Answer to Question #173713 in Statistics and Probability for Krisha

Question

In a time and motion study, it was found that a certain manual work can be finished at an average time of 40 minutes with a standard deviation of 8 minutes. A group of 16 students is given a special training and then found to average only 35 minutes. Can we conclude that the special training can speed up the work using 0.01 significance level?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\geq40$

$H_1:\mu<40$

This corresponds to a left-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is $\alpha=0.01,$ and the critical value for a left-tailed test is $z_c=-2.3263.$

The rejection region for this left-tailed test is $R=\{z:z<-2.3263\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{X}-\mu}{\sigma/\sqrt{n}}=\dfrac{35-40}{8/\sqrt{16}}=-2.5$$

Since it is observed that $z=-2.5<-2.3263=z_c,$ it is then 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 0.01 significance level.

Using the P-value approach: The p-value is $p=P(Z<-2.5)=0.0062,$ and sice $p=0.0062<0.01=\alpha,$ 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 0.01 significance level.

We can conclude that the special training speed up the work using $\alpha=0.01$.