Question

A certain health maintenance organization (HMO) wishes to study why patients leave the HMO. A SRS of 429 patients was taken. Data was collected on whether a patient had filed a complaint and, if so, whether the complaint was medical or nonmedical in nature. After a year, a tally from these patients was collected to count number who left the HMO voluntarily. Here are the data on the total number in each group and the number who voluntarily left the HMO:

	No complaint	Medical complaint	Nonmedical complaint
Total	138	195	96
Left	53	61	29

If the null hypothesis is H0:p1=p2=p3 and using α=0.05, then do the following:

(a) Find the expected number of people with no complaint who leave the HMO:

(b) Find the expected number of people with a medical complaint who leave the HMO:

(c) Find the expected number of people with a nonmedical complaint who leave the HMO:

(d) Find the test statistic:

(e) Find the degrees of freedom:

(f) Find the critical value:

(g) The final conclusion is

A. We can reject the null hypothesis that the proportions are equal.
B. There is not sufficient evidence to reject the null hypothesis.

Answer 1

Here we need to use chi square test.

Following is the completed test :

	No Complaint	Medical complaint	Non medical complaint	Total
Left	53	61	29	143
Not left	85	134	67	286
Total	138	195	96	429

Expected frequencies will be calculated as follows:

Following table shows the expected frequencies:

		Expected frequencies (E)
	No Complaint	Medical complaint	Non medical complaint	Total
Left	46	65	32	143
Not left	92	130	64	286
Total	138	195	96	429

(a) Find the expected number of people with no complaint who leave the HMO:

Answer: 46

(b) Find the expected number of people with a medical complaint who leave the HMO:

Answer: 65

(c) Find the expected number of people with a non medical complaint who leave the HMO:

Answer: 32

(d)

Following table shows the calculations for test statistics:

O	E	(O-E)^2/E
53	46	1.065217391
85	92	0.532608696
61	65	0.246153846
134	130	0.123076923
29	32	0.28125
67	64	0.140625
Total		2.388931856

So,

(e)

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

(f)

The critical value is; 5.991

(g)

Since test statistics is less than critical value so we fail to reject the null hypothesis.

B. There is not sufficient evidence to reject the null hypothesis.