Question

3) Consider the following hypothetical data which explores whether educational attainment influences divorce.

	Highest Degree
Ever Been Divorced	<High School	High School	Junior College	Bachelors Degree	Graduate Degree
Yes	107	362	55	89	52
No	285	867	149	374	195

3a) State your null and alternative hypotheses.

3b) Use the data above to construct a bivariate table by handwith cells containing the appropriate percentages for each cell.

3c) Using the method of comparing percentage differences, what can you conclude about these data?

Answer 1

3a)

The hypotheses are:

H0: Educational attainment does not influence divorce.

Ha: Educational attainment influences divorce.

b)

Following table shows the row total and column total:

	Highest Degree
Ever Been Divorced	<High	High	Junior	Bachelors	Graduate Degree
	School	School	College	Degree		Total
Yes	107	362	55	89	52	665
No	285	867	149	374	195	1870
Total	392	1229	204	463	247	2535

To find the percentage of each cell we need to divide the each cell by grand total 2535. Following table shows the percentage of each cell:

	Highest Degree
Ever Been Divorced	<High	High	Junior	Bachelors	Graduate Degree
	School	School	College	Degree		Total
Yes	4.22	14.28	2.17	3.51	2.05	26.23
No	11.24	34.2	5.88	14.75	7.69	73.77
Total	15.46	48.48	8.05	18.26	9.74	100

3c)

Here we need to use chi square test. For that we need to find the expected frequencies. Expected frequencies will be calculated as follows:

Following table shows the expected frequencies:

	Highest Degree
Ever Been Divorced	<High	High	Junior	Bachelors	Graduate Degree
	School	School	College	Degree		Total
Yes	102.832	322.4	53.515	121.458	64.795	665
No	289.168	906.6	150.485	341.542	182.205	1870
Total	392	1229	204	463	247	2535

Following table shows the calculations for chi square test statistics:

O	E	(O-E)^2/E
107	102.832	0.168937918
362	322.4	4.864019851
55	53.515	0.041207605
89	121.458	8.673959426
52	64.795	2.526615094
285	289.168	0.060076578
867	906.6	1.72971542
149	150.485	0.014654118
374	341.542	3.084603838
195	182.205	0.898504569
Total		22.06229442

Following is the test statistics:

Degree of freedom: df =( number of rows -1)*(number of columns-1) = (5-1)*(2-1)=4

The p-value is: 0.0002

Excel function used for p-value: "=CHIDIST(22.062,4)"

Since p-value is less than 0.05 so we reject the null hypothesis.