Question

A sample of 1300 computer chips revealed that 50% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature claimed that more than 47% do not fail in the first 1000 hours of their use. Is there sufficient evidence at the 0.05 level to support the company's claim

Answer 1

Answer:

The p-value of the test is 0.015 < 0.05, which means that there is sufficient evidence at the 0.05 level to support the company's claim.

Step-by-step explanation:

The company's promotional literature claimed that more than 47% do not fail in the first 1000 hours of their use.

At the null hypothesis, we test if the proportion is of 47% or less, that is:

$H_0: p \leq 0.47$

At the alternative hypothesis, we test if the proportion is of more than 47%, that is:

$H_1: p > 0.47$

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.

0.47 is tested at the null hypothesis:

This means that $\mu = 0.47, \sigma = \sqrt{0.47*0.53}$

A sample of 1300 computer chips revealed that 50% of the chips do not fail in the first 1000 hours of their use.

This means that $n = 1300, X = 0.5$

Value of the statistic:

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

$z = \frac{0.5 - 0.47}{\frac{\sqrt{0.47*0.53}}{\sqrt{1300}}}$

$z = 2.17$

P-value of the test and decision:

The p-value of the test is the probability of finding a sample proportion above 0.5, which is 1 subtracted by the p-value of z = 2.17.

Looking at the z-table, z = 2.17 has a p-value of 0.9850

1 - 0.985 = 0.015

The p-value of the test is 0.015 < 0.05, which means that there is sufficient evidence at the 0.05 level to support the company's claim.