Answer to Question #342353 in Statistics and Probability for Mac

Question

In a study made on time and motion, it was found out that a certain manual work can be finished at an average time of 40 minutes with a standard deviation of 18 minutes. A group of 16 workers 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 a 0.05 level? 

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge40$

$H_1:\mu<40$

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 $\alpha = 0.05,$ and the critical value for a left-tailed test is $z_c = -1.6449.$

The rejection region for this left-tailed test is $R = \{z:z<-1.6449\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{35-40}{18/\sqrt{16}}\approx-1.111111$$

Since it is observed that $z=-1.111111>-1.6449=z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p=P(z<-1.111111)=0.13326,$ and since $p=0.13326>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$

is less than 40, at the $\alpha = 0.05$ significance level.