Question

Refer to Exhibit 8-1. If the sample mean is 9 hours, then the 95% confidence interval is approximately

Exhibit 8-1: In order to estimate the average time spent on the computer terminals per student at a local university, data were collected from a sample of 81 business students over a one-week period. Assume the population standard deviation is 1.2 hours.

a. 7.04 to 110.96 hours

b. 7.36 to 10.64 hours

c. 7.80 to 10.20 hours

d. 8.74 to 9.26 hours

Answer 1

Solution :

Given that,

Point estimate = sample mean =     = 9

Population standard deviation =    =1.2

Sample size n =81

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

Margin of error = E =   Z/2    * ( /n)
= 1.96 * ( 1.2 / 81)

= 0.26
At 95% confidence interval estimate of the population mean
is,

- E < < + E

9 - 0.26 <   < 9+ 0.26

8.74 <   < 9.26

d. 8.74 to 9.26 hours