Stress and heart attacks 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 $3$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, $7$of the $34$people in the exercise group and $12$out of the $40$patients in usual care suffered such events.

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

b. Compare the success rates of the three treatments in preventing cardiac events.

c. Do the data provide convincing evidence at the $\alpha =0.05$level that the true success rates for patients like these are not the same for the three treatments?

Step by step solution

01

Part (a) Step 1: Given information

We need to find the two way table.

02

Part (a) Step 2: Explanation

We know that

• Each combination of care and the cardiac event is counted in the two-way table (or not).
• The total of each row is shown in the last column.
• The total of each column is shown in the last row.

	Cardiac event	No cardiac event
Stress Management	$3$	$30$	$33$
Exercise	$7$	$27$	$34$
Usual care	$12$	$28$	role="math" $40$
	$22$	$85$	$107$

03

Part (b) Step 1: Given information

We need to find the comparison of the success rates of the three treatments.

04

Part (b) Step 2: Explanation

The success rate is calculated by dividing the number of "no cardiac events" by the entire row count:

• Stress management is $\frac{33}{100}=0.9091$
• Exercise is $\frac{27}{34}=0.7941$
• Usual care is$\frac{28}{40}=0.7$

05

Part (c) Step 1: Given information

We need to find out whether the data is convincing evidence or not.

06

Part (c) Step 2: Explanation

We know that

The null hypothesis asserts that the variables are unrelated, whereas the alternative hypothesis asserts that they are.

$H_0$is the true success rate for patients having the same for three treatments.

$H_{\alpha }$ is the true success rate for patients not having the same for three treatments.

And expected frequencies are a product of row and column total divided by table total.

And The squared differences between the actual and predicted frequencies, divided by the expected frequency, make up the chi-square subtotals.

Therefore, the data is not proving convincing evidence for the true success rates in patient treatment.

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