# Answer in Statistics and Probability for Carrie #251395

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)

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

“H_0: muleq110”

“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.

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