Question

A manager will switch to a new technology if the production process exceeds 80 units per hour. The manager asks the company statistician to test the null hypothesis: H0: μ = 80 against the alternative hypothesis: H1: μ >80 If there is strong evidence to reject the null hypothesis then the new technology will be adopted. Past experience has shown that the standard deviation is 8. A data set with n = 25 for the new technology has a sample mean of 83. Does this justify adoption of the new technology?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0: \mu=80$

$H_1: \mu>80$

This corresponds to a right-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 right-tailed test is $z_c=1.6449.$

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

The z-statistic is computed as follows:

$$z=\dfrac{\bar{X}-\mu}{\sigma/\sqrt{n}}=\dfrac{83-80}{8/\sqrt{25}}=1.875$$

Since it is observed that $z=1.875>1.6449=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $p=P(Z>1.875)=0.030396,$ and since $p=0.030396<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is greater than 80, at the $\alpha=0.05$ significance level.

Therefore, there is enough evidence to adopt the new technology , at the $\alpha=0.05$ significance level.