Question

Random samples of 50 women and 50 men are taken at Norwich University. They are asked their reaction to increased tuition fees. Of the women, 23 favored the increase. Of the men, 19 favor the increase. At a 10% significance level, does this indicate that a larger proportion of women favor the increase than men?

Answer 1

`z=\frac{0.46-0.38}{\sqrt{0.42(1-0.42)((1)/(50)+(1)/(50))}}=0.8104`

`p_v =P(Z>0.8104)=0.209`

If we compare the p value and using any significance level for example
`\alpha=0.1` always
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the proportion of women favored the increase is not significantly higher than the proportion of men favored the increase at 10% of significance.

Explanation:

1) Data given and notation

`X_(1)=23` represent the number of women favored the increase

`X_(2)=19` represent the number of men favored the increase

`n_(1)=50` sample 1 selected

`n_(2)=50` sample 2 selected

`p_(1)=(23)/(50)=0.46` represent the proportion of women favored the increase

`p_(2)=(19)/(50)=0.38` represent the proportion of men favored the increase

z would represent the statistic (variable of interest)

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

`\alpha=0.1` represent the significance level

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if that a larger proportion of women favor the increase than men, the system of hypothesis would be:

Null hypothesis:
`p_(1) \leq p_(2)`

Alternative hypothesis:
`p_(1) > p_(2)`

We need to apply a z test to compare proportions, and 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))=(23+19)/(50+50)=0.42`

3) Calculate the statistic

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

`z=\frac{0.46-0.38}{\sqrt{0.42(1-0.42)((1)/(50)+(1)/(50))}}=0.8104`

4) Statistical decision

We can calculate the p value for this test.

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

`p_v =P(Z>0.8104)=0.209`

If we compare the p value and using any significance level for example
`\alpha=0.1` always
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the proportion of women favored the increase is not significantly higher than the proportion of men favored the increase at 10% of significance.