Chapter 24, Problem 25E

(a)

To determine

The two-way table that explains the study results of stress and heart attacks.

Answer to Problem 25E

Solution: The two-way table that explains the study results of stress and heart attacks is obtained as:

Groups	Cardiac events	No cardiac events	Total
Stress Management	3	30	33
Exercise Program	7	27	34
Usual Heart care	12	28	40
Total	22	85	107

Explanation of Solution

Calculation:

The newspaper article describes three groups and hence there are three rows for the two-way table. There are two types of events for the provided three groups. Hence, there are two columns for the two-way table. So, the two-way table that explains the study results of stress and heart attacks is obtained as:

Groups	Cardiac events	No cardiac events	Total
Stress Management	3	30	33
Exercise Program	7	27	34
Usual Heart care	12	28	40
Total	22	85	107

(b)

To determine

The success rates for the three treatments to prevent cardiac events.

Answer to Problem 25E

Solution: The success rates for the three treatments to prevent cardiac events are obtained as:

$\begin{array}{c}\text{Stress Management}=90.9\%\\ Exercise=79.4\%\\ \text{Usual Heart care}=\text{70\%}\end{array}$

Explanation of Solution

Calculation:

The success of the program is defined as “No cardiac events.” Hence, the success rate of three treatments is the percentage of no cardiac events in three groups. So, the success rate of stress management program is calculated as:

$\begin{array}{c}\text{Stress Management program}=\frac{30}{33}\times 100\\ =\frac{3000}{33}\\ =90.9\%\end{array}$

Now, calculate the success rate of Exercise program. So, the success rate of exercise program is calculated as:

$\begin{array}{c}\text{Exercise program}=\frac{27}{34}\times 100\\ =\frac{2700}{33}\\ =79.4\%\end{array}$

And, the success rate of usual heart care is calculated as:

$\begin{array}{c}\text{Usual heart care}=\frac{28}{40}\times 100\\ =\frac{2800}{40}\\ =70\%\end{array}$

(c)

To determine

To find: The expected cells count for the three treatment groups and also verify that the obtained expected counts satisfy the guidelines to use chi-square test.

Answer to Problem 25E

Solution: The expected cell counts for the three treatment groups are obtained as:

Groups	Expected
		Cardiac events	No cardiac events
Stress Management	6.79	26.21
Exercise Management	6.99	27.01
Usual heart care	8.22	31.78

All the obtained expected counts are greater than 5, so it satisfies the guidelines to use chi-square test.

Explanation of Solution

Calculation:

The two-way table is obtained in part (a) which is shown below:

Groups	Cardiac events	No cardiac events	Total
Stress Management	3	30	33
Exercise Program	7	27	34
Usual Heart care	12	28	40
Total	22	85	107

Now, calculate the expected cell counts of each cell. The expected cell count is defined by the formula:

$\text{Expected count}=\frac{\text{row total}\times \text{column total}}{\text{table total}}$

The expected count for cardiac events for stress management group is calculated as:

$\begin{array}{c}\text{Expected count}=\frac{\text{row total}\times \text{column total}}{\text{table total}}\\ =\frac{33\times 22}{107}\\ =\frac{726}{107}\\ =6.79\end{array}$

The expected count for no cardiac events for stress management is calculated as:

$\begin{array}{c}\text{Expected count}=\frac{\text{row total}\times \text{column total}}{\text{table total}}\\ =\frac{33\times 85}{107}\\ =\frac{2805}{107}\\ =26.21\end{array}$

Similarly calculate the expected counts for all cells and the table that shows the observed and expected counts is obtained as:

Groups	Observed		Expected
		Cardiac events	No cardiac events	Cardiac events	No cardiac events
Stress Management	3	30	6.79	26.21
Exercise Management	7	27	6.99	27.01
Usual heart care	12	28	8.22	31.78

The requirement to satisfy to use the chi-square test is that all the expected counts should be greater than 5.

The obtained expected counts shows that all the counts are greater than 5. Hence, the guideline to use chi-square test is satisfied.

(d)

To determine

To Test: Whether there is a significant difference between the success rates for the three treatment groups.

Answer to Problem 25E

Solution: There is a significant difference between success rates for the three treatment groups at the significance level of 10%, but not significant at significance level of 5%.

Explanation of Solution

Calculation:

The steps followed for a significance test are as provided below:

Step 1: Formulate the hypotheses.

The null hypothesis $\left(H_0\right)$ and alternative hypothesis $\left(H_a\right)$ are formulated as:

$H_0$: There is no significant difference among the success rates for the three treatments.

$H_a$: There is some significant difference among the success rates for the three treatments.

Step 2: Define the sampling distribution.

The chi-square test can be used if the expected cell counts are larger than 5. The obtained two-way table shows that there are three rows and two columns. Therefore, the degrees of freedom are obtained as:

$\begin{array}{c}\text{Degrees of freedom}=\left(r-1\right)\times \left(c-1\right)\\ =\left(3-1\right)\times \left(2-1\right)\\ =2\times 1\\ =2\end{array}$

Step 3: Find the expected cell counts.

The expected cell counts are obtained in part (c). So, the table that shows the observed counts and expected counts is obtained as:

Groups	Observed		Expected
		Cardiac events	No cardiac events	Cardiac events	No cardiac events
Stress Management	3	30	6.79	26.21
Exercise Management	7	27	6.99	27.01
Usual heart care	12	28	8.22	31.78

Step 4: Determine the chi-square statistic.

Since all the expected counts are greater than 5. The chi-square test statistic can be used. The chi-square statistic is the measure of the distance of the observed counts from the expected counts in a two-way table. The formula for the χ² statistic is defined as:

${\chi }^2={{\displaystyle \sum \frac{\left(\text{observed count}-\text{expected count}\right)}{\text{expected count}}}}^2$

Substitute the obtained observed and expected counts for each cell to determine the chi-square statistic. So, the chi-square statistic is calculated as:

$\begin{array}{c}{\chi }^2={{\displaystyle \sum \frac{\left(\text{observed count}-\text{expected count}\right)}{\text{expected count}}}}^2\\ =\left(\begin{array}{c}\frac{{\left(3-6.79\right)}^2}{6.79}+\frac{{\left(30-26.21\right)}^2}{26.21}+\frac{{\left(7-6.99\right)}^2}{6.99}\\ +\frac{{\left(27-27.01\right)}^2}{27.01}+\frac{{\left(12-8.22\right)}^2}{8.22}+\frac{{\left(28-31.78\right)}^2}{31.78}\end{array}\right)\\ =2.115+0.548+0.000+0.000+1.738+0.450\\ =4.851\end{array}$

Thefore, the chi-square statistic is obtained as 4.851.

Step 5: Test the significance.

The degrees of freedom are obtained as two. Use Table 24.1 of critical values for chi-square test which is provided in the textbook. The obtained chi-square value is 4.851, which is smaller than the critical value of 5.99 at the significance level of 0.05 and greater than the critical value of 4.61 at the significance level of 0.10 for two degrees of freedom.

This shows that the result is significant at 10% level of significance, as the calculated 4.851 is less than the critical value 5.99, but not at 5% significance level as the calculated 4.851 is less than the critical value 4.61.

Conclusion:

From the report, it can be concluded that there is some difference between the success rates of the three experiment groups at 10% significance level and shows that there is no difference between the success rates at 5% significance level.