Question

CCN and ActMedia provided a television channel targeted to individuals waiting in supermarket checkout lines. The channel showed news, short features, and advertisements. The length of the program was based on the assumption that the population mean time a shopper stands in a supermarket checkout line is 8 minutes. A sample of actual waiting times will be used to test this assumption and determine whether actual mean waiting time differs from this standard. A sample of 100 shoppers showed a sample mean waiting time of 8.4 minutes. Assume that only the sample standard deviation is given and that is 3.2 minutes.

Compute a 99% confidence interval for the population mean and out the value of the upper limit  and the lower limit .

Answer 1

Solution :

t _/2,df = 2.626

Margin of error = E = t_/2,df * (s /n)

= 2.626 * (3.2 / 100)

Margin of error = E = 0.84

The 99% confidence interval estimate of the population mean is,

- E < < + E

8.4 - 0.84 < < 8.4 + 0.84

7.56 < < 9.24

(7.56,9.24)