Answer to Question #251395 in Statistics and Probability for Carrie

Question

A new manufacturing method is supposed to increase the average life span of electronic com-

ponents, while the variance of the life span is expected to stay the same. Using the previous

manufacturing method, the average life span was 110 hours with a variance of 9 hours. The manu-

facturer measures the life spans of a sample of components manufactured using the new method.

The sample of 20 yields a sample mean of 125 for the life spans of the components. Does the data

provide sufficient evidence for the claim at the 1% level of significance? Clearly show how you draw

a conclusion when you test this hypothesis, using both methods, i.e.

(a) critical value approach, and (15)

(b) p-value approach. (5)

Answer 1

a. The following null and alternative hypotheses need to be tested:

$H_0: \mu\leq110$

$H_1:\mu>110$

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 $\alpha=0.01,$ and the critical value for a right-tailed test is $z_c=2.3263.$

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

(a) The z-statistic is computed as follows:

$$z=\dfrac{x-\mu}{\sigma/\sqrt{n}}=\dfrac{125-110}{9/\sqrt{20}}=7.45356$$

Since it is observed that $z=7.45356>2.3263=z_c,$ it is then concluded that the null hypothesis is rejected.

(b) Using the P-value approach: The p-value is $p=P(z>7.45356)\approx0,$ and since $p=0<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is not evidence to claim that the population mean $\mu$ is greater than $110,$ at the $\alpha = 0.01$ significance level.