Question

Clark Heter is an industrial engineer at Lyons Products. He would like to determine whether there are more units produced on the night shift than on the day shift. The mean number of units produced by a sample of 54 day-shift workers was 345. The mean number of units produced by a sample of 60 night-shift workers was 351. Assume the population standard deviation of the number of units produced on the day shift is 21 and 28 on the night shift. At the 0.05 significance level, is the number of units produced on the night shift larger?

Answer 1

Answer:

No, at the 0.05 significance level, the number of units produced on the night shift is not larger.

Step-by-step explanation:

We are given that the mean number of units produced by a sample of 54 day-shift workers was 345. The mean number of units produced by a sample of 60 night-shift workers was 351.

Assume the population standard deviation of the number of units produced on the day shift is 21 and 28 on the night shift.

Let $\mu_1$ = population mean number of units produced on the day shift

      $\mu_2$ = population mean number of units produced on the night shift

So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2\geq0$  or  $\mu_1\geq\mu_2$    {means that the mean number of units produced on the night shift is same or lesser on the day shift}

Alternate Hypothesis, $H_A$ : $\mu_1-\mu_2$  or  $\mu_1$    {means that the mean number of units produced on the night shift is larger}

The test statistics that will be used here is Two-sample z test statistics as we know about population standard deviations;

              T.S.  =  $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^{2} }{n_1}+\frac{\sigma_2^{2} }{n_2} } }$  ~ N(0,1)

where, $\bar X_1$ = sample mean number of units produced by a sample of 54 day-shift workers = 345

        $\bar X_2$ = sample mean number of units produced by a sample of 60 night-shift workers = 351

       $\sigma_1$  = population standard deviation of the number of units produced on the day shift = 21

        $\sigma_2$ = population standard deviation of the number of units produced on the day shift = 28

        $n_1$ = sample of day-shift workers = 54

        $n_2$ = sample of night-shift workers = 60

So, test statistics  =  $\frac{(345-351)-(0)}{\sqrt{\frac{21^{2} }{54}+\frac{28^{2} }{60} } }$

                              =  -1.302

Now at 0.05 significance level, the z table gives critical value of -1.6449 for left-tailed test. Since our test statistics is more than the critical value of z as -1.302 > -1.6449 so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the mean number of units produced on the night shift is same or lesser than those produced on the day shift.