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20\. Does Aspirin Prevent Heart Disease? In a trial designed to test the effectiveness of aspirin in preventing heart disease, 11,037 male physicians were treated with aspirin and 11,034 male physicians were given placebos. Among the subjects in the aspirin treatment group, 139 experienced myocardial infarctions (heart attacks). Among the subjects given placebos, 239 experienced myocardial infarctions (based on data from "Final Report on the Aspirin Component of the Ongoing Physicians' Health Study," New England Journal of Medicine, Vol. 321 : \(129-135)\). Use a \(0.05\) significance level to test the claim that aspirin has no effect on myocardial infarctions. a. Test the claim using a hypothesis test. b. Test the claim by constructing an appropriate confidence interval. c. Based on the results, does aspirin appear to be effective?In a trial designed to test the effectiveness of aspirin in preventing heart disease, 11,037 male physicians were treated with aspirin and 11,034 male physicians were given placebos. Among the subjects in the aspirin treatment group, 139 experienced myocardial infarctions (heart attacks). Among the subjects given placebos, 239 experienced myocardial infarctions (based on data from "Final Report on the Aspirin Component of the Ongoing Physicians' Health Study," New England Journal of Medicine, Vol. 321 : \(129-135)\). Use a \(0.05\) significance level to test the claim that aspirin has no effect on myocardial infarctions. a. Test the claim using a hypothesis test. b. Test the claim by constructing an appropriate confidence interval. c. Based on the results, does aspirin appear to be effective?

Step by step solution

01

Set Up Hypotheses

State the null hypothesis (H_0) and the alternative hypothesis (H_1).\(\displaystyle H_0\) : Aspirin has no effect on myocardial infarctions (heart attacks) \(\displaystyle H_1\) : Aspirin has an effect on myocardial infarctions (heart attacks).

02

Determine the Test Statistic

Use the formula for the test statistic for two independent proportions:\(\displaystyle Z=\dfrac{\hat{p_1}-\hat{p_2}}{\sqrt {\hat{p}(1-\hat{p})(\dfrac{1}{n_1} + \dfrac{1}{n_2}) }}\)where:\(\displaystyle \hat{p_1}\) is the proportion of myocardial infarctions in the aspirin group.\(\displaystyle \hat{p_1}=\dfrac{139}{11037}=0.0126$$\displaystyle \hat{p_2}\) is the proportion of myocardial infarctions in the placebo group.\(\displaystyle \hat{p_2}=\dfrac{239}{11034}=0.0217\)The combined proportion (\(\displaystyle \hat{p}\)) is \(\displaystyle \hat{p}=\dfrac{139+239}{11037+11034}=\dfrac{378}{22071}=0.0171\). Substituting these values into the formula gives\(\displaystyle Z=\dfrac{0.0126-0.0217}{\sqrt {0.0171(1-0.0171)(\dfrac{1}{11037} + \dfrac{1}{11034}) }}\)

03

Compute the Test Statistic

Calculate the Z score:\(\displaystyle Z=\dfrac{0.0126-0.0217}{\sqrt {0.0171(0.9829)(0.00009085) }}$$\displaystyle Z=\dfrac{-0.0091}{\sqrt{0.0170859 * 0.00009085}}$$\displaystyle Z=\dfrac{-0.0091}{0.0012398} \approx -7.34 \)

04

Determine the P-value

Using the Z table, find the P-value for Z = -7.34. The P-value is extremely small, much smaller than any typical significance level such as 0.05.

05

Make a Decision

Compare the P-value with the significance level:Since the P-value is much smaller than \(\displaystyle \alpha=0.05\), reject the null hypothesis. There is enough evidence to support the claim that aspirin has an effect on myocardial infarctions.

06

Constructing the Confidence Interval

Construct a 95% confidence interval for the difference of proportions:\(\displaystyle (\hat{p_1}-\hat{p_2}) \pm Z \sqrt{\hat{p}(1-\hat{p})(\dfrac{1}{n_1} + \dfrac{1}{n_2})}$$Z\) for a 95% confidence level is 1.96. Substituting the values in:\(\displaystyle (0.0126-0.0217) \pm 1.96 \sqrt{0.0171(1-0.0171)(\dfrac{1}{11037} + \dfrac{1}{11034})}$$\displaystyle -0.0091 \pm 1.96 \times 0.0012398$$\displaystyle -0.0091 \pm 0.0024\)The confidence interval is \(\displaystyle [-0.0115, -0.0067]\).

07

Interpret the Results

Since the confidence interval does not contain 0 and is entirely negative, it indicates that the proportion of myocardial infarctions is reduced in the aspirin group compared to the placebo group.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

significance level

In hypothesis testing, the significance level, denoted as \( \alpha \), is a measure of how much evidence you require to reject the null hypothesis. It's a threshold set before conducting a test that determines when you would consider the results statistically significant. Common choices for significance level are 0.05, 0.01, and 0.10.

For example, in the aspirin study, we used \( \alpha = 0.05 \), meaning we were willing to accept a 5% chance of wrongly rejecting the null hypothesis. If our computed p-value is less than 0.05, we reject the null hypothesis and conclude that the results are statistically significant. This is important in ensuring that our findings are not due to random chance.

confidence interval

A confidence interval gives a range of values within which we expect the true population parameter to lie, with a certain level of confidence. In our case, we constructed a 95% confidence interval for the difference in proportions of myocardial infarctions between the aspirin group and the placebo group.

The formula used is: \( (\hat{p_1} - \hat{p_2}) \pm Z \sqrt{\hat{p} (1 - \hat{p}) \left( \dfrac{1}{n_1} + \dfrac{1}{n_2} \right)} \). The Z value for a 95% confidence level is 1.96. Calculating this provides the interval \( [-0.0115, -0.0067] \).

Since this interval does not contain zero and is entirely negative, it suggests that aspirin significantly reduces the proportion of myocardial infarctions compared to the placebo.

test statistic

The test statistic is a standardized value calculated to test hypotheses. In proportion tests, for detecting differences between two proportions (like in our aspirin study), we compute the Z-score. The formula used is: \( Z = \dfrac{\hat{p_1} - \hat{p_2}}{\sqrt{\hat{p} (1 - \hat{p}) \left( \dfrac{1}{n_1} + \dfrac{1}{n_2} \right)}} \).

Here, \( \hat{p_1} \) and \( \hat{p_2} \) are sample proportions from both groups. It measures how far a point estimate (like the difference between sample proportions) is from its hypothesized value under the null hypothesis, standardized by its standard error. In our study, we calculated Z to be approximately -7.34.

A high magnitude test statistic, resulting in a very small p-value, suggests the result is statistically significant.

proportion test

A proportion test is used to determine if there is a significant difference between the proportions of two groups. Specifically, in the aspirin study, we compared the proportion of myocardial infarctions between those who took aspirin (treatment group) and those who took a placebo (control group).

The null hypothesis (\( H_0 \)) stated that there was no difference in proportions, while the alternative hypothesis (\( H_1 \)) claimed a difference exists. To perform the test, we calculated the proportions \( \hat{p_1} \) and \( \hat{p_2} \), pooled proportion \( \hat{p} \), and the Z statistic using the formula: \( Z = \dfrac{\hat{p_1} - \hat{p_2}}{\sqrt{\hat{p} (1 - \hat{p}) \left( \dfrac{1}{n_1} + \dfrac{1}{n_2} \right)}} \).

After computation, the Z value helped us determine the p-value, which was compared against the significance level (\( \alpha \)). In this case, the p-value was significantly small, leading to the rejection of the null hypothesis and supporting the claim that aspirin does affect myocardial infarctions.