Question

I need to run a chi square and f test that compares the variances of the following:

Does the proportion of women who study more than 4 hours a week differ from the proportion of men that study more than 4 hours a week?

Please include null and alternative hypothesis, the results, and show all work. Please identify if any tests in excel were used. Here is the data

Women

1

7

6

5

6

8

6

5

2

7

8

9

3

1

2

9

5

7

Men

1

6

4

3

1

8

2

2

8

6

7

4

Answer 1

The data is copied in Excel :

Women	Men
1	1
7	6
6	4
5	3
6	1
8	8
6	2
5	2
2	8
7	6
8	7
9	4
3
1
2
9
5
7

Formula used for number of men/women exceeding 4 hours : “=COUNTIF(<Range of values>,”>4”)”
Formula used for proportion : “=Number of men/women exceeding 4 hours/COUNT(<Range of values>”

Number of women exceeding 4 hours	13
Proportion	0.722222
Number of men exceeding 4 hours	5
Proportion	0.416667

n1	18
n2	12

Here, our null hypothesis is H0 : p1=p2 and our alternative hypothesis is H1: p1 p2.
Test Statistic :

= 1.724333
​Critical value = Z _0.05/2 = Z _0.025 = 1.96
Since |Z| < Critical value, we fail to reject H0 and conclude that there is no significant difference in the proportion of men and women studying for more than 4 hours.

Steps for F-test : Click on DATA tab -> Click on DATA ANALYSIS Toolpack -> Choose F-Test Two-Sample for Variances -> Click ok -> Choose data for women in the input range1 and data for men in the input range2 ->Tick labels -> Click Ok.

Null Hypothesis H0:variances are equal and H1 : variances are not equal

F-Test Two-Sample for Variances
	Women	Men
Mean	5.388889	4.333333
Variance	6.839869	6.787879
Observations	18	12
df	17	11
F	1.007659
P(F<=f) one-tail	0.510288
F Critical one-tail	2.6851

Since F<Critical value=2.6851, we fail to reject H0 and conclude tht there is no significant difference in the variances for samples of men and women.