Question

sample of 1800 computer chips revealed that 25% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature claimed that 28% do not fail in the first 1000 hours of their use. Is there sufficient evidence at the 0.02 level to dispute the company's claim

Answer 1

Answer:

The p-value of the test is 0.0023 < 0.02, which means that there is sufficient evidence at the 0.02 level to dispute the company's claim.

Step-by-step explanation:

The company's promotional literature claimed that 28% do not fail in the first 1000 hours of their use.

At the null hypothesis, we test that at least 28% do not fail, that is:

$H_0: p \geq 0.28$

At the alternative hypothesis, we test if the proportion is of less than 28%, that is:

$H_1: p < 0.28$

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.

0.28 is tested at the null hypothesis:

This means that $\mu = 0.28, \sigma = \sqrt{0.28*0.72}$

Sample of 1800 computer chips revealed that 25% of the chips do not fail in the first 1000 hours of their use.

This means that $n = 1800, X = 0.25$

Value of the test statistic:

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

$z = \frac{0.25 - 0.28}{\frac{\sqrt{0.28*0.72}}{\sqrt{1800}}}$

$z = -2.83$

P-value of the test and decision:

The p-value of the test is the probability of finding a sample proportion below 0.25, which is the p-value of Z = -2.83.

Looking at the z-table, z = -2.83 has a p-value of 0.0023.

The p-value of the test is 0.0023 < 0.02, which means that there is sufficient evidence at the 0.02 level to dispute the company's claim.