A sample of 1700 computer chips revealed that 35% - id17003513984669, halliehedman

Ответ:

haileyparrill703

22.08.2021

The p-value of the test is 0.004 < 0.02, which means that there is sufficient evidence at the 0.02 level to support the company's claim.

Step-by-step explanation:

The company's promotional literature claimed that over 32% do not fail in the first 1000 hours of their use.

At the null hypothesis, we test if the proportion is of at most 32%, that is:

$H_0: p \leq 0.32$

At the alternative hypothesis, we test if the proportion is more than 32%, that is:

H_1: p 0.32

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.

0.32 is tested at the null hypothesis:

This means that $\mu = 0.32 \sigma = \sqrt{0.32*0.68}$

A sample of 1700 computer chips revealed that 35% of the chips do not fail in the first 1000 hours of their use.

This means that n = 1700, X = 0.35

Value of the test statistic:

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

$z = \frac{0.35 - 0.32}{\frac{\sqrt{0.32*0.68}}{\sqrt{1700}}}$

z = 2.65

P-value of the test and decision:

The p-value of the test is the probability of finding a sample proportion above 0.35, which is 1 subtracted by the p-value of z = 2.65.

Looking at the z-table, z = 2.65 has a p-value of 0.9960.

1 - 0.9960 = 0.004.

The p-value of the test is 0.004 < 0.02, which means that there is sufficient evidence at the 0.02 level to support the company's claim.

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