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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

User Jeremiahs
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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.

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 = (X - \mu)/((\sigma)/(√(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 = √(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 = (X - \mu)/((\sigma)/(√(n)))


z = (0.25 - 0.28)/((√(0.28*0.72))/(√(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.

User Ian Roke
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