/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 181 Refer to a study on hormone repl... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Refer to a study on hormone replacement therapy. Until \(2002,\) hormone replacement therapy (HRT), taking hormones to replace those the body no longer makes after menopause, was commonly prescribed to post-menopausal women. However, in 2002 the results of a large clinical trial \({ }^{48}\) were published, causing most doctors to stop prescribing it and most women to stop using it, impacting the health of millions of women around the world. In the experiment, 8506 women were randomized to take HRT and 8102 were randomized to take a placebo. Table 6.8 shows the observed counts for several conditions over the five years of the study. (Note: The planned duration was 8.5 years. If Exercises 6.181 through 6.184 are done correctly, you will notice that several of the intervals just barely exclude zero. The study was terminated as soon as some of the intervals included only positive values, because at that point it was unethical to continue forcing women to take HRT.) $$ \begin{array}{lcc} \hline \text { Condition } & \text { HRT Group } & \text { Placebo Group } \\ \hline \text { Cardiovascular Disease } & 164 & 122 \\ \text { Invasive Breast Cancer } & 166 & 124 \\ \text { Cancer (all) } & 502 & 458 \\ \text { Fractures } & 650 & 788 \\ \hline \end{array} $$ Find a \(95 \%\) confidence interval for the difference in proportions of women who get cardiovascular disease taking HRT vs taking a placebo.

Short Answer

Expert verified
The 95% confidence interval for the difference in proportions of women who get cardiovascular disease taking HRT vs taking a placebo is approximately (0.0015, 0.0071).

Step by step solution

01

Calculate the Proportions

Calculate the proportion of women who developed cardiovascular disease for each group. For the HRT group, it's \( p1 = 164/8506 \approx 0.0193 \), and for the placebo group, it's \( p2 = 122/8102 \approx 0.0150 \).
02

Calculate the Difference in Proportions

Find the difference in proportions. This can be done by subtracting the proportion of the placebo group from the HRT group, giving us \( p1 - p2 \approx 0.0193 - 0.0150 = 0.0043 \).
03

Calculate the Standard Error

Calculate the standard error using the formula \[ SE = \sqrt{ \frac{p1 * (1 - p1)} {n1} + \frac{p2 * (1 - p2)} {n2} } \], where \(n1\) and \(n2\) are sizes of the HRT and placebo groups respectively. Substituting the values, we get \( SE \approx \sqrt{ \frac{0.0193 * 0.9807}{8506} + \frac{0.0150 * 0.9850}{8102} } \approx 0.0014 \).
04

Calculate the Confidence Interval

Calculate the 95% confidence interval using the formula \[ CI = (p1 - p2) \pm Z * SE \], where Z represents the Z score for a 95% confidence level, which is approximately 1.96. Substituting our values, we get \[ CI \approx 0.0043 \pm 1.96 * 0.0014 = (0.0015, 0.0071) \].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

HRT (Hormone Replacement Therapy)
Hormone Replacement Therapy (HRT) involves taking hormones to replace those the body no longer produces after menopause. For many years, HRT was widely used to manage menopausal symptoms such as hot flashes and to prevent long-term health issues such as osteoporosis. The use of HRT changed dramatically after a landmark study in 2002 showed potential risks, including an increased incidence of cardiovascular disease and certain cancers.

Understanding the implications of HRT on women's health requires rigorous clinical trials and statistical analysis to determine its benefits and risks. Healthcare decisions often hinge on such evidence, which can impact the lives of millions of women around the world.
Cardiovascular Disease in Clinical Trials
Clinical trials are essential for establishing the relationship between medical interventions, like HRT, and health outcomes, such as cardiovascular disease (CVD). In the study referenced, the occurrence of CVD in both the HRT group and placebo group was meticulously recorded. Cardiovascular disease remains a leading cause of mortality in women, and understanding how HRT could influence its incidence is crucial.

In clinical trials, researchers often look for differences in disease occurrence between groups to assess the efficacy or risks associated with a treatment. A confidence interval for the difference in proportions, which was calculated based on the study data, helps quantify the uncertainty around the observed effect of HRT on CVD rates.
Statistics in Medical Research
Statistics play a pivotal role in medical research, allowing for the rigorous evaluation of clinical trial data. In the context of the HRT study, calculating the confidence interval for the difference in proportions of women developing CVD while on HRT versus a placebo is an essential statistical process.

A confidence interval provides a range of values that are believed, with a certain level of confidence, to contain the true difference in proportions. In this case, the 95% confidence interval suggests that we can be 95% confident that the true difference in the proportion of CVD cases between the HRT and placebo groups lies within the calculated interval. The process involves several steps: calculating the proportions in each group, finding the difference, determining standard error, and finally constructing the confidence interval using a Z score for the desired confidence level. By providing clarity on the magnitude and precision of the effect, statistics support informed decision-making in healthcare.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A sample with \(n=75, \bar{x}=18.92,\) and \(s=10.1\)

Involve scores from the high school graduating class of 2010 on the SAT (Scholastic Aptitude Test). The average score on the Mathematics part of the SAT exam for males is 534 with a standard deviation of 118 , while the average score for females is 500 with a standard deviation of 112 . (a) If random samples are taken with 40 males and 60 females, find the mean and standard deviation of the distribution of differences in sample means, \(\bar{x}_{m}-\bar{x}_{f},\) where \(\bar{x}_{m}\) represents the sample mean for the males and \(\bar{x}_{f}\) represents the sample mean for the females. (b) Repeat part (a) if the random samples contain 400 males and 600 females. (c) What effect do the different sample sizes have on center and spread of the distribution?

Green Tea and Prostate Cancer A preliminary study suggests a benefit from green tea for those at risk of prostate cancer. \(^{55}\) The study involved 60 men with PIN lesions, some of which turn into prostate cancer. Half the men, randomly determined, were given \(600 \mathrm{mg}\) a day of a green tea extract while the other half were given a placebo. The study was double-blind, and the results after one year are shown in Table \(6.14 .\) Does the sample provide evidence that taking green tea extract reduces the risk of developing prostate cancer? $$ \begin{array}{lcc} \hline \text { Treatment } & \text { Cancer } & \text { No Cancer } \\ \hline \text { Green tea } & 1 & 29 \\ \text { Placebo } & 9 & 21 \end{array} $$

Exercise 4.86 on page 263 introduces a matched pairs study in which 47 participants had cell phones put on their ears and then had their brain glucose metabolism (a measure of brain activity) measured under two conditions: with one cell phone turned on for 50 minutes (the "on" condition) and with both cell phones off (the "off" condition). Brain glucose metabolism is measured in \(\mu \mathrm{mol} / 100 \mathrm{~g}\) per minute, and the differences of the metabolism rate in the on condition minus the metabolism rate in the off condition were computed for all participants. The mean of the differences was 2.4 with a standard deviation of \(6.3 .\) Find and interpret a \(95 \%\) confidence interval for the effect size of the cell phone waves on mean brain metabolism rate.

According to the 2006 Australia Census, \(^{43} 25.5 \%\) of Australian women over the age of 25 had a college degree, while the percentage for Australian men was \(21.4 \% .\) Suppose we select random samples of 200 women and 200 men from this population and look at the differences in proportions with college degrees, \(\hat{p}_{f}-\hat{p}_{m}\), in those samples. (a) Describe the distribution (center, spread,shape) for the difference in sample proportions. Include a rough sketch of the distribution with values labeled on the horizontal axis. (b) What is the chance that the proportion with college degrees in the men's sample is actually more than the proportion in the women's sample? (Hint: Think about what must be true about \(\hat{p}_{f}-\hat{p}_{m}\) when this happens.)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.