Question 1058514: Playbill magazine reported that the mean annual household income of its readers is $120,750.
(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 = $31,000.
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.
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! sample size(n) is 80
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sample mean is 120,750 and population standard deviation is 31,000
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since we are given the standard deviation of the population and the sample size > 30, we use the z-score tables for a normal distribution
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a) critical value(CV) calculation for 90% confidence interval(CI)
alpha(a) = 1 - (90/100) = 0.10
cumulative probability(p*) = 1 - (a/2) = 1 - 0.05 = 0.95
CV is 1.64
margin of error(ME) = 1.64 * 31000 = 50840
90% confidence interval(CI) = 120750 + or - 50840 = (69910, 171590)
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b) CV calculation for 95% CI
a = 1 - (95/100) = 0.05
p* = 1 - (0.05/2) = 0.975
CV is 1.96
ME = 1.96 * 31000 = 60760
95% CI = 120750 + or - 60760 = (59990, 181510)
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c) CV calculations for 99% CI
a = 1 - (99/100) = 0.01
p* = 1 - (0.01/2) = 0.995
CV is 2.57
ME = 2.57 * 31000 = 79670
99% CI = 120750 + or - 79670 = (41080, 200420)
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Note For our problem a 90% confidence level means that we can expect 90% of the interval estimates to include the population mean; A 95% confidence level means that 95% of the intervals would include the population mean; and so on - we see that as the CI % increases the interval expands
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