Question

The National Assessment of Educational Progress interviewed a random sample of 1917 people 21 to 25 years old. The sample contained 840 men, of whom 775 were fully employed. Of the 1077 women 680 were fully employed. Is there sufficient evidence to support the claim that the employment levels for men compared to woman were different

Answer 1

`z=\frac{0.923-0.631}{\sqrt{0.759(1-0.759)((1)/(840)+(1)/(1077))}}=14.83`

The p value for this case would be:

`p_v =2*P(Z>14.83)\approx 0`

For this case the p value is very low so we have enough evidence to reject the null hypothesis and we can conclude that the true proportion of employment levels for men compared to woman were different

Explanation:

Information given

`X_(1)=775` represent the number of people fully employed men

`X_(2)=680` represent the number of people fully employed women

`n_(1)=840` sample 1 selected

`n_(2)=1077` sample 2 selected

`p_(1)=(775)/(840)=0.923` represent the proportion estimated for male employed

`p_(2)=(680)/(1077)=0.631` represent the proportion estimated for women employed

`\hat p` represent the pooled estimate of p

z would represent the statistic

`p_v` represent the value for the test

System of hypothesis

We want to test the claim that the employment levels for men compared to woman were different, the system of hypothesis would be:

Null hypothesis:
`p_(1) = p_(2)`

Alternative hypothesis:
`p_(1) \\eq p_(2)`

For this case the statistic is given by:

`z=\frac{p_(1)-p_(2)}{\sqrt{\hat p (1-\hat p)((1)/(n_(1))+(1)/(n_(2)))}}` (1)

Where
`\hat p=(X_(1)+X_(2))/(n_(1)+n_(2))=(775+680)/(840+1077)=0.759`

Replacing the info given we got:

`z=\frac{0.923-0.631}{\sqrt{0.759(1-0.759)((1)/(840)+(1)/(1077))}}=14.83`

The p value for this case would be:

`p_v =2*P(Z>14.83)\approx 0`

For this case the p value is very low so we have enough evidence to reject the null hypothesis and we can conclude that the true proportion of employment levels for men compared to woman were different.