Can someone please help me with number 2

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%.

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3 years ago