Respuesta :
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:
[tex]H_0: p \geq 0.4[/tex]
At the alternative hypothesis, we test if the proportion is of less than 0.4, that is:
[tex]H_1: p < 0.4[/tex]
The test statistic is:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.
0.4 is tested at the null hypothesis:
This means that [tex]\mu = 0.4, \sigma = \sqrt{0.4*0.6}[/tex]
74 out of the 200 workers sampled said they would return to work
This means that [tex]n = 200, X = \frac{74}{200} = 0.37[/tex]
Value of the test statistic:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{0.37 - 0.4}{\frac{\sqrt{0.4*0.6}}{\sqrt{200}}}[/tex]
[tex]z = -0.87[/tex]
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%.
