A sample of 900 computer chips revealed that 61% of the chips fail in the first 1000 hours of their use. The company's promotional literature claimed that under 64% fail in the first 1000 hours of their use. Is there sufficient evidence at the 0.02 level to support the company's claim

The p-value of the test is 0.0301 > 0.02, which means that there is not sufficient evidence at the 0.02 level to support the company's claim.

Step-by-step explanation:

The company's promotional literature claimed that under 64% fail in the first 1000 hours of their use.

At the null hypothesis, we test if the proportion is of at least 64%, that is:

At the alternative hypothesis, we test if the proportion is of less than 64%, 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.

64% is tested at the null hypothesis:

This means that

A sample of 900 computer chips revealed that 61% of the chips 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 0.61, which is the p-value of z = -1.88.

Looking at the z-table, z = -1.88 has a p-value of 0.0301.

The p-value of the test is 0.0301 > 0.02, which means that there is not sufficient evidence at the 0.02 level to support the company's claim.

step-by-step explanation:

hans did not square the 40-unit side; he merely multiplied it by 2.   incorrect.

well we know there are 4 clubs in 52 cards so it’s replaced so it should be 3/51

step-by-step explanation: