 3. (a) Check Requirements Is it appropriate to use a Students t distribution to compute a confidence interval for the population mean ? Explain. (b) Find a 95% confidence interval for . (c) Interpretation Explain the meaning of the confidence interval you computed.In Problems 1319, assume that the population of x values has an approximately normal distribution. Archaeology: Tree Rings At Burnt Mesa Pueblo, the method of tree-ring dating gave the following years A.D. for an archaeological excavation site (Bandelier Archaeological Excavation Project: Summer 1990 Excavations at Burnt Mesa Pueblo, edited by Kohler, Washington State University): (a) Use a calculator with mean and standard deviation keys to verify that the sample mean year is x 1272, with sample standard deviation 37years. (b) Find a 90% confidence interval for the mean of all tree-ring dates from this archaeological site. (c) Interpretation What does the confidence interval mean in the context of this problem?In Problems 1319, assume that the population of x values has an approximately normal distribution. Camping: Cost of a Sleeping Bag How much does a sleeping bag cost? Lets say you want a sleeping bag that should keep you warm in temperatures from 20F to 45F. A random sample of prices () for sleeping bags in this temperature range was taken from Backpacker Magazine: Gear Guide (Vol. 25, Issue 157, No. 2). Brand names include American Camper, Cabelas, Camp 7, Caribou, Cascade, and Coleman. (a) Use a calculator with mean and sample standard deviation keys to verify that x 83.75 and 28.97. (b) Using the given data as representative of the population of prices of all summer sleeping bags, find a 90% confidence interval for the mean price of all summer sleeping bags. (c) Interpretation What does the confidence interval mean in the context of this problem?In Problems 1319, assume that the population of x values has an approximately normal distribution. Wildlife: Mountain Lions How much do wild mountain lions weigh? The 77th Annual Report of the New Mexico Department of Game and Fish, edited by Bill Montoya, gave the following information. Adult wild mountain lions (18 months or older) captured and released for the first time in the San Andres Mountains gave the following weights (pounds): (a) Use a calculator with mean and sample standard deviation keys to verify that x = 91.0 pounds and s 30.7 pounds. (b) Find a 75% confidence interval for the population average weight m of all adult mountain lions in the specified region. (c) Interpretation What does the confidence interval mean in the context of this problem?In Problems 1319, assume that the population of x values has an approximately normal distribution. Franchise: Candy Store Do you want to own your own candy store? With some interest in running your own business and a decent credit rating, you can probably get a bank loan on startup costs for franchises such as Candy Express, The Fudge Company, Karmel Corn, and Rocky Mountain Chocolate Factory. Startup costs (in thousands of dollars) for a random sample of candy stores are given below (Source: Entrepreneur Magazine, Vol. 23, No. 10). (a) Use a calculator with mean and sample standard deviation keys to verify that x 106.9 thousand dollars and s 29.4 thousand dollars. (b) Find a 90% confidence interval for the population average startup costs m for candy store franchises. (c) Interpretation What does the confidence interval mean in the context of this problem?In Problems 1319, assume that the population of x values has an approximately normal distribution. Diagnostic Tests: Total Calcium Over the past several months, an adult patient has been treated for tetany (severe muscle spasms). This condition is associated with an average total calcium level below 6 mg/dl (Reference: Manual of Laboratory and Diagnostic Tests by F. Fischbach). Recently, the patients total calcium tests gave the following readings (in mg/dl). (a) Use a calculator to verify that x = 9.95 and s 1.02. (b) Find a 99.9% confidence interval for the population mean of total calcium in this patients blood. (c) Interpretation Based on your results in part (b), does it seem that this patient still has a calcium deficiency? Explain.18P 19P 20P 21P 22P 23P For all these problems, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding. Statistical Literacy For a binomial experiment with r successes out of n trials, what value do we use as a point estimate for the probability of success p on a single trial?2P 3P 4P For all these problems, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding. Critical Thinking Jerry tested 30 laptop computers owned by classmates enrolled in a large computer science class and discovered that 22 were infected with keystroke-tracking spyware. Is it appropriate for Jerry to use his data to estimate the proportion of all laptops infected with such spyware? Explain.6P 7P 8P 9P For all these problems, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding. Basic Computation: Sample Size What is the minimal sample size needed for a 99% confidence interval to have a maximal margin of error of 0.06 (a) if a preliminary estimate for p is 0.8? (b) if there is no preliminary estimate for p?11P 12P 13P For all these problems, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding. Archaeology: