Question

Suppose that you work for a newly restructured automotive company with nearly 100,000 employees. You are in charge of purchasing engines from an overseas supplier. Company policy is that you purchase 500 engines each month to be placed into cars on the assembly line. Your overseas supplier of engines guarantees that no more than 0.9% of the new engines shipped to you will fail a simple electrical test. To check out the monthly shipment of the 500 engines you randomly select and test 50 of these engines, and you find that 1 is defective. Do you think that the supplier has met the guarantee

Answer 1

Answer:

The p-value of the test is 0.0764 > 0.05, which means that there is not enough evidence to reject the null hypothesis that the proportion is of at most 0.009, and thus we can conclude that the supplier has met the guarantee.

Step-by-step explanation:

Your overseas supplier of engines guarantees that no more than 0.9% of the new engines shipped to you will fail a simple electrical test.

At the null hypothesis, we test if the proportion is of at most 0.9% = 0.009, that is:

$H_0: p \leq 0.009$

At the alternative hypothesis, we test if the proportion is of more than 0.009, that is:

$H_1: p > 0.009$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

0.009 is tested at the null hypothesis:

This means that $\mu = 0.009, \sigma = \sqrt{0.009*0.991}$

50 of these engines, and you find that 1 is defective.

This means that $n = 50, X = \frac{1}{50} = 0.02$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.02 - 0.009}{\frac{\sqrt{0.009*0.991}}{\sqrt{50}}}$

$z = 1.43$

P-value of the test and decision:

The p-value of the test is the probability of finding a sample proportion above 0.02, which is 1 subtracted by the p-value of z = 1.43.

Looking at the z-table, z = 1.43 has a p-value of 0.9236.

1 - 0.9236 = 0.0764.

The p-value of the test is 0.0764 > 0.05, which means that there is not enough evidence to reject the null hypothesis that the proportion is of at most 0.009, and thus we can conclude that the supplier has met the guarantee.