Answer to Question #206484 in Statistics and Probability for Cam

Question

The average length of time for students to register for fall classes at a certain college

has been 50 minutes with a standard deviation of 10 minutes. A new registration

procedure using modern computing machines is being tried. If a random sample of

12 students had an average registration time of 42 minutes with a standard

deviation of 11.9 minutes under the new system, test the hypothesis that the

population mean is now less than 50, using a level of significance of (a) 0.05, and (b)

0.01. Assume the population of time to be normal.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\geq50$

$H_1:\mu<50$

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 $\alpha=0.05,$ and the critical value for a left-tailed test for $\alpha=0.05$ and $df=n-1=12-1=11$ degrees of freedom is $t_c=-1.795885.$ The rejection region for this left-tailed test is $R=\{t: t<-1.795885\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{42-50}{11.9/\sqrt{12}}\approx-2.328808$$

Since it is observed that $t=-2.328808<-1.795885=t_c,$ it is then concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population mean $\mu$ is less than 50, at the $\alpha=0.05$ significance level.

Using the P-value approach: The p-value for left-tailed, $t=-2.328808, df=11,$ $\alpha=0.05$ is $p=0.01997,$ and since $p=0.01997<0.05,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is less than 50, at the $\alpha=0.05$ significance level.

Based on the information provided, the significance level is $\alpha=0.01,$ and the critical value for a left-tailed test for $\alpha=0.01$ and $df=n-1=12-1=11$ degrees of freedom is $t_c=-2.718078.$ The rejection region for this left-tailed test is $R=\{t: t<-2.718078\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{42-50}{11.9/\sqrt{12}}\approx-2.328808$$

Since it is observed that $t=-2.328808>-2.718078=t_c,$ it is then concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population mean $\mu$ is less than 50, at the $\alpha=0.01$ significance level.

Using the P-value approach: The p-value for left-tailed, $t=-2.328808, df=11,$ $\alpha=0.01$ is $p=0.01997,$ and since $p=0.01997>0.01,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$ is less than 50, at the $\alpha=0.01$ significance level.