Question

2. The average length of time for students to

have their subjects controlled is 40 minutes. A

new controlling procedure using modern

computing machines is being tried. If a random

sample of 15 students has an average

controlling time of 25 minutes with a standard

deviation of 12.9 minutes under the new

system, test the hypothesis that the average

length of time to control student’s subjects is

less than 40 minutes. Use a level of

significance of 0.10 and assume the

population of controlling times to be normally

distributed.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge 40$

$H_a:\mu<40$

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 $\alpha = 0.1,$ $df=n-1=14$ degrees of freedom, and the critical value for a left-tailed test is $t_c = -1.34503.$

The rejection region for this left-tailed test is $R = \{t: t < -1.34503\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{X}-\mu}{s/\sqrt{n}}=\dfrac{25-40}{12.9/\sqrt{15}}\approx-4.50347$$

Since it is observed that $t = -4.50347 <-1.34503= t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for left-tailed $df=14$ degrees of freedom, $t=-4.50347$ is $p = 0.000248,$ and since $p=0.000248<0.10=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is less than 40, at the $\alpha = 0.10$ significance level.