Answer to Question #337628 in Statistics and Probability for Shey

Question

The purchasing director of an industrial part factory is investigating the possibility of purchasing a new type of milling machine. He determines that the new machine will be bought if there is evidence that the parts produced have a higher mean breaking strength than those from the old machine. The population standard deviation of the breaking strength for the old machine is 10kg and for the new machine is 9 kg. A sample of 100 parts was taken from each machine. Old machine indicates a sample mean of 65kg while new machine indicates a sample mean of 72 kg. Using the 0.01 level of significance, is there evidence that the purchasing director should buy the new machine? (critical value = + 2.575 )

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1\ge\mu_2$

$H_a:\mu_1<\mu_2$

This corresponds to a left-tailed test, and a z-test for two means, with known population standard deviations 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:

$$z=\dfrac{\bar{X}_1-\bar{X}_2}{\sqrt{\sigma_1^2/n_1+\sigma_2^2/n_2}}$$

$$=\dfrac{65-72}{\sqrt{10^2/100+9^2/100}}=-5.20306$$

Since it is observed that $z = -5.20306<-2.3263= z_c,$

it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $p=P(Z<-5.20306)=0,$ and since $p=0<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu_1$

is less than $\mu_2,$ at the $\alpha = 0.01$ significance level.

Therefore, there is enough evidence to claim that the purchasing director should buy the new machine  at the $\alpha = 0.01$ significance level.