Question

The average time customers spent on the American Greetings Web site has been 11.85 minutes (Top Web Properties, USAToday, April 27, 2000). You want to show that the average time spent on that web site has now changed. Assume that the population of site visit times is normally distributed with a population standard deviation of 4.0 minutes. In order to test the hypothesis at 0.05 level of significance, the critical value(s) is(are):

Answer 1

Here we want to test

$Ho :\mu =11.85$ vs  $Ha :\mu \neq11.85$

Here to test the mean we will use the z statistic as the population standard deviation is known .

And the z statistic will be

$Z =\frac{\sqrt{n}(M-\mu)}{\sigma}$

n is the sampl size

M is the sample mean

Mu is the population mean. Here 11.85

Sigma is the population standard deviation.

here 4 minutes

As here our test is two tailed. The critical Value will be = $\pm Z_{\frac{\alpha}{2}}$ = $\pm Z_{\frac{0.05}{2}}$ =

41.96

So critical values are -1.96 and +1.96