Answer to Question #349537 in Statistics and Probability for Van

Question

A nationwide survey found out that the average time that college students spent on their personal computer is 10.5 hours per week. A random sample of 28 college students showed that they spent 8.5 hours per week using their computers with a standard deviation of 1.2 hours. Test whether the average number of hours spent by the 28 college students is significantly lower than the national average of 10.5 hours. Use a level of significance of 5%

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge10.5$

$H_1:\mu<10.5$

This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha = 0.05,$ $df=n-1=27$ and the critical value for a left-tailed test is $t_c =-1.703288.$

The rejection region for this left-tailed test is $R = \{t:t<-1.703288\}.$

The t-statistic is computed as follows:

$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{8.5-10.5}{1.2/\sqrt{28}}=-8.8192$

Since it is observed that $t=-8.8192<-1.703288=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for left-tailed, $df=27$ degrees of freedom, $t=-2.7106$ is $p=0,$ and since $p=0<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is less than 10.5, at the $\alpha = 0.05$ significance level.