Chapter 11, Problem R11.4RE

a)

To determine

To make a two-way table.

Explanation of Solution

In the two-table, there will be a count of combination of are and cardiac event. Therefore, two-way table becomes,

	Cardiac event	No cardiac event	Total
Stress management	3	30	33
Exercise	7	27	34
Usual care	12	28	40
Total	22	85	107

b)

To determine

To compare the success rate of three treatments that prevent the cardiac event.

Answer to Problem R11.4RE

The success rate of Stress management is higher and success rate of usual care is less.

Explanation of Solution

Given:

	Cardiac event	No cardiac event	Total
Stress management	3	30	33
Exercise	7	27	34
Usual care	12	28	40
Total	22	85	107

Calculation:

The success rate will be calculated as,

Stress management:

  $\frac{30}{33}=0.9091$

Exercise:

  $\frac{27}{34}=0.7941$

Usual care:

  $\frac{28}{40}=0.7$

Therefore, success rate of Stress management is higher and success rate of usual care is less.

c)

To determine

To explain whether data provide a convincing evidence that the true success rate for patients like these are not the same for the three treatments.

Answer to Problem R11.4RE

There is not convincing evidence that the true success rates for patients like these are not same for the three treatments.

Explanation of Solution

Given:

	Cardiac event	No cardiac event	Total
Stress management	3	30	33
Exercise	7	27	34
Usual care	12	28	40
Total	22	85	107

Calculation:

The null and alternative hypotheses:

H₀: The true success rates for patients like these are the same for the three treatments.

H_a: The true success rates for patients like these are not same for the three treatments.

Using, excel, this chi-square test is,

  

Decision: P-value = 0.0889 > 0.05, fail to reject H₀.

Conclusion: There is not convincing evidence that the true success rates for patients like these are not same for the three treatments.