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Chapter 10: Problem 58

Statin drugs are used to decrease cholesterol levels, and therefore hopefully to decrease the chances of a heart attack. In a British study ("MRC/BHF Heart Protection Study of Cholesterol Lowering with Simvastin in 20,536 High-Risk Individuals: A Randomized Placebo-Controlled Trial,"Lancet, 2002: 7-22) 20,536 at-risk adults were assigned randomly to take either a 40 -mg statin pill or placebo. The subjects had coronary disease, artery blockage, or diabetes. After 5 years there were 1328 deaths (587 from heart attack) among the 10,269 in the statin group and 1507 deaths (707 from heart attack) among the 10,267 in the placebo group. a. Give a \(95 \%\) confidence interval for the difference in population death proportions. b. Give a \(95 \%\) confidence interval for the difference in population heart attack death proportions. c. Is it reasonable to say that most of the difference in death proportions is due to heart attacks, as would be expected?

Short Answer

Expert verified
Yes, most of the difference in death proportions is likely due to heart attacks, as evidenced by similar confidence intervals for both total and heart attack deaths.

Step by step solution

01

Identify the Data

We need to identify the relevant data for each group. For the statin group, there are 10,269 individuals with 1,328 deaths (587 from heart attack). For the placebo group, there are 10,267 individuals with 1,507 deaths (707 from heart attack).
02

Calculate Death Proportions

The death proportion for the statin group is \( \frac{1328}{10269} \), and for the placebo group, it is \( \frac{1507}{10267} \). We can convert these fractions to decimals to find the proportions.
03

Calculate Heart Attack Death Proportions

The heart attack death proportion for the statin group is \( \frac{587}{10269} \), and for the placebo group, it is \( \frac{707}{10267} \). Again, convert these to decimals to find the proportions.
04

Calculate Difference in Proportions

Subtract the statin group proportions from the placebo group proportions both for total deaths and heart attack deaths. This gives us the observed differences in the sample.
05

Standard Error Calculation

Use the formula for the standard error (SE) of the difference between two proportions: \[ SE = \sqrt{ \frac{{\hat{p}_1 (1 - \hat{p}_1)}}{n_1} + \frac{{\hat{p}_2 (1 - \hat{p}_2)}}{n_2} } \]where \( \hat{p}_1 \) and \( \hat{p}_2 \) are the proportions of the statin and placebo groups, respectively, and \( n_1, n_2 \) are their sample sizes.
06

Construct 95% Confidence Interval

Find the critical z-value for a 95% confidence interval (usually \(1.96\) for large samples). Multiply the standard error by this z-value to find the margin of error, and then construct the confidence interval as:\[ \text{Difference} \pm \text{Margin of Error} \]
07

Analyze Results for Heart Attack Deaths

Compare the confidence intervals for total deaths and for heart attack deaths. If most of the difference in total death proportions is reflected in the heart attack death proportions, this supports the statement that most of the difference is due to heart attacks.

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Key Concepts

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

Understanding Hypothesis Testing
Hypothesis testing plays a crucial role in determining if there is a significant difference between two groups, in this case, those who received statin pills and those who received a placebo. It involves setting up two competing statements called hypotheses:
  • The null hypothesis (H鈧�): This is the default position that there is no difference between the groups in terms of proportions.
  • The alternative hypothesis (H鈧�): This suggests that there is a difference in proportions between the two groups.
After forming these hypotheses, statistical tests help decide whether to reject the null hypothesis. Rejecting the null hypothesis implies that the observed difference is unlikely due to random chance.
To systematically approach this, we calculate the test statistic, usually a z-score or t-score, which measures how far the sample statistic is from the null hypothesis value. By selecting a confidence level, such as 95%, we determine the critical value that the test statistic must exceed to reject the null hypothesis.
This structured process is essential for making objective conclusions about whether statin drugs influenced mortality rates.
Performing Statistical Analysis
Statistical Analysis provides a framework for interpreting data and making informed decisions based on statistical evidence. It involves several steps:
  • Data Collection: Gathering accurate and relevant data is fundamental. In the study, mortality and heart attack rates were collected for the statin and placebo groups over five years.
  • Data Cleaning and Summarization: Before analysis, the data must be checked for consistency and errors. This is followed by summarizing the data using descriptive statistics like mean, proportion, etc.
  • Data Visualization: Charts and graphs can illuminate patterns in the data, making them easier to understand and analyze.
  • Statistical Testing: Tools like confidence intervals and significance tests, such as z-tests, help draw conclusions about the broader population based on this sample data.
  • Interpretation and Analysis: Finally, interpreting results in context, including confidence intervals, involves assessing what the data says about the relationship between statins and mortality rates.
Ultimately, statistical analysis provides a pathway from raw data to actionable insights about real-world outcomes.
Exploring Proportion Differences
Proportion differences are central to this type of analysis, especially when comparing outcomes between two groups such as the statin and placebo groups. This involves calculating the proportion of deaths or specific outcomes in each group and then determining the difference between these proportions.
For a thorough comparison, it's crucial to compute the confidence interval for the proportion differences. The confidence interval offers a range in which the true population parameter is expected to lie, considering the sample data. For example, in the study, by calculating the 95% confidence interval for heart attack death proportions, we determine a range of values where the true difference in heart attack death rates between the groups likely exists.
Calculating the difference and performing confidence interval analysis gives an understanding of the effect size and significance of the treatment. This analysis helps to determine if drugs like statins have a real-world, significant impact on reducing heart attack risks. Understanding whether observed differences reflect real effects or mere chance is essential in medical research and policymaking.

