Question

A certain health maintenance organization (HMO) wishes to study why patients leave the HMO. A SRS of 396 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	159	102	135
Left	56	29	47

If the null hypothesis is H0:p1=p2=p3 and using α=0.05 then do the following: (Use two decimal places when appropriate)

(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

from above:

Observed		No complaint	Medical	Nonmedical	Total
	left	56	29	47	132
	right	103	73	88	264
	total	159	102	135	396

a)

expected number of people with no complaint who leave the HMO =row total*column total/grand total

=159*132/396 =53.00

b)

expected number of people with a medical complaint who leave the HMO=102*132/396=34.00
c)

expected number of people with a nonmedical complaint who leave the HMO =135*132/396=45.00

d)

Applying chi square test of independence:

Expected	Ei=row total*column total/grand total	No complaint	Medical	Nonmedical	Total
	left	53.00	34.00	45.00	132.00
	right	106.00	68.00	90.00	264.00
	total	159.00	102.00	135.00	396.00
chi square    χ2	=(Oi-Ei)2/Ei	No complaint	Medical	Nonmedical	Total
	left	0.170	0.735	0.089	0.9940
	right	0.085	0.368	0.044	0.4970
	total	0.2547	1.1029	0.1333	1.491
test statistic X2 =		1.49

e)

degree of freedom(df) =(rows-1)*(columns-1)=

2

f)

for 2 df and 0.05 level , critical value       χ2=

5.99 (from excel function chiinv(0.05,2)

g)

since test statistic < critical value:

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