(1 point) A certain health maintenance organization (HMO) wishes to study why pa
ID: 3310880 • Letter: #
Question
(1 point) A certain health maintenance organization (HMO) wishes to study why patients leave the HMO. A SRS of 237 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 complaintNonmedical complaint Total 57 60 31 130 10 If the null hypothesis is H0 : Pl =p2=p3 and using = 0.05, then do the following: Use at least 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: () 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.Explanation / Answer
here applying chi square test:
a) expected number =19
b) =40
c) =20
d) test statistic =15.6197
e) degree of freedom =(row-1)*(column-1) =2
f)crtiical value =5.9915
g)
option A is correct
Observed O no complaint medical complaint nonmedical complaint Total left 10 38 31 79 not left 47 82 29 158 Total 57 120 60 237 Expected E=rowtotal*column total/grand total no complaint medical complaint nonmedical complaint Total left 19.000 40.000 20.000 79 not left 38.000 80.000 40.000 158 Total 57 120 60 237 chi square =(O-E)^2/E no complaint medical complaint nonmedical complaint Total left 4.2632 0.1000 6.0500 10.413 not left 2.1316 0.0500 3.0250 5.207 Total 6.395 0.150 9.075 15.6197Related Questions
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