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Chapter 6: Problem 202

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? $$ \begin{array}{lcc} \hline \text { Treatment } & \text { Cancer } & \text { No Cancer } \\ \hline \text { Green tea } & 1 & 29 \\ \text { Placebo } & 9 & 21 \end{array} $$

Short Answer

Expert verified
Based on the z-test conducted, there is strong evidence at a 0.05 significance level to suggest that taking a green tea extract significantly reduces the risk of developing prostate cancer.

Step by step solution

01

State the Hypotheses

First, the null and alternative hypotheses must be stated. The null hypothesis \(H_0\) is that the proportion of men who develop prostate cancer is the same for both the green tea group and the placebo group. The alternative hypothesis \(H_1\) is that the proportion of men who develop prostate cancer in the green tea group is less than that in the placebo group. Mathematically, this can be represented as: \(H_0: P_{tea} = P_{placebo}\) and \(H_1: P_{tea} < P_{placebo}\).
02

Calculate Proportions

Next, calculate the observed proportions of developing cancer for both the green tea group and the placebo group. \(P_{tea}= 1/30 = 0.033\) and \(P_{placebo} = 9/30 = 0.3\).
03

Calculate Test Statistic

Now, a test statistic needs to be computed. For this two-proportion z-test, the test statistic is calculated by the formula: \[Z = \frac{(p_{1} - p_{2}) - 0}{\sqrt{p(1-p) [(1/n_{1}) + (1/n_{2})]}}\] where \(p\) is the pooled sample proportion calculated as \[(x_{1} + x_{2}) / (n_{1} + n_{2}) = (1 + 9) / (30 + 30) = 0.1667\] Substituting, the test statistic \(Z\) equals -3.169.
04

Test Hypothesis

Using a significance level of 0.05, check the critical value of Z from a standard normal distribution table. If the calculated Z is outside this range, reject the null hypothesis. The critical Z for a one-tailed test and a significance level of 0.05 is -1.645. The calculated Z equals -3.169, hence the null hypothesis can be rejected.

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Key Concepts

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

Two-Proportion Z-Test
The Two-Proportion Z-Test is a statistical method used to determine if there is a significant difference between the proportions of two groups. It's especially useful when you want to compare two independent samples with binary outcomes, like "cancer" or "no cancer." In our green tea study example, we are comparing the proportion of men developing cancer in the green tea group to those in the placebo group.

This test involves several steps:
  • Stating the null and alternative hypotheses.
  • Calculating the observed proportions for each group.
  • Computing the test statistic using the calculated proportions and a pooled proportion.
  • Interpreting the test statistic by comparing it to standard normal distribution values.
Through these steps, the Two-Proportion Z-Test helps us draw conclusions about the difference in proportions between the two groups, and whether it is statistically significant or merely due to chance.
Null and Alternative Hypotheses
In statistical hypothesis testing, the null and alternative hypotheses are foundational concepts. They represent opposing statements about the parameters of a population based on sample data.

The **null hypothesis (H鈧�)** is typically a statement of no effect or no difference. In our example, it states that the proportion of men developing prostate cancer is the same in both the green tea and placebo groups: \[ H_0: P_{tea} = P_{placebo} \]The **alternative hypothesis (H鈧�)**, on the other hand, is what you want to prove. It suggests an effect or difference exists. For this study, the alternative hypothesis asserts that the proportion of cancer cases is less in the green tea group than in the placebo group:\[ H_1: P_{tea} < P_{placebo} \]By testing these hypotheses, researchers aim to determine whether the collected data provides sufficient evidence to reject the null hypothesis in favor of the alternative.
Statistical Significance
Statistical significance is a concept that tells us whether an observed effect or difference is likely due to chance or if it reflects a true underlying relationship.

In hypothesis testing, this is often determined by comparing a test statistic to a critical value, which is derived from a standard distribution. For the green tea study, we use a significance level of 0.05, which means we are willing to accept a 5% chance that our results are due to random variation alone.

The critical value for a one-tailed test at this level is -1.645. In the solution, the calculated Z statistic was -3.169. Since -3.169 is less than -1.645, we exceed our threshold for significance. This tells us the observed difference between the two groups is unlikely to be due to chance, thus allowing us to reject the null hypothesis.

Statistical significance does not always imply practical relevance, but it is a powerful indicator that further exploration is warranted.

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Most popular questions from this chapter

Standard Error from a Formula and Simulation In Exercises 6.15 to \(6.18,\) find the mean and standard error of the sample proportions two ways: (a) Use StatKey or other technology to simulate at least 1000 sample proportions. Give the mean and standard error and comment on whether the distribution appears to be normal. (b) Use the formulas in the Central Limit Theorem to compute the mean and standard error. Are the results similar to those found in part (a)? Sample proportions of sample size \(n=40\) from a population with \(p=0.5\)

In Exercises 6.159 and \(6.160,\) situations comparing two proportions are described. In each case, determine whether the situation involves comparing proportions for two groups or comparing two proportions from the same group. State whether the methods of this section apply to the difference in proportions. (a) Compare the proportion of students who use a Windows-based \(\mathrm{PC}\) to the proportion who use a Mac. (b) Compare the proportion of students who study abroad between those attending public universities and those at private universities. (c) Compare the proportion of in-state students at a university to the proportion from outside the state. (d) Compare the proportion of in-state students who get financial aid to the proportion of outof-state students who get financial aid.

Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1} \neq \mu_{2}\) using the paired difference sample results \(\bar{x}_{d}=-2.6, s_{d}=4.1\) \(n_{d}=18\)

Number of Bedrooms in Houses in New York and New Jersey The dataset HomesForSale has data on houses available for sale in three Mid-Atlantic states (NY, NJ, and PA). For this exercise we are specifically interested in homes for sale in New York and New Jersey. We have information on 30 homes from each state and observe the proportion of homes with more than three bedrooms. We find that \(26.7 \%\) of homes in NY \(\left(\hat{p}_{N Y}\right)\) and \(63.3 \%\) of homes in NJ \(\left(\hat{p}_{N J}\right)\) have more then three bedrooms. (a) Is the normal distribution appropriate to model this difference? (b) Test for a difference in proportion of homes with more than three bedrooms between the two states and interpret the result.

We see in the AllCountries dataset that the percent of the population that is elderly (over 65 years old) is 17.0 in Austria and 15.9 in Denmark. Suppose we take random samples of size 200 from each of these countries and compute the difference in sample proportions \(\hat{p}_{A}-\hat{p}_{D}\), where \(\hat{p}_{A}\) represents the proportion of the sample that is elderly in Austria and \(\hat{p}_{D}\) represents the proportion of the sample that is elderly in Denmark. Find the mean and standard deviation of the differences in sample proportions.
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