Question

A research group conducted a survey about church attendance. The survey respondents were asked about their church attendance and asked to indicate their age. The results are as follows:

Church Attendance	Age
	20 to 29	30 to 39	40 to 49	50 to 59
Yes	60	30	68	100
No	87	107	80	50

Answer the following four questions. Use 5% level of significance throughout.

1.We use the sample data above to determine whether church attendance is independent of age. Calculate the value of your test statistic. (4 decimal places) (5)

2. What is the critical value of test statistic? (1)

3. What is your conclusion? Interpret in terms of the null and alternative hypothesis. (2)

4. Using the multiple comparison procedure, find out if there is a significant difference between age groups 20-29 and 30-39?

Answer 1

1:

Hypotheses are:

H0: Church attendance is independent of age.

Ha: Church attendance is not independent of age.

Following table shows the row total and column total:

	Age
	20 to 29	30 to 39	40 to 49	50 to 59	Total
Yes	60	30	68	100	258
No	87	107	80	50	324
Total	147	137	148	150	582

Expected frequencies will be calculated as follows:

Following table shows the expected frequencies:

	20 to 29	30 to 39	40 to 49	50 to 59	Total
Yes	65.165	60.732	65.608	66.495	258
No	81.835	76.268	82.392	83.505	324
Total	147	137	148	150	582

Following table shows the calculations for chi square test statistics:

O	E	(O-E)^2/E
60	65.165	0.409379652
30	60.732	15.55120569
68	65.608	0.087209852
100	66.495	16.88224716
87	81.835	0.325987964
107	76.268	12.3833826
80	82.392	0.069444412
50	83.505	13.44332705
Total		59.15218438

The test statistics is:

2:

Degree of freedom: df =( number of rows -1)*(number of columns-1) = (2-1)*(3-1)=2

Critical value of chi square for 5% level of significance is : 5.9915

3:

Since test statistics is greater than critical value so we reject the null hypothesis.

That is we can conclude that "Church attendance is not independent of age.".

4:

Here the claim (alternative hypothesis) is

So test is two tailed. Following is the null hypothesis:

Here we have

Pooled sample proportion is

Standard error of the test is:

So test statistics will be

Test is tao twaild so p-value is:

p-value = 2P(z > 3.42) = 0.0006

Since p-value is less than 0.05 so we reject the null hypothesis. That is we can conclude that there is a significant difference between age groups 20-29 and 30-39.