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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.
Confidence interval = mean ± z critical * std dev /n
Z critical at 90% interval = 1.645
Confidence interval = 39 ± 1.645 * 9.5/ 1000
= 39 ± 0.494
= (38.506, 39.494)
2) construct an 80% interval estimate
Confidence interval = mean ± z critical * std dev /n
Z critical at 80% interval = 1.282
Confidence interval = 39 ± 1.282 * 9.5/ 1000
= 39 ± 0.385
= (38.615, 39.385)
3) .Suppose sample size was 2000. Show the 90% and the 80% intervals.
90% Interval
Confidence interval = mean ± z critical * std dev /n
Z critical at 90% interval = 1.645
Confidence interval = 39 ± 1.645 * 9.5/ 2000
= 39 ± 0.35
= (38.65, 39.35)
80% interval
Confidence interval = mean ± z critical * std dev /n
Z critical at 80% interval = 1.282
Confidence interval = 39 ± 1.282 * 9.5/ 2000

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= 39 ± 0.272
= (38.728, 39.272)
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.
Standard error = std dev/ n
= 1230/√600
= 50.2145
2) margin of error in a 90% confidence interval estimate of the mean annual grocery expense for all
American households.
Margin of error = z critical * standard error
= 1.645 * 50.2145
= 82.603
3) Suppose sample size was 1200 rather than 600. Compute the margin of error and the standard error
for a 90% confidence interval.
Standard error = std dev/ n
= 1230/√1200
= 35.507
Margin of error = z critical * standard error
= 1.645 * 50.2145
= 58.409
31. Use the t table to determine the proper boundaries for the
1) 80% interval, where df = 9.
= ±1.383
2) 95% interval, where df = 17.
= ± 2.110

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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. Confidence interval = mean ± z critical * std dev /√n Z critical at 90% interval = 1.645 Confidence interval = 39 ± 1.645 * 9.5/ √1000 = 39 ± 0.494 = (38.506, 39.494) 2) construct an 80% interval estimate Confidence interval = mean ± z critical * std dev /√n Z critical at 80% interval = 1.282 Confidence interval = 39 ± 1.282 * 9.5/ √1000 = 39 ± 0.385 = (38.615, 39.385) 3) .Suppose sample size was 2000. Show the 90% and the 80% intervals. 90% Interval Confidence interval = mean ± z critical * std dev /√n Z critical at 90% interval = 1.645 Confidence interval = 39 ± 1.645 * 9.5/ √2000 = 39 ± 0.35 = (38.65, 39.35) 80% interval Confidence interval = mean ± z critical * std dev /√n Z critical at 80% interval = 1.282 Confidence interval = 39 ± 1.282 * 9.5/ √2000 = 39 ± 0.272 = (38.728, 39.272) 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. Standard error = std dev/ ? ...
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Just what I was looking for! Super helpful.

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