Answer to Question #338610 in Statistics and Probability for Keith gitonga

Question

It has been found that the average time internet users spend online per week is 18.3 hours.

A random sample of 48 teenagers indicated that their mean amount of internet time per week was 20.9 hours with a population variance of 32.49. At the 0.02 level of significance, can it be concluded that the mean time differs from 18.3 hours per week?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=18.3$

$H_1:\mu\not=18.3$

This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is $\alpha = 0.02,$ and the critical value for a two-tailed test is $z_c = 2.3263.$

The rejection region for this two-tailed test is $R = \{z: |z| > 2.3263\}.$

The z-statistic is computed as follows:

$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{20.9-18.3}{5.7/\sqrt{48}}=3.16$

Since it is observed that $|z| = 3.16>2.3263=z_c ,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=2P(Z>3.16)= 0.001578,$  and since $p = 0.001578< 0.02=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is different than 18.3, at the $\alpha = 0.02$ significance level.