1. A recent survey showed that from a sample of 500 packages delivered by a Postal Service, 480 were delivered on time. a) Construct a 95% confidence interval for the proportion of all packages that are delivered on time by the Postal Service.

Q: 1) A sample of 532 Peanut Butter M&M's contains 74 yellow candies; write a 95% confidence statement…

A: In this case, the confidence level is 95%. The 95% confidence interval for the proportion/percentage…

Q: 3- Car and Driver Magazine tested a random sample of 9 hybrid cars and found their MPG to be 42, 51,…

A:

Q: According to the February 2008 Federal Trade Commission report on consumer fraud and identity theft,…

A: Given n=1750 X=436

Q: If a retailer would like to estimate the proportion of their customers who bought an item after…

A: Given data,     P=58%=0.58     E=7%=0.07     z-value at 95% confidence is  zc =1.98     n=?

Q: 12. In a sample of 600 Canadians, 265 say they would rather retire in the US than in Canada.…

A:

Q: Suppose we want to estimate the proportion of teenagers (aged 13-18) who are lactose intolerant. If…

A: Given information Error = 4% = 0.04 Confidence level = 0.95

Q: 2. A nutritionist wants to know the population proportion of Grade 1 learners who eat vegetables.…

A: We have given thatn = 1200x = 200p̂ = x/n = 200/1200 = 0.167Significance level (α) = 1 - 0.95 = 0.05

Q: According to Scarborough Research, more than 85% of working adults commute by car. Of all U.S.…

A: Given that, According to Scarborough Research, more than 85% of working adults commute by car. Of…

Q: In a study of 200 accidents that required treatment in an emergency room, 80 occurred at work. Find…

A: Obtain the 90% confidence interval for the true proportion of accidents that occurred at work. The…

Q: 17. A researcher surveys 50 individuals in Smithville and 40 in Amherst, finding that 30% of…

A: State the hypotheses. That is, there is no evidence to conclude that the mobility rates in the two…

Q: Suppose in a recent poll of 1250 randomly selected Californians who took vacations, 132 of them…

A:

Q: 2. A national health organization warns that 30% of the middle school students nationwide have been…

A: Given : n = 110 X = 21

Q: Suppose we want to estimate the proportion of teenagers (aged 13-18) who are lactose intolerant. If…

A: Given information Error = 2% = 0.02 Confidence level = 0.95

Q: 1. In a recent poll 62 out of 149 randomly chosen students at Allen High School say they plan to buy…

A:

Q: 11) In a Gallup poll, 1011 adults were asked if they consume alcoholic beverages and 64% of them…

A: (11) Obtain the 90% confidence interval estimate of the proportion of all adults who consume…

Q: Suppose we want to estimate the proportion of teenagers (aged 13-18) who are lactose intolerant. If…

A:

Q: In a study of 120 randomly selected families, 50 said they streamed their television content. Find…

A:

Q: 1) A consumer magazine polls cell-phone-users to see if they are happy with their phones and would…

A:

Q: The production manager in an automobile plant is concerned with the number of cars that do not pass…

A:

Q: If n=160 and ˆp (p-hat) =0.48, find the margin of error at a 99% confidence level

A: Given Data: p^=0.48n=160   We know that, p^-pp^(1-p^)n~ Normal(0,1)

Q: In a study of 150 randomly selected accidents that required treatment in an emergency room, 36%…

A:

Q: Suppose we want to estimate the proportion of teenagers (aged 13-18) who are lactose intolerant. If…

A: According to the provided information, The z score at 95% confidence interval from the standard…

Q: In the Framingham Offspring study 3630 subjects were on anti-hypertensive medication out of 4657…

A: Note: Hi, thank you for the question. As per our company guideline we are supposed to answer only…

Q: 1. The Hartford Insurance Company wants to estimate the percentage of drivers who change tapes or…

A: Please see the attachment..