Pottery Santa Fe black-on-white is a type of pottery commonly found at archaeological excavations in Bandelier National Monument. At one excavation site a sample of 592 potsherds was found, of which 360 were identified as Santa Fe black-on-white (Bandelier Archaeological Excavation Project: Summer 1990 Excavations at Burnt Mesa Pueblo and Casa del Rito, edited by Kohler and Root, Washington State University). (a) Let p represent the population proportion of Santa Fe black-on-white potsherds at the excavation site. Find a point estimate for p. (b) Find a 95% confidence interval for p. Give a brief statement of the meaning of the confidence interval. (c) Check Requirements Do you think the conditions np 5 and nq 5 are satisfied in this problem? Why would this be important?15P 16P 17P 18P 19P 20P 21P 22P 23P 24P 25P 26P 27P 28P 1P 2P 3P 4P 5P 6P 7P 8P 9P 10P 11P 12P 13P 14P 15P 16P Answers may vary slightly due to rounding. MyersBriggs: Marriage Counseling Isabel Myers was a pioneer in the study of personality types. She identified four basic personality preferences, which are described at length in the book A Guide to the Development and Use of the MyersBriggs Type Indicator by Myers and McCaulley (Consulting Psychologists Press). Marriage counselors know that couples who have none of the four preferences in common may have a stormy marriage. Myers took a random sample of 375 married couples and found that 289 had two or more personality preferences in common. In another random sample of 571 married couples, it was found that only 23 had no preferences in common. Let p1 be the population proportion of all married couples who have two or more personality preferences in common. Let p2 be the population proportion of all married couples who have no personality preferences in common. (a) Check Requirements Can a normal distribution be used to approximate the p1p2 distribution? Explain. (b) Find a 99% confidence interval for p1 p2. (c) Interpretation Explain the meaning of the confidence interval in part (a) in the context of this problem. Does the confidence interval contain all positive, all negative, or both positive and negative numbers? What does this tell you (at the 99% confidence level) about the proportion of married couples with two or more personality preferences in common compared with the proportion of married couples sharing no personality preferences in common?18P 19P 20P 21P 22P 23P 24P 25P 26P 27P 28P 29P 30P 31P 1CRP Critical Thinking Suppose you are told that a 95% confidence interval for the average price of a gallon of regular gasoline in your state is from 3.15 to 3.45. Use the fact that the confidence interval for the mean has the form x E to x 1 E to compute the sample mean and the maximal margin of error E.3CRP 4CRP 5CRP For Problems 419, categorize each problem according to the parameter being estimated: proportion p, mean m, difference of means 1 2, or difference of proportions p1 p2. Then solve the problem. Psychology: Closure How large a sample is needed in Problem 5 if we wish to be 99% confident that the sample mean score is within 2 points of the population mean score for students who are high on the need for closure?7CRP For Problems 419, categorize each problem according to the parameter being estimated: proportion p, mean m, difference of means 1 2, or difference of proportions p1 p2. Then solve the problem. Archaeology: Pottery Shards of clay vessels were put together to reconstruct rim diameters of the original ceramic vessels found at the Wind Mountain archaeological site (see source in Problem 7). A random sample of ceramic vessels gave the following rim diameters (in centimeters): (a) Use a calculator with mean and sample standard deviation keys to verify that x 15.8 cm and s 3.5 cm. (b) Compute an 80% confidence interval for the population mean m of rim diameters for such ceramic vessels found at the Wind Mountain archaeological site.9CRP For Problems 419, categorize each problem according to the parameter being estimated: proportion p, mean , difference of means 1 2, or difference of proportions p1 p2. Then solve the problem. Telephone Interviews: Survey How large a sample is needed in Problem 9 if we wish to be 95% confident that the sample percentage of those equating success with personal satisfaction is within 1% of the population percentage? Hint: Use p 0.52 as a preliminary estimate.For Problems 419, categorize each problem according to the parameter being estimated: proportion p, mean m, difference of means 1 2, or difference of proportions p1 p2. Then solve the problem. Archaeology: Pottery Three-circle, red-on-white is one distinctive pattern painted on ceramic vessels of the Anasazi period found at the Wind Mountain archaeological site (see source for Problem 7). At one excavation, a sample of 167 potsherds indicated that 68 were of the three-circle, redfion-white pattern. (a) Find a point estimate p for the proportion of all ceramic potsherds at this site that are of the three-circle, red-on-white pattern. (b) Compute a 95% confidence interval for the population proportion p of all ceramic potsherds with this distinctive pattern found at the site.For Problems 419, categorize each problem according to the parameter being estimated: proportion p, mean m, difference