Concept explainers
(a)
To find: The row and column totals.
(a)
Answer to Problem 29RE
Solution: The result obtained is as follows:
No complaint | Medical complaint | Non-medical complaint | Total | |
Stayed | 721 | 173 | 412 | 1306 |
Left | 22 | 26 | 28 | 76 |
Total | 743 | 199 | 440 | Grand Total = 1382 |
Explanation of Solution
Calculation:
Compute the row and column total as follows:
No complaint | Medical complaint | Non-medical complaint | Total | |
Stayed | 721 | 173 | 412 | 721+173+412 = 1306 |
Left | 22 | 26 | 28 | 22+26+28 = 76 |
Total | 721+22 = 743 | 173+26 = 199 | 412+28 = 440 | Grand Total = 1382 |
Interpretation: The row total of people who stayed HMO is 1306 and who left HMO is 76. The column total of people with no complaint is 743, people with medical complaint is 199 and people with nonmedical complaint is 440. The grand total is 1382.
(b)
To find: The percentage of each group who left.
(b)
Answer to Problem 29RE
Solution: The percentage of people who left HMO with no complaint, with medical complaint and with nonmedical complaint is 2.96%, 13.07%, and 6.36%, respectively.
Explanation of Solution
Calculation:
Compute the percentage of each group who left as follows:
Similarly,
(c)
To find: The expected counts.
(c)
Answer to Problem 29RE
Solution: Yes, chi-square test can be used safely and the table of expected counts is as follows:
Excepted Counts Table | ||||
No complaint | Medical complaint | Non-medical complaint | Total | |
Stayed | 702.14 | 188.06 | 415.8 | 1306 |
Left | 40.86 | 10.94 | 24.20 | 76 |
Total | 743 | 199 | 440 | 1382 |
Explanation of Solution
Calculation:
Calculate the expected count using the formula as follows:
Now, compute the expected count for all the rows and column total as follows:
Excepted Counts Table | ||||
No complaint | Medical complaint | Non-medical complaint | Total | |
Stayed | 1306 | |||
Left | 76 | |||
Total | 743 | 199 | 440 | 1382 |
Interpretation: From the above table, it can be seen that all the expected counts are greater than 5. Therefore, it can be concluded that chi-square test can be used safely.
(d)
Section 1:
The null and alternate hypothesis.
(d)
Section 1:
Answer to Problem 29RE
Solution: The null and alternate hypothesis is as follows:
Explanation of Solution
Section 2:
To find: The degrees of freedom of the test.
Section 2:
Answer to Problem 29RE
Solution: The degrees of freedom is 2.
Explanation of Solution
Calculation:
Compute the degree of freedom as follows:
Therefore, it can be concluded that the degrees of freedom of the test are 2.
Section 3:
To Test: The significance of the chi-square test statistic.
Solution: Yes, the result is significant.
Explanation:
Calculation:
The
Conclusion:
Since, the null hypothesis is rejected, it can be concluded that the test is significant.
Section 4:
To explain: The relationship between complaints and people leaving HMO.
Solution: There is a significant relationship between complaints and people leaving HMO.
Explanation: The null and alternate hypothesis stated in section 1 of part (d) is as follows:
In addition, according to the results in section 3 of part (d), the null hypothesis was rejected.
Since the null hypothesis is rejected, it can be stated that it is wrong to say that there is no relationship between members complaining and leaving HMO.
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Chapter 24 Solutions
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