Q.3 - The Practice of Statistics for AP

Question

Stress and heart attack You read a newspaper article that describes a study of whether stress management can help reduce heart attacks. The $107$ subjects all had reduced blood flow to the heart and so were at risk of a heart attack. They were assigned at random to three groups. The article goes on to say: One group took a four-month stress management program, another underwent a four-month exercise program, and the third received usual heart care from their personal physicians. In the next three years, only three of the $33$ people in the stress management group suffered “cardiac events,” defined as a fatal or non-fatal heart attack or a surgical procedure such as a bypass or angioplasty. In the same period, seven of the $34$ people in the exercise group and $12$ out of the 40 patients in usual care suffered such events.36

(a) Use the information in the news article to make a two-way table that describes the study results. 

(b) What are the success rates of the three treatments in preventing cardiac events? 

(c) Is there a significant difference in the success rates for the three treatments? Give appropriate statistical evidence to support your answer. 

Step-by-Step Solution

Answer

a). The table is

b).  Stress management $=0.9091$,

Exercise program $=0.7941$,

Usual care $=0.70$.

c). There is no significant difference among the success rates for the three treatments. 

Part (a) Step 1: Given Information

Total number of people $=107$.

Number of people in stress management program $=33$.

Number of people suffered cardiac events in stress management program $=3$.

Number of people in the exercise group $=34$.

Number of people suffered cardiac events in the exercise group $=7$.

Number of people in usual care $=40$.

Number of people suffered cardiac events in usual care $=12$.

Part (a) Step 2: Explanation

The two-way table will be 

Part (b) Step 3: Given Information

In the same period, seven of the $34$ people in the exercise group and $12$ out of the $40$ patients in usual care suffered such events.

Part (b) Step 4: Explanation

$\text{ Success rate }=\frac{\text{ Number of people who did not suffered cardiac events in the group }}{\text{ total number of people in the group }}$

Success rates

For stress management:

$=\frac{30}{33}$

$=0.9091$

For exercise program:

$=\frac{27}{34}$

$=0.7941$

For usual care:

$=\frac{28}{40}$

$=0.7$

Part (c) Step 5: Given Information

In the same period, seven of the $34$ people in the exercise group and $12$ out of the $40$ patients in usual care suffered such events.

Part (c) Step 6: Explanation

The expected count of the number of people can be calculated as follows 

Part (c) Step 7: Explanation

The null and alternate hypotheses are

$H_0$ : There is no difference in the actual success rates for the three treatments.

$H_1$ : There is a difference in the actual success rates for the three treatments.

Test statistic $x^2=\sum \frac{(0-E{)}^2}{E}$

Here,$0\to Observed Frequency$

$E\to Expected frequency$

Under null hypothesis

$\frac{(3-6.79{)}^2}{6.79}+\frac{(7-6.99{)}^2}{6.99}+\frac{(12-8.22{)}^2}{8.22}+\frac{(30-26.21{)}^2}{26.21}+\frac{(27-27.01{)}^2}{27.01}+\frac{(28-31.78{)}^2}{31.78}~{\chi }_2$

The test value is $4.8514$.

$P$ value is taken from Chi square table.

P value:

$P\left({\chi }_2>4.8514\right)=0.0884$

Conclusion:

As p value is greater than $0.05$ (level of significance), so there is insufficient evidence to reject null hypothesis at $5\%$ level of significance and thus conclude that there is no difference in the actual success rates for the treatments.