Answer to Question #343320 in Statistics and Probability for adam

Question

The average length of time for students to enroll in college is 6 hours and 10 minutes. Computerized enrolment is being tested. In a random sample of 180 students, it was found out that the average time for them to enroll is 2 hours and 45 minutes with a standard deviation of 20 minutes. At = 0.01, does this indicate that the average time for students to enroll using computerized enrolment is less than the regular enrolment?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge370$

$H_a:\mu<370$

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 $\alpha = 0.01,$ $df=n-1=179$ degrees of freedom, and the critical value for a left-tailed test is $t_c =-2.34736.$

The rejection region for this left-tailed test is $R = \{t:t< -2.34736\}.$

The t-statistic is computed as follows:

$$t=\dfrac{165-370}{20/\sqrt{180}}=-137.518$$

Since it is observed that $t =-137.518<-2.34736=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for left-tailed $df=179$ degrees of freedom, $t=-137.518$ is $p=0,$ and since $p=0<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is less than 190, at the $\alpha = 0.01$ significance level.