of means 1 2, or difference of proportions p1 p2. Then solve the problem. Archaeology: Pottery Consider the three-circle, red-on-white pattern discussed in Problem 11. How many ceramic potsherds must be found and identified if we are to be 95% confident that the sample proportion p of such potsherds is within 6% of the population proportion of three-circle, red-on-white patterns found at this excavation site? Hint: Use the results of Problem 11 as a preliminary estimate.13CRP 14CRP 15CRP For Problems 419, categorize each problem according to the parameter being estimated: proportion p, mean m, difference of means 1 2, or difference of proportions p1 p2. Then solve the problem. Wildlife: Wolves A random sample of 17 wolf litters in Ontario, Canada, gave an average of x1 = 4.9 wolf pups per litter, with estimated sample standard deviation s1 = 1.0. Another random sample of 6 wolf litters in Finland gave an average of x2 = 2.8 wolf pups per litter, with sample standard deviation s2 = 1.2 (see source for Problem 15). (a) Find an 85% confidence interval for 1 2, the difference in population mean litter size between Ontario and Finland. (b) Interpretation Examine the confidence interval and explain what it means in the context of this problem. Does the interval consist of numbers that are all positive? all negative? of different signs? At the 85% level of confidence, does it appear that the average litter size of wolf pups in Ontario is greater than the average litter size in Finland?17CRP 18CRP 19CRP 1DH 2DH 3DH 1LC 2LC 3LC 1UT 2UT 3UT Statistical Literacy Discuss each of the following topics in class or review the topics on your own. Then write a brief but complete essay in which you answer the following questions. (a) What is a null hypothesis H0? (b) What is an alternate hypothesis H1? (c) What is a type I error? a type II error? (d) What is the level of significance of a test? What is the probability of a type II error?Statistical Literacy In a statistical test, we have a choice of a left-tailed test, a right-tailed test, or a two-tailed test. Is it the null hypothesis or the alternate hypothesis that determines which type of test is used? Explain your answer.Statistical Literacy If we fail to reject (i.e., accept) the null hypothesis, does this mean that we have proved it to be true beyond all doubt? Explain your answer.Statistical Literacy If we reject the null hypothesis, does this mean that we have proved it to be false beyond all doubt? Explain your answer.Statistical Literacy What terminology do we use for the probability of rejecting the null hypothesis when it is true? What symbol do we use for this probability? Is this the probability of a type I or a type II error?6P Statistical Literacy If the P-value in a statistical test is greater than the level of significance for the test, do we reject or fail to reject H0?Statistical Literacy If the P-value in a statistical test is less than or equal to the level of significance for the test, do we reject or fail to reject H0?Statistical Literacy Suppose the P-value in a right-tailed test is 0.0092. Based on the same population, sample, and null hypothesis, what is the P-value for a corresponding two-tailed test?Statistical Literacy Suppose the P-value in a two-tailed test is 0.0134. Based on the same population, sample, and null hypothesis, and assuming the test statistic z is negative, what is the P-value for a corresponding left-tailed test?Basic Computation: Setting Hypotheses Suppose you want to test the claim that a population mean equals 40. (a) State the null hypothesis. (b) State the alternate hypothesis if you have no information regarding how the population mean might differ from 40. (c) State the alternate hypothesis if you believe (based on experience or past studies) that the population mean may exceed 40. (d) State the alternate hypothesis if you believe (based on experience or past studies) that the population mean may be less than 40.Basic Computation: Setting Hypotheses Suppose you want to test the claim that a population mean equals 30. (a) State the null hypothesis. (b) State the alternate hypothesis if you have no information regarding how the population mean might differ from 30. (c) State the alternate hypothesis if you believe (based on experience or past studies) that the population mean may be greater than 30. (d) State the alternate hypothesis if you believe (based on experience or past studies) that the population mean may not be as large as 30.Basic Computation: Find Test Statistic, Corresponding P-value, and Conclude Test A random sample of size 20 from a normal distribution with = 4 produced a sample mean of 8. (a) Check Requirements Is the x distribution normal? Explain. (b) Compute the sample test statistic z under the null hypothesis H0: = 7. (c) For H1: 7, estimate the P-value of the test statistic. (d) For a level of significance of 0.05 and the hypotheses of parts (b) and (c), do you reject or fail to reject the null hypothesis? Explain.Basic Computation: Find the Test Statistic, Corresponding P-value, and Conclude Test A random sample of size 16 from a normal distribution with = 3 produced a sample mean of 4.5. (a) Check Requirements Is the x distribution normal? Explain. (b) Compute the sample test statistic z under the null hypothesis H0: = 6.3. (c) For H1: 6.3, estimate the P-value of the test statistic. (d) For a level of significance of 0.01 and the hypotheses of parts (b) and (c), do you reject or fail to reject the null hypothesis? Explain.Veterinary Science: Colts The body weight of a healthy 3-month-old colt should be about 60 kg (Source: The Merck Veterinary Manual, a standard reference manual used in most veterinary colleges). (a) If you want to set up a statistical test to challenge the claim that = 60 kg, what would you use for the null hypothesis H0? (b) In Nevada, there are many herds of wild horses. Suppose you want to test the claim that the average weight of a wild Nevada colt (3 months old) is less than 60 kg. What would you use for the alternate hypothesis H1? (c) Suppose you want to test the claim that the average weight of such a wild colt is greater than 60 kg. What would you use for the alternate hypothesis? (d) Suppose you want to test the claim that the average weight of such a wild colt is different from 60 kg. What would you use for the alternate hypothesis? (e) For each of the tests in parts (b), (c), and (d), would the area corresponding to the P-value be on the left, on the right, or on both sides of the mean? Explain your answer in each case.Marketing: Shopping Time How much customers buy is a direct result of how much time they spend in a store. A study of average shopping times in a large national housewares store gave the following information (Source: Why We Buy: The Science of Shopping by P. Underhill): Women with female companion: 8.3 min. Women with male companion: 4.5 min. Suppose you want to set up a statistical test to challenge the claim that a woman with a female friend spends an average of 8.3 minutes shopping in such a store. (a) What would you use for the null and alternate hypotheses if you believe the average shopping time is less than 8.3 minutes? Is this a right-tailed, left-tailed, or two-tailed test? (b) What would you use for the null and alternate hypotheses if you believe the average shopping time is different from 8.3 minutes? Is this a right-tailed, left-tailed, or two-tailed test? Stores that sell mainly to women should figure out a way to engage the interest of menperhaps comfortable seats and a big TV with sports programs! Suppose such an entertainment center was installed and you now wish to challenge the claim that a woman with a male friend spends only 4.5 minutes shopping in a housewares store. (c) What would you use for the null and alternate hypotheses if you believe the average shopping time is more than 4.5 minutes? Is this a right-tailed, left-tailed, or two-tailed test? (d) What would you use for the null and alternate hypotheses if you believe the average shopping time is different from 4.5 minutes? Is this a right-tailed, left-tailed, or two-tailed test?Meteorology: Storms Weatherwise magazine is published in association with the American Meteorological Society. Volume 46, Number 6 has a rating system to classify Noreaster storms that frequently hit New England states and can cause much damage near the ocean coast. A severe storm has an average peak wave height of 16.4 feet for waves hitting the shore. Suppose that a Noreaster is in progress at the severe storm class rating. (a) Let us say that we want to set up a statistical test to see if the wave action (i.e., height) is dying down or getting worse. What would be the null hypothesis regarding average wave height? (b) If you wanted to test the hypothesis that the storm is getting worse, what would you use for the alternate hypothesis? (c) If you wanted to test the hypothesis that the waves are dying down, what would you use for the alternate hypothesis? (d) Suppose you do not know whether the storm is getting worse or dying out. You just want to test the hypothesis that the average wave height is different (either higher or lower) from the severe storm class rating. What would you use for the alternate hypothesis? (e) For each of the tests in parts (b), (c), and (d), would the area corresponding to the P-value be on the left, on the right, or on both sides of the mean? Explain your answer in each case.Chrysler Concorde: Acceleration Consumer Reports stated that the mean time for a Chrysler Concorde to go from 0 to 60 miles per hour is 8.7 seconds. (a) If you want to set up a statistical test to challenge the claim of 8.7 seconds, what would you use for the null hypothesis? (b) The town of Leadville, Colorado, has an elevation over 10,000 feet. Suppose you wanted to test the claim that the average time to accelerate from 0 to 60 miles per hour is longer in Leadville (because of less oxygen). What would you use for the alternate hypothesis? (c) Suppose you made an engine modification and you think the average time to accelerate from 0 to 60 miles per hour is reduced. What would you use for the alternate hypothesis? (d) For each of the tests in parts (b) and (c), would the P-value area be on the left, on the right, or on both sides of the mean? Explain your answer in each case.For Problems 1924, please provide the following information. (a) What is the level of significance? State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test? (b) Check Requirements What sampling distribution will you use? Explain the rationale for your choice of sampling distribution. Compute the z value of the sample test statistic. (c) Find (or estimate) the P-value. Sketch the sampling distribution and show the area corresponding to the P-value. