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23. MPA Worldwide Market Research found the average age in a random sample of adult moviegoers was 39 (source: commercialalert.org/moviemadem.htm). If the sample size was 1000 and the population standard deviation is 9.5 years,

1) construct a 90% confidence interval estimate of the mean age of all adult moviegoers.

2) construct an 80% interval estimate

3) .Suppose sample size was 2000. Show the 90% and the 80% intervals.

90% Interval

80% interval

27. The average annual expense for groceries in a 2012 random sample of 600 US households is $8562. If the standard deviation of grocery expenses in the population of US households is $1230,

1) compute the standard error of the sampling distribution of the sample mean that could be used here to estimate the population mean.

2) margin of error in a 90% confidence interval estimate of the mean annual grocery expense for all American households.

31. Use the t table to determine the proper boundaries for the

1) 80% interval, where df = 9.

2) 95% interval, where df = 17.

3) 98 % interval, where df = 24.

35. Land's End wants to determine the average age of customers who purchase products through its catalog sales department. The company has contacted a simple random sample of 5 customers, with the following results:

Assume the population distribution of customer ages is normal.

1)      Compute the sample average, , and the sample standard deviation, s.


2)      Show the 90% confidence interval estimate of the average age of this customer population.

43. To explore how unemployment compensation returns money to the economy, the state of Washington annually conducts a study of household expenditures for claimants who received unemployment compensation from the state. In a recent study, a random sample of 1049 claimants was surveyed from a population of approximately 350,000 people who had received unemployment benefits during the year. Average household expenditure for the sample during the month of June was $2,754 (source: State of Washington, Employment Security Department, Claimant Expenditure Survey). If the sample standard deviation for June household expenditures in the sample was $243, show the 99% confidence interval estimate of average June expenditures for the population of 350,000 claimants.

49. You plan to take a random sample from a population of 500 items and build a 95% confidence interval estimate of the population mean.

1) You want a margin of error no bigger than ±5. You estimate the population standard deviation to be 100. How large a sample is needed?

2) Suppose you want the margin of error to be no bigger than ±4. How large a sample would be needed?

  73. It has been reported that American taxpayers spend 6.6 billion hours filling out tax forms each year (source: all-headlinenews.com). You take a random sample of 1500 taxpayers and find the average time filling out tax forms for the sample is 28.4 hours, with a sample standard deviation of 5.7 hours.

1) Use sample results to build a 95% confidence interval estimate of the average time taken filling out tax forms for the population of all American taxpayers.


2) If there are 150 million taxpayers in the US, show the 95% confidence interval for the total hours taken filling out tax forms for this population.


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