Q.3 Stress and heart attack You read... [FREE SOLUTION]

91影视

The Practice of Statistics for AP

David Moore,Daren Starnes,Dan Yates

$Math Studyset 91影视 Explanations$ Math

4th Edition 路 809 pages

ISBN: 9781319113339

Chapter 11: Q.3 (page 732)

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 鈥渃ardiac events,鈥� 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.

Short Answer

Expert verified

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.

Step by step solution

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

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

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 $x^{2} = 鈭� \frac{(0 - E)^{2}}{E}$

Here, $0 鈫� O b s e r v e d F r e q u e n c y$

$E 鈫� E x p e c t e d f r e q u e n c y$

Under null hypothesis

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

The test value is $4.8514$ .

$P$ value is taken from Chi square table.

P value:

$P (蠂_{2} > 4.8514) = 0.0884$

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.