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?
The following null and alternative hypotheses need to be tested:
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 degrees of freedom, and the critical value for a left-tailed test is
The rejection region for this left-tailed test is
The t-statistic is computed as follows:
Since it is observed that it is then concluded that the null hypothesis is rejected.
Using the P-value approach:
The p-value for left-tailed degrees of freedom, is and since it is concluded that the null hypothesis is rejected.
Therefore, there is enough evidence to claim that the population mean is less than 190, at the significance level.
Comments