Question

The average length of time for student to register summer classes at a certain university has been 50 minutes with a standard deviation of 10 minutes. A new registration procedure using modern computing machine is being tried. If a random sample of 12 students had a n average registration 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. (critical value is +/- 1.796)

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge50$

$H_1:\mu<50$

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 $\alpha = 0.05,$ $df=n-1=11$ degrees of freedom, and the critical value for a left-tailed test is $t_c = -1.796.$

The rejection region for this left-tailed test is $R = \{t:t<-1.796\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{42-50}{11.9/\sqrt{12}}\approx-2.3288$$

Since it is observed that $t=-2.3288<-1.796=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for left-tailed, $df=11$ degrees of freedom, $t=-2.3288,$ is $p= 0.019977 ,$ and since $p= 0.019977 <0.05=\alpha,$ 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 $\alpha = 0.05$ significance level.