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.

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 $=12$.

The two-way table will be

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.

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

Success rates

For stress management:

$=0.9091$

For exercise program:

$=0.7941$

For usual care:

$=0.7$

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.

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

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,$0\to ObservedFrequency$

$E\to Expectedfrequency$

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.