Answer to Question #269770 in Statistics and Probability for ALI

Question

The average amount of time that visitors spent looking at a retail company’s old home page was 23.6 seconds. The company commissions a new home page. On its first day in place the mean time spent at the new page by 7,628 visitors was 23.5 seconds with standard deviation 5.1 seconds.  Test at the 5% level of significance whether the mean visit time for the new page is less than the former mean of 23.6 seconds.

Answer 1

$H_0:\mu=23.6$ , mean visit time for the new page is not less than the former mean of 23.6 seconds

$H_a:\mu<23.6$ , mean visit time for the new page is less than the former mean of 23.6 seconds

$z=\frac{\overline{x}-\mu}{\sigma/\sqrt n}=\frac{23.5-23.6}{5.1/\sqrt{7628}}=-1.71$

critical value for 5% level of significance:

$z_{crit}=-1.645$

Since $z<z_{crit}$ we reject the null hypothesis. Mean visit time for the new page is less than the former mean of 23.6 seconds.