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Playbill magazine reported that the mean annual household income of its readers is $120,255. (Playbill, January...

Playbill magazine reported that the mean annual household income of its readers is $120,255. (Playbill, January 2006). Assume this estimate of the mean annual household income is based on a sample of 80 households, and based on past studies, the population standard deviation is known to be σ = $33,225.

a. Develop a 90% confidence interval estimate of the population mean.

b. Develop a 95% confidence interval estimate of the population mean.

c. Develop a 99% confidence interval estimate of the population mean.

d. Discuss what happens to the width of the confidence interval as the confidence level is increase. Does this result seem reasonable? Explain.

Solutions

Expert Solution

a)

90% confidence interval for is

- Z/2 * / sqrt(n) < < + Z/2 * / sqrt(n)

120255 - 1.645 * 33225 / sqrt(80) < < 120255 + 1.645 * 33225 / sqrt(80)

119643 < < 126366

b)

95% confidence interval for is

- Z/2 * / sqrt(n) < < + Z/2 * / sqrt(n)

120255 - 1.96 * 33225 / sqrt(80) < < 120255 + 1.96 * 33225 / sqrt(80)

112974 < < 1127536

c)

99% confidence interval for is

- Z/2 * / sqrt(n) < < + Z/2 * / sqrt(n)

120255 - 2.576 * 33225 / sqrt(80) < < 120255 + 2.576 * 33225 / sqrt(80)

110686 < < 129824

d)

When confidence level increases, the width of confidence interval increase.

milcah answered 1 year ago
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