A manufacturer considers his production process to be out of control when defects exceed 3%. In a random sample of 100 items, the defect rate is 4% but the manager claims that this is only a sample fluctuation and production is not really out of control. At the 0.01 level of significance, test the manager's claim.

The p-value of the test is of 0.2776 > 0.01, which means that the we accept the null hypothesis, that is, the manager's claim that this is only a sample fluctuation and production is not really out of control.

Step-by-step explanation:

A manufacturer considers his production process to be out of control when defects exceed 3%.

At the null hypothesis, we test if the production process is in control, that is, the defective proportion is of 3% or less. So

At the alternate hypothesis, we test if the production process is out of control, that is, the defective proportion exceeds 3%. So

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.

0.03 is tested at the null hypothesis

This means that

In a random sample of 100 items, the defect rate is 4%.

This means that

Value of the test statistic:

P-value of the test

The p-value of the test is the probability of finding a sample proportion above 0.04, which is 1 subtracted by the p-value of z = 0.59.

Looking at the z-table, z = 0.59 has a p-value of 0.7224

1 - 0.7224 = 0.2776

The p-value of the test is of 0.2776 > 0.01, which means that the we accept the null hypothesis, that is, the manager's claim that this is only a sample fluctuation and production is not really out of control.

hope it

step-by-step explanation:

x^(1/6) · x^(1/6) · x^(1/6)   ·x^(1/6) = x^(4/6)   since we add the exponents

4/6 can be simplified to 2/3

the simplified form is x^(2/3)

yes the total will be same because in both of them the expression.is rearranged but their sign remains the same hence it hss just been shuffled