Chapter 24, Problem 29RE

(a)

To determine

To find: The row and column totals.

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 determine

To find: The percentage of each group who left.

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:

$\begin{array}{c}\left(\begin{array}{l}\text{Percentage of people}\\ \text{left without compaint}\end{array}\right)=\frac{\left(\begin{array}{l}\text{Number of people}\\ \text{left without compaint}\end{array}\right)}{\left(\begin{array}{l}\text{Total number of people}\\ \text{with no compaint}\end{array}\right)}\\ =\frac{22}{743}\times 100\\ =0.0296\times 100\\ =2.96\%\end{array}$

Similarly,

$\begin{array}{c}\left(\begin{array}{l}\text{Percentage of people left}\\ \text{with medical compaint}\end{array}\right)=\frac{\left(\begin{array}{l}\text{Number of people left}\\ \text{with medical compaint}\end{array}\right)}{\left(\begin{array}{l}\text{Total number of people}\\ \text{with medical compaint}\end{array}\right)}\\ =\frac{26}{199}\times 100\\ =0.1307\times 100\\ =13.07\%\end{array}$

$\begin{array}{c}\left(\begin{array}{l}\text{Percentage of people left}\\ \text{with nonmedical compaint}\end{array}\right)=\frac{\left(\begin{array}{l}\text{Number of people left}\\ \text{with nonmedical compaint}\end{array}\right)}{\left(\begin{array}{l}\text{Total number of people}\\ \text{with nonmedical compaint}\end{array}\right)}\\ =\frac{28}{440}\times 100\\ =0.0636\times 100\\ =6.36\%\end{array}$

(c)

To determine

To find: The expected counts.

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:

$\text{Expected total}=\frac{\text{Row total}\times \text{Column total}}{\text{Grand total}}$

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	$\frac{1306\times 743}{1382}=702.14$	$\frac{1306\times 199}{1382}=188.06$	$\frac{1306\times 440}{1382}=415.80$	1306
Left	$\frac{76\times 743}{1382}=40.86$	$\frac{76\times 199}{1382}=10.94$	$\frac{76\times 440}{1382}=24.20$	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:

To determine

The null and alternate hypothesis.

Answer to Problem 29RE

Solution: The null and alternate hypothesis is as follows:

$\begin{array}{c}{H}_0:\left(\begin{array}{l}\text{There is no relationship between the}\\ \text{members complaining and leaving HMO}\end{array}\right)\\ H_1:\left(\begin{array}{l}\text{There is a relationship between the}\\ \text{members complaining and leaving HMO}\end{array}\right)\end{array}$

Explanation of Solution

The Null hypothesis states that there is no difference or relationship between any two factors, whereas the alternate hypothesis suggests that there is a significant relationship between the two factors.

Section 2:

To determine

To find: The degrees of freedom of the test.

Answer to Problem 29RE

Solution: The degrees of freedom is 2.

Explanation of Solution

Calculation:

Compute the degree of freedom as follows:

$\begin{array}{c}\text{Degrees of freedom}=(\text{Number of rows}-1)\times (Numberofcolumns-1)\\ =(2-1)\times (3-1)\\ =2\end{array}$

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 ${\chi }^2$ statistic is provided as 31.765. The degrees of freedom obtained in section 2 of part (d) is 2. The critical value of the chi-square statistic at 2 degrees of freedom is 5.991. As the provided test statistic value is greater than the critical value so the null hypothesis will be rejected at 5% level of significance.

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:

$\begin{array}{c}{H}_0:\left(\begin{array}{l}\text{There is no relationship between the}\\ \text{members complaining and leaving HMO}\end{array}\right)\\ H_1:\left(\begin{array}{l}\text{There is a relationship between the}\\ \text{members complaining and leaving HMO}\end{array}\right)\end{array}$

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.