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level a? (e) Interpret your conclusion in the context of the application. 19. Dividend Yield: Australian Bank Stocks Let x be a random variable representing dividend yield of Australian bank stocks. We may assume that x has a normal distribution with s 5 2.4%. A random sample of 10 Australian bank stocks gave the following yields. 5.7 4.8 6.0 4.9 4.0 3.4 6.5 7.1 5.3 6.1 The sample mean is x = 5.38%. For the entire Australian stock market, the mean dividend yield is = 4.7% (Reference: Forbes). Do these data indicate that the dividend yield of all Australian bank stocks is higher than 4.7%? Use = 0.01.For Problems 1924, please provide the following information. (a) What is the level of significance? State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test? (b) Check Requirements What sampling distribution will you use? Explain the rationale for your choice of sampling distribution. Compute the z value of the sample test statistic. (c) Find (or estimate) the P-value. Sketch the sampling distribution and show the area corresponding to the P-value. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level a? (e) Interpret your conclusion in the context of the application. 20. Glucose Level: Horses Gentle Ben is a Morgan horse at a Colorado dude ranch. Over the past 8 weeks, a veterinarian took the following glucose readings from this horse (in mg/100 ml). 93 88 82 105 99 110 84 89 The sample mean is x 93.8. Let x be a random variable representing glucose readings taken from Gentle Ben. We may assume that x has a normal distribution, and we know from past experience that = 12.5. The mean glucose level for horses should be = 85 mg/100 ml (Reference: Merck Veterinary Manual). Do these data indicate that Gentle Ben has an overall average glucose level higher than 85? Use = 0.05.For Problems 1924, please provide the following information. (a) What is the level of significance? State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test? (b) Check Requirements What sampling distribution will you use? Explain the rationale for your choice of sampling distribution. Compute the z value of the sample test statistic. (c) Find (or estimate) the P-value. Sketch the sampling distribution and show the area corresponding to the P-value. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level a? (e) Interpret your conclusion in the context of the application. 21. Ecology: Hummingbirds Bill Alther is a zoologist who studies Annas hummingbird (Calypte anna) (Reference: Hummingbirds by K. Long and W. Alther). Suppose that in a remote part of the Grand Canyon, a random sample of six of these birds was caught, weighed, and released. The weights (in grams) were 3.7 2.9 3.8 4.2 4.8 3.1 The sample mean is x = 3.75 grams. Let x be a random variable representing weights of Annas hummingbirds in this part of the Grand Canyon. We assume that x has a normal distribution and = 0.70 gram. It is known that for the population of all Annas hummingbirds, the mean weight is = 4.55 grams. Do the data indicate that the mean weight of these birds in this part of the Grand Canyon is less than 4.55 grams? Use = 0.01.For Problems 1924, please provide the following information. (a) What is the level of significance? State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test? (b) Check Requirements What sampling distribution will you use? Explain the rationale for your choice of sampling distribution. Compute the z value of the sample test statistic. (c) Find (or estimate) the P-value. Sketch the sampling distribution and show the area corresponding to the P-value. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level a? (e) Interpret your conclusion in the context of the application. Finance: P/E of Stocks The price-to-earnings (P/E) ratio is an important tool in financial work. A random sample of 14 large U.S. banks (J.P. Morgan, Bank of America, and others) gave the following P/E ratios (Reference: Forbes). The sample mean is x 17.1. Generally speaking, a low P/E ratio indicates a value or bargain stock. A recent copy of The Wall Street Journal indicated that the P/E ratio of the entire SP 500 stock index is = 19. Let x be a random variable representing the P/E ratio of all large U.S. bank stocks. We assume that x has a normal distribution and = 4.5. Do these data indicate that the P/E ratio of all U.S. bank stocks is less than 19? Use = 0.05.23P 24P Statistical Literacy For the same sample data and null hypothesis, how does the P-value for a two-tailed test of compare to that for a one-tailed test?Statistical Literacy To test for an x distribution that is mound-shaped using sample size n 30, how do you decide whether to use the normal or the Students t distribution?Statistical Literacy When using the Students t distribution to test , what value do you use for the degrees of freedom?Critical Thinking Consider a test for . If the P-value is such that you can reject H0 at the 5% level of significance, can you always reject H0 at the 1% level of significance? Explain.

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