Q: In an article in Marketing Science, Silk and Berndt investigate the output of advertising agencies.…

A: Givenx=115n=390p^=xn=115390=0.2949confidence interval = 90%

Q: 4. The Department of the Treasury is interested in finding out what proportion of people support…

A: Given that  Population proportion p=15%=0.15 Margin of error =E =2%=0.02 95% confidence.  NOTE:-…

Q: The following table shows the results of a survey in which 1017 adults from the United States, 1060…

A: (1)

Q: in a random sample of 800 austin automobile owners, it was found that 480 would like to see the size…

A: We have to find given ci.

Q: It was observed that 30 of the 100 randomly selected students smoked. Find the confidence interval…

A: It is given that, 30 out of 100 randomly selected students smoked i.e. proportion of the student who…

Q: 6) An Internet service provider sampled 590 customers and found that 85 of them experienced an…

A: According to the given information, we have Total number of customers = 590 85 of them experienced…

Q: 9. A survey found that out of 900 workers, 150 said they were interrupted by frank phone calls. Find…

A:

Q: In a sports poll, 142 of 500 randomly selected Americans indicated that they consider themselves to…

A: Given that, In a sports poll, 142 of 500 randomly selected Americans indicated that they consider…

Q: In a sample of 1901 patients who underwent a certain type of surgery, 22% experienced complications.…

A:

Q: If 64% of a sample of 550 people leaving a shopping mall claims to have spent over $25 create and…

A: Given,   Sample size = 550 Sample proportion = 0.64   Critical value: Using z-table, the z-critical…

Q: wed that 122 involved uninsured drivers. Construct a 96% confidence interval for the proportion of…

A: A survey of 500 non-fatal accidents showed that 122 involved uninsured drivers. Let X be the number…

Q: In a random sample of 184 college students, 97 had part-time jobs. Find the margin of error for the…

A: Given : Sample Size, n = 184 Number of successes, x = 97 confidence level,c=0.99

Q: 1. In a poll of 2,500 likely voters, 966 say that the Philippines spend too little on education.…

A:

Q: A survey of 80 recent arcanum students showed that 46 were ready for summer break. Find the 95%…

A: Obtain the 95% confidence interval for the true proportion of students ready for summer break. The…

Q: Suppose that a safety group surveyed 1,200 drivers. Among those​ surveyed,78​% said that careless or…

A: Obtain the 99% confidence interval estimate for the true proportion in the population of all drivers…

Q: 1. A random sample of 15 households in Cebu City provided the following data on the number of…

A: Obtain the 99% confidence interval for the true average monthly electricity consumption in kWh.…

Q: 1. Construct a 90% confidence interval that would contain the Rampo College population mean of…

A: Let xi ,(i=1,2,...n) be a random sample of n observations from a population involving a single…

Q: b. Construct a 90% confidence interval for the percentage of all judges who are in favor of the…

A: a. A random sample of 15 judges and asked them whether or not they favor the death penalty in which…

Q: The Universal Music Group is the music industry leader worldwide in sales according to the company…

A: The random variable music sales follow normal distribution. The sample size is 1003. The sample…

Q: Construct a 99% confidence interval for the true proportion of all car license applications that…

A: Given Information:x=25n=155c=99%=0.99

Q: 1. Construct a 90% confidence interval for the proportion of employees requiring size 3.

A: here given  A civil engineering firm issues every new employee with a safety helmet. Four different…

Q: Data collected by a behavioral risk factor surveillance group revealed that in a state, 27.9% of…

A: Givena)Let p1 and p2 be the proportion of whites and blacks who were cigarette smokers…

Q: 5. It is observed that randomly chosen 40 students out of 100 have a daily reading habit. Construct…

A:

Q: 2. A survey found that out of 900 workers, 150 said that they were interrupted by frank phone calls.…

A: In question, We have given the number of favorable cases 150 and sample size 900 then we'll find 95%…

Q: 17. A researcher surveys 50 individuals in Smithville and 40 in Amherst, finding that 30% of…

A: Let n1 be the sample size  in smithville surveys : 50 individuals Let n2 be the sample size in…

Q: If n=360 and ˆp(p-hat) =0.69, find the margin of error at a 95% confidence level

A: