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

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:

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

The test statistic is:

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

0.28 is tested at the null hypothesis:

This means that

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

Value of the test statistic:

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.

x -

median = 1.25

range = 12-14.5

interquartile = q12-q14.5

y -

median = 1

range = 15-17

interquartile = q15-q17

idk

step-by-step explanation: