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

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:

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

0.47 is tested at the null hypothesis:

This means that

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

Value of the statistic:

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.

the dimensions are:

step-by-step explanation:

we want to find the dimensions of a rectangular box with volume of

we factor to obtain;

we rewrite the rightmost factor to obtain;

recall that:

we apply difference of two squares formula to obtain;

hence the dimensions are

idk the answer but just put a

step-by-step explanation: