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Chapter 11: Problem 64

Botox side effects An advertisement for Botox Cosmetic by Allergan, Inc. for treating wrinkles appeared in several magazines. The back page of the ad showed the results of a randomized clinical trial to compare 405 people receiving Botox injections to 130 people receiving placebo in terms of the frequency of various side effects. The side effect of pain in the face was reported by 9 people receiving Botox and 0 people receiving placebo. Use software (or a web app) to conduct a two-sided test of the hypothesis that the probability of this side effect is the same for each group. Report all steps of the test and interpret results.

Short Answer

Expert verified
Conduct a 2-proportion z-test; if p-value < 0.05, facial pain differs between groups.

Step by step solution

01

Define Hypotheses

The null hypothesis (H0) states that the probability of experiencing facial pain is the same in both groups (Botox and placebo). The alternative hypothesis (H1) is that the probabilities are different. Formally: \( H_0: p_1 = p_2 \) versus \( H_1: p_1 eq p_2 \), where \( p_1 \) is the probability for the Botox group and \( p_2 \) for the placebo group.
02

Set Significance Level

Choose a significance level (\( \alpha \)). Common practice is to use \( \alpha = 0.05 \). This means we will reject the null hypothesis if the p-value is less than 0.05.
03

Calculate Test Statistic

Use a 2-proportion z-test to calculate the test statistic. First, find the proportion of people reporting side effects in each group: \( \hat{p}_1 = \frac{9}{405} \) for Botox and \( \hat{p}_2 = \frac{0}{130} = 0 \) for placebo. Then calculate the pooled proportion: \( \hat{p} = \frac{9 + 0}{405 + 130} \). Compute the test statistic using: \[ z = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_1} + \frac{1}{n_2})}} \] where \( n_1 = 405 \) and \( n_2 = 130 \).
04

Calculate P-value

Using the calculated z-value, find the p-value from the standard normal distribution. This p-value represents the probability of observing a test statistic as extreme as, or more extreme than, the observed statistic under the null hypothesis.
05

Make a Decision

Compare the p-value to the significance level (\( \alpha = 0.05 \)). If the p-value is less than 0.05, reject the null hypothesis; otherwise, do not reject it.
06

Interpret Results

If you rejected the null hypothesis, it means there is sufficient evidence to conclude that the probability of experiencing facial pain differs between the Botox and placebo groups. If not rejected, there is not sufficient evidence to claim a difference.

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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 determine whether there is a significant difference between the proportions of two groups.
This test is useful when you want to compare the likelihood of a particular event occurring in two separate groups.
For example, in the clinical trial for Botox Cosmetic, researchers were interested in whether the proportion of participants experiencing facial pain differed between those who received Botox and those who received a placebo.
Key steps in conducting a two-proportion z-test include:
  • Determining the sample proportions - For Botox: \( \hat{p}_1 = \frac{9}{405} \) - For placebo: \( \hat{p}_2 = \frac{0}{130} \)
  • Calculating a pooled proportion, which combines the probabilities of both groups together.
  • Using this information to find the z-value that reflects how far apart the sample proportions are.
Null Hypothesis
The null hypothesis is an assumption that there is no effect or no difference between groups in an experiment.
It's the default position predicting that any observed effects are due to random variation rather than a true difference.
For the Botox study, the null hypothesis (\( H_0 \)) states that the probability of experiencing facial pain is the same for both the Botox group and the placebo group. Formally, this is expressed as:\( H_0: p_1 = p_2 \).The essence of the null hypothesis is to maintain an assumption of equality until proven otherwise, indicating the role of randomness in any observed differences.
Alternative Hypothesis
The alternative hypothesis is a statement that contradicts the null hypothesis. It suggests that there is a genuine effect or a difference between groups.
It's what researchers aim to support by conducting the test.
In the Botox clinical trial, the alternative hypothesis (\( H_1 \)) posits that there is a difference in the probability of facial pain between those injected with Botox and those given a placebo.
Formally, it's described as:\( H_1: p_1 eq p_2 \). While the null hypothesis is what you attempt to disprove, the alternative hypothesis offers an explanation for observed effects.
Significance Level
The significance level, denoted as \( \alpha \), is a threshold that determines when you should reject the null hypothesis.
A commonly used significance level is 0.05.
This means you are willing to accept a 5% chance that the observed difference is due to sampling variability rather than a real effect. Choosing a lower significance level, like 0.01, requires stronger evidence to reject the null hypothesis.
In the Botox example, if the calculated p-value is less than 0.05, it indicates that the differences observed in facial pain reports are statistically significant, prompting a rejection of the null hypothesis.
Adopting this level signifies the balance between Type I errors (falsely rejecting a true null) and Type II errors (failing to reject a false null).

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Most popular questions from this chapter

In a study conducted by a pharmaceutical company, 605 out of 790 smokers and 122 out of 434 nonsmokers were diagnosed with lung cancer. a. Construct a \(2 \times 2\) contingency table relating smoking (SMOKING, categories smoker and nonsmoker) as the rows to lung cancer (LUNGCANCER, categories present and absent) as the columns. b. Find the four expected cell counts when assuming independence. Compare them to the observed cell counts, identifying cells having more observations than expected. c. For this data, \(X^{2}=272.89 .\) Verify this value by plugging into the formula for \(X^{2}\) and computing the sum.

Karl Pearson devised the chi-squared goodness-of-fit test partly to analyze data from an experiment to analyze whether a particular roulette wheel in Monte Carlo was fair, in the sense that each outcome was equally likely in a spin of the wheel. For a given European roulette wheel with 37 pockets (with numbers \(0