Answer to Question #106130 in Statistics and Probability for Geraldine

Question

The average length of time for students to register for classes at a certain college has been 50 minutes. A new registration procedure is being tried. A random sample of 12 students resulted in an average registration time of 42 minutes with a standard deviation of 11.9. Assume the distribution of registration times are normally distributed and test the hypothesis that the population mean with the new procedure is less than 50 minutes. Using a .05 level of significance.

Answer 1

We have our hypotheses

$H_0: \; \mu=50 \text{ vs. }H_1:\;\mu<50$

In order to test this hypothesis, we must constuct a test statistic. As we know the registration times follow a Normal distribution, we can create the following test statistic

$T=\frac{\hat{\mu}-50}{s/\sqrt{n}}$

where, $s$ is the sample standard deviation, and $\hat{\mu}$ is the sample mean. This follows a t-distribution with degrees of freedom $n-1=11$

Thus, our critical region is $<t_{n-1,\alpha}=-t_{n-1,1-\alpha}$ , which denotes the corresponding quantile of the t-distribution, as it is symmetric

We calculate $T=\frac{-8\sqrt{12}}{11.9}\approx -2.3288$ and from empirical tables, we have $t_{11,0.05}=1.795885$

Thus, we have $T<-1.795885$ . Thus, there is enough evidence to reject the null hypothesis.