Question #343320

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?

1
Expert's answer
2022-05-23T12:18:46-0400

The following null and alternative hypotheses need to be tested:

H0:μ370H_0:\mu\ge370

Ha:μ<370H_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 α=0.01,\alpha = 0.01, df=n1=179df=n-1=179 degrees of freedom, and the critical value for a left-tailed test is tc=2.34736.t_c =-2.34736.

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

The t-statistic is computed as follows:



t=16537020/180=137.518t=\dfrac{165-370}{20/\sqrt{180}}=-137.518

Since it is observed that t=137.518<2.34736=tc,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=179df=179 degrees of freedom, t=137.518t=-137.518 is p=0,p=0, and since p=0<0.01=α,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 μ\mu is less than 190, at the α=0.01\alpha = 0.01 significance level.


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