Question

The Adecco workplace insights survey sampled men and women workers and asked if they expected to get a raise or promotion this year (USA Today, February 16, 2012). Suppose the survey sampled 200 men and 200 women. If 104 of the men replied yes and 74 of the women replied yes, are the results statistically significant so that you can conclude a greater proportion of men expect to get a raise or a promotion this year?

a. State the hypothesis test in terms of the population proportion of men and the population proportion of women.
b. What is the sample proportion for men? For women?
c. Use a 0.1 level of significance. What is the p-value and what is your conclusion?

Answer 1

Answer:

a) Null hypothesis:$p_{M} \leq p_{W}$  

Alternative hypothesis:$p_{M} > p_{W}$  

b) $p_{M}=\frac{104}{200}=0.52$ represent the proportion of men that replied yes

$p_{W}=\frac{74}{200}=0.37$ represent the proportion of women that replied yes

c) $z=\frac{0.52-0.37}{\sqrt{0.445(1-0.445)(\frac{1}{200}+\frac{1}{200})}}=3.018$  

$p_v =P(Z>3.018)= 0.0013$  

So the p value is a very low value and using the significance level given $\alpha=0.1$ always $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion of men is significant higher than the proportion of female .  

Step-by-step explanation:

1) Data given and notation  

$X_{M}=104$ represent the number of men that replied yes

$X_{W}=74$ represent the number of women that replied yes

$n_{M}=200$ sample of male selected

$n_{W}=200$ sample of demale selected

$p_{M}=\frac{104}{200}=0.52$ represent the proportion of men that replied yes

$p_{W}=\frac{74}{200}=0.37$ represent the proportion of women that replied yes

z would represent the statistic (variable of interest)  

$p_v$ represent the value for the test (variable of interest)

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the proportion for men that replied yes is higher than the proportion of women that replied yes:  

Null hypothesis:$p_{M} \leq p_{W}$  

Alternative hypothesis:$p_{M} > p_{W}$  

We need to apply a z test to compare proportions, and the statistic is given by:  

$z=\frac{p_{M}-p_{W}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{M}}+\frac{1}{n_{W}})}}$   (1)

Where $\hat p=\frac{X_{M}+X_{W}}{n_{M}+n_{W}}=\frac{104+74}{200+200}=0.445$

3) Calculate the statistic

Replacing in formula (1) the values obtained we got this:  

$z=\frac{0.52-0.37}{\sqrt{0.445(1-0.445)(\frac{1}{200}+\frac{1}{200})}}=3.018$  

4) Statistical decision

For this case we don't have a significance level provided $\alpha$, but we can calculate the p value for this test.  

Since is a one right tailed test the p value would be:  

$p_v =P(Z>3.018)= 0.0013$  

So the p value is a very low value and using the significance level given $\alpha=0.1$ always $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion of men is significant higher than the proportion of female .