Question

The claim that 40% of those persons who retired from an industrial job before the age of 60 would return to work if a suitable job was available is to be investigated at the 0.02 significance level. If 74 out of the 200 workers sampled said they would return to work, what is our decision

Answer 1

Answer:

The p-value of the test is of 0.1922 > 0.02, which means that there is not significant evidence to reject the null hypothesis, that is, there is not significant evidence to conclude that the proportion is of less than 40%.

Step-by-step explanation:

Test if the proportion is less than 40%:

At the null hypothesis, we test if the proportion is of at least 0.4, that is:

$H_0: p \geq 0.4$

At the alternative hypothesis, we test if the proportion is of less than 0.4, that is:

$H_1: p < 0.4$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

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

0.4 is tested at the null hypothesis:

This means that $\mu = 0.4, \sigma = \sqrt{0.4*0.6}$

74 out of the 200 workers sampled said they would return to work

This means that $n = 200, X = \frac{74}{200} = 0.37$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

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

$z = -0.87$

P-value of the test and decision:

The p-value of the test is the probability of finding a sample proportion below 0.37, which is the p-value of z = -0.87.

Looking at the z-table, z = -0.87 has a p-value of 0.1922.

The p-value of the test is of 0.1922 > 0.02, which means that there is not significant evidence to reject the null hypothesis, that is, there is not significant evidence to conclude that the proportion is of less than 40%.