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Most popular questions from this chapter

Let \(\mu_{1}\) and \(\mu_{2}\) denote true average tread lives for two competing brands of size P205/65R 15 radial tires. Test \(H_{0}: \mu_{1}-\mu_{2}=0\) versus \(H_{\mathrm{a}}: \mu_{1}-\) \(\mu_{2} \neq 0\) at level \(.05\) using the following data: \(m=45, \quad \bar{x}=42,500, \quad s_{1}=2200, \quad n=45\), \(\bar{y}=40,400\), and \(s_{2}=1900\).

Suppose a level \(.05\) test of \(H_{0}: \mu_{1}-\mu_{2}=0\) versus \(H_{2}: \mu_{1}-\mu_{2}>0\) is to be performed, assuming \(\sigma_{1}=\sigma_{2}=10\) and normality of both distributions, using equal sample sizes \((m=n)\). Evaluate the probability of a type II error when \(\mu_{1}-\mu_{2}=1\) and \(n=25,100,2500\), and 10,000 . Can you think of real problems in which the difference \(\mu_{1}-\mu_{2}=1\) has little practical significance? Would sample sizes of \(n=10,000\) be desirable in such problems?

The level of monoamine oxidase (MAO) activity in blood platelets \((\mathrm{nm} / \mathrm{mg}\) protein/ \(\mathrm{h})\) was determined for each individual in a sample of 43 chronic schizophrenics, resulting in \(\bar{x}=2.69\) and \(s_{1}=2.30\), as well as for 45 normal subjects, resulting in \(\bar{y}=6.35\) and \(s_{2}=4.03\). Does this data strongly suggest that true average MAO activity for normal subjects is more than twice the activity level for schizophrenics? Derive a test procedure and carry out the test using \(\alpha=.01\). [Hint: \(H_{0}\) and \(H_{\mathrm{a}}\) here have a different form from the three standard cases. Let \(\mu_{1}\) and \(\mu_{2}\) refer to true average MAO activity for schizophrenics and normal subjects, respectively, and consider the parameter \(\theta=2 \mu_{1}-\mu_{2}\). Write \(H_{0}\) and \(H_{\mathrm{a}}\) in terms of \(\theta\), estimate \(\theta\), and derive \(\hat{\sigma}_{\hat{\theta}}\) ("Reduced Monoamine Oxidase Activity in Blood Platelets from Schizophrenic Patients," Nature, July 28, 1972: 225-226).]

Can the right diet help us cope with diseases associated with aging such as Alzheimer's disease? A study ("Reversals of Age-Related Declines in Neuronal Signal Transduction, Cognitive, and Motor Behavioral Deficits with Blueberry, Spinach, or Strawberry Dietary Supplement," \(J\). Neurosci., 1999; 8114-8121) investigated the effects of fruit and vegetable supplements in the diet of rats. The rats were 19 months old, which is aged by rat standards. The 40 rats were randomly assigned to four diets, of which we will consider just the blueberry diet and the control diet here. After 8 weeks on their diets, the rats were given a number of tests. We give the data for just one of the tests, which measured how many seconds they could walk on a rod. Here are the times for the ten control rats (C) and ten blueberry rats (B): The objective is to obtain a \(95 \%\) confidence interval for the difference of population means. a. Determine a \(95 \%\) confidence interval for the difference of population means using the method based on the Theorem of Section 10.2. b. Obtain a bootstrap sample of 999 differences of means. Check the bootstrap distribution for normality using a normal probability plot. c. Use the standard deviation of the bootstrap distribution along with the mean and \(t\) critical value from (a) to get a \(95 \%\) confidence interval for the difference of means. d. Use the bootstrap sample and the percentile method to obtain a \(95 \%\) confidence interval for the difference of means. e. Compare your three confidence intervals. If they are very similar, why do you think this is the case? If you had used a critical value from the normal table rather than the \(t\) table, would the result of (c) agree better with the result of (d)? Why? f. Interpret your results. Do the blueberries make a substantial difference?

Give as much information as you can about the \(P\)-value of the \(F\) test in each of the following situations: a. \(v_{1}=5, v_{2}=10\), upper-tailed test, \(f=4.75\) b. \(v_{1}=5, v_{2}=10\), upper-tailed test, \(f=2.00\) c. \(v_{1}=5, v_{2}=10\), two-tailed test, \(f=5.64\) d. \(v_{1}=5, v_{2}=10\), lower-tailed test, \(f=.200\) e. \(v_{1}=35, v_{2}=20\), upper-tailed test, \(f=3.24\)
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