Question

In: Statistics and Probability

A certain health maintenance organization (HMO) wishes to study why patients leave the HMO. A SRS...

A certain health maintenance organization (HMO) wishes to study why patients leave the HMO. A SRS of 459 patients was taken. Data was collected on whether a patient had filed a complaint and, if so, whether the complaint was medical or non-medical 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 non-medical complaint
total 186 195 78
left 65 68 20

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

a) Find the expected number of people with no complain 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 non-medical 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.

Solutions

Expert Solution

Ans:

Observed(fo)
no complaint medical complaint non-medical complaint Total
not left 121 127 58 306
left 65 68 20 153
Total 186 195 78 459
Expected(fe)
no complaint medical complaint non-medical complaint Total
not left 124 130 52 306
left 62 65 26 153
Total 186 195 78 459
Chi square=(fo-fe)^2/fe
no complaint medical complaint non-medical complaint Total
not left 0.07 0.07 0.69 0.83
left 0.15 0.14 1.38 1.67
Total 0.22 0.21 2.08 2.50

a) the expected number of people with no complain who leave the HMO=186*153/459=62

b) expected number of people with a medical complaint who leave the HMO=195*153/459=65

c) expected number of people with a non-medical complaint who leave the HMO=78*153/459=26

d) Test statistic=2.50

e) degrees of freedom=(2-1)*(3-1)=2

f) critical value=CHIINV(0.05,2)=5.99

g) the final conclusion is

There is not sufficient evidence to reject the null hypothesis.(as test statistic is less than 5.99)


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