Question

This claim is to be investigated at .01 levels. “Forty percent or more of those persons who retired from an industrial job before the age of 60 would return to work if a suitable job were available. “Seventy-four persons out of the 200 sampled said they would return to work

- State the null hypothesis.
- What is the decision rule?
- Compute the value of the test statistic.​

Answer 1

According to the described situation, we have that:

• The null hypothesis is $H_0: p < 0.4$

The decision rule is:

• z < 2.327: Do not reject the null hypothesis.

• z > 2.327: Reject the null hypothesis.

The value of the test statistic is of z = -0.866.

What is the null hypothesis?

The claim is:

"Forty percent or more of those persons who retired from an industrial job before the age of 60 would return to work if a suitable job were available"

At the null hypothesis, we consider that the claim is false, that is, the proportion is of less than 40%, hence:

$H_0: p < 0.4$

What is the decision rule?

We have a right-tailed test, as we are testing if a proportion is less/greater than a value. Since we are working with a proportion, the z-distribution is used.

Using a z-distribution calculator, the critical value for a right-tailed test with a significance level of 0.01 is of z = 2.327, hence, the decision rule is:

• z < 2.327: Do not reject the null hypothesis.

• z > 2.327: Reject the null hypothesis.

What is the test statistic?

The test statistic is given by:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

In which:

• $\overline{p}$ is the sample proportion.

• p is the proportion tested at the null hypothesis.

• n is the sample size.

In this problem, the parameters are:

$p = 0.4, n = 200, \overline{p} = \frac{74}{200} = 0.37$

Hence:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

$z = \frac{0.37 - 0.4}{\sqrt{\frac{0.4(0.6)}{200}}}$

$z = -0.866$

The value of the test statistic is of z = -0.866.

You can learn more about hypothesis tests at https://brainly.com/question/16313918