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The Women's Health Initiative conducted a randomized experiment to see whether hormone therapy was helpful for postmenopausal women. The women were randomly assigned to receive the estrogen plus progestin hormone therapy or a placebo. After five years, 107 of the 8506 on the hormone therapy developed cancer and 88 of the 8102 in the placebo group developed cancer. Is this a significant difference? a. Set up notation and state assumptions and hypotheses. b. Find the test statistic and P-value and interpret. (If you prefer, use software, such as MINITAB or a web app, for which you can conduct the analysis by entering summary counts.) c. What is your conclusion for a significance level of \(0.05 ?\) (The study was planned to be eight years long but was stopped after five years because of increased heart and cancer problems for the therapy group. This shows a benefit of doing two-sided tests, because results sometimes have the opposite direction from the expected one.)

Step by step solution

01

Define Notation and Hypotheses

Let's define our notation: - Let \( p_1 \) be the probability of developing cancer for those in the hormone therapy group.- Let \( p_2 \) be the probability of developing cancer for those in the placebo group.Our null hypothesis \( H_0 \) is that there is no difference between the probabilities: \( H_0: p_1 = p_2 \).The alternative hypothesis \( H_a \) is that there is a difference: \( H_a: p_1 eq p_2 \).

02

Find Test Statistic

To calculate the test statistic for a two proportion z-test, we first calculate the sample proportions:- Proportion for hormone therapy group: \( \hat{p}_1 = \frac{107}{8506} \approx 0.0126 \).- Proportion for placebo group: \( \hat{p}_2 = \frac{88}{8102} \approx 0.0109 \).The pooled sample proportion \( \hat{p} \) is:\[ \hat{p} = \frac{107 + 88}{8506 + 8102} = \frac{195}{16608} \approx 0.0117 \]The z-test statistic is:\[ z = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p}(1 - \hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \]Substitute the values to calculate.

03

Calculate P-value

Calculate the z-value using the numbers from Step 2:\[ z \approx \frac{0.0126 - 0.0109}{\sqrt{0.0117(1 - 0.0117)\left(\frac{1}{8506} + \frac{1}{8102}\right)}} \approx 1.136 \]Using a standard normal distribution table or software, find the P-value for \( z = 1.136 \). The P-value for a two-tailed test with \( z = 1.136 \) is approximately 0.256.

04

Draw Conclusion

At a significance level of \( \alpha = 0.05 \), we compare the P-value to \( \alpha \).Since the P-value (0.256) is greater than \( 0.05 \), we fail to reject the null hypothesis.This means there is not a statistically significant difference in cancer rates between the hormone therapy group and the placebo group.

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Key Concepts

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

Two-Proportion Z-Test

A two-proportion z-test is a statistical method used to compare the proportions of two independent groups. In simpler terms, it's like checking if two different groups have similar rates or not. In the scenario of the Women's Health Initiative, the two groups of interest are women receiving hormone therapy and those on a placebo.

Here's how it works:

• Calculate the sample proportions: This involves finding the proportion of women who developed cancer in each group. For the hormone therapy group, it's \( \hat{p}_1 = \frac{107}{8506} \approx 0.0126 \), and for the placebo group, it's \( \hat{p}_2 = \frac{88}{8102} \approx 0.0109 \).
• Pooled sample proportion: This is a weighted average of both groups' proportions and is calculated as \( \hat{p} = \frac{107 + 88}{8506 + 8102} \approx 0.0117 \).
• Z-statistic: This tells how far apart the sample proportions are, relatively. It's given by:\[ z = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p}(1 - \hat{p})(\frac{1}{n_1} + \frac{1}{n_2})}} \]

This kind of test helps determine if proportion differences are statistically significant or if they could have happened by random chance.

Randomized Experiment

A randomized experiment is a type of scientific study design that is crucial for understanding cause-and-effect relationships. In the Women's Health Initiative, researchers randomly assigned participants into one of two groups: hormone therapy or placebo. This randomization is vital because it helps eliminate selection bias, ensuring that the groups are as similar as possible at the start. Why was randomization used?

• To prevent biases: By randomly assigning individuals, we aim to distribute any known and unknown confounding variables equally across the groups.
• To strengthen causal inference: Randomization increases the reliability of the study outcomes, making it possible to confidently attribute differences in outcomes to the treatment rather than some other variable.

Randomized experiments like this are often considered the gold standard in research because they provide robust and credible results.

P-Value Interpretation

The p-value is a statistical measure that helps us determine the strength of the results from a hypothesis test. In simple terms, it tells us how likely it is to observe the data, or something more extreme, if the null hypothesis is true. For the Women's Health Initiative, the calculated p-value was approximately 0.256. Here's how to interpret it:

• If the p-value is low (usually less than 0.05), it suggests that the observed data is unlikely under the null hypothesis, and thus we may consider rejecting the null hypothesis.
• If the p-value is high (greater than 0.05), like in this scenario (0.256), it indicates that the data is quite plausible under the null hypothesis, and we fail to reject it.

In this case, the p-value suggests there's no significant difference between cancer rates in the hormone therapy and placebo groups, meaning the data doesn't strongly contradict the assumption of no difference.

Statistical Significance

Statistical significance is a concept that tells us when an effect observed in a sample is unlikely to have occurred just by chance, given a threshold (usually set at 0.05 for many scientific studies). It implies a meaningful difference that is not explained by variability in the data alone. In the Women's Health Initiative example:

• We set the significance level (b1) at 0.05. This means we were willing to accept a 5% chance of being wrong if we conclude there's a significant difference when there isn't one.
• Our calculated p-value was 0.256, which is greater than 0.05. This indicates that the observed differences in cancer rates are not statistically significant at the 5% level.
• Because the p-value is higher than the significance level, we fail to reject the null hypothesis, concluding that any observed differences may be due to chance.

Statistical significance helps researchers interpret the results, but it's important to remember that it doesn't measure the importance or size of the effect.