Answer to Question #337627 in Statistics and Probability for Shey

Question

The average length of time for students to register for summer classes at a certain college has been 50 minutes. A new registration procedure using modern computing machines is being tried. If a random sample of 35 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 minutes, using a level of signficance of 0.01. (critical value = +/- 2.33)

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge 50$

$H_a:\mu<50$

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 $\alpha = 0.01,$ $df=n-1=34$ degrees of fredom, and the critical value for a left-tailed test is $t_c = -2.44115.$

The rejection region for this left-tailed test is $R = \{t: t < -2.44115\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{42-50}{11.9/\sqrt{35}}\approx-3.9772$$

Since it is observed that $t = -3.9772< -2.44115=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for left-tailed, $df=34$ degrees of freedom, $t=-3.9772$ is $p =0.000173,$ and since $p=0.000173<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is less than 50, at the $\alpha = 0.01$ significance level.