In: Statistics and Probability
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.
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)