Question

An article in the National Geographic News ("U.S. Racking Up Huge Sleep Debt," February 24, 2005) argues that Americans are increasingly skimping on their sleep. A researcher in a small Midwestern town wants to estimate the mean weekday sleep time of its adult residents. He takes a random sample of 80 adult residents and records their weekday mean sleep time as 6.4 hours. Assume that the population standard deviation is fairly stable at 1.8 hours. Calculate the upper bound of the 95% confidence interval for the population mean weekday sleep time of all adult residents of this Midwestern town. (Round intermediate calculations to 2 decimal places. Round the z-value and final answer to 2 decimal places.)

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 6.4

Population standard deviation = = 1.8

Sample size = n = 80

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z_/2* ( /n)

= 1.96 * ( 1.8/ 80 )

= 0.39

At 95% confidence interval estimate of the population mean is,

- E < < + E

6.4 - 0.39 < < 6.4 + 0.39

6.01 < < 6.79

(6.01 , 6.79)