Respuesta :
Answer:
We conclude that there is no difference between this metropolitan area and the larger U.S. population.
Step-by-step explanation:
We are given that Labor statistics indicate that 77% of all U.S. cashiers and servers are women.
A random sample of cashiers and servers in a particular metropolitan area found that 112 of 150 cashiers and 150 of 200 servers were women.
Let p = proportion of all cashiers and servers who are women in metropolitan area.
So, Null Hypothesis, [tex]H_0[/tex] : p = 77% {means that there is no difference between this metropolitan area and the larger U.S. population}
Alternate Hypothesis, [tex]H_A[/tex] : p [tex]\neq[/tex] 77% {means that there is a difference between this metropolitan area and the larger U.S. population}
The test statistics that would be used here One-sample z proportion statistics;
T.S. = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = sample proportion of all cashiers and servers who are women in metropolitan area = [tex]\frac{112+150}{150+200}[/tex] = 0.75
n = sample of cashiers and servers = 150 + 200 = 350
So, test statistics = [tex]\frac{0.75-0.77}{\sqrt{\frac{0.75(1-0.75)}{350} } }[/tex]
= -0.86
The value of z test statistics is -0.86.
Now, at 0.05 significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.
Since our test statistic lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.
Therefore, we conclude that there is no difference between this metropolitan area and the larger U.S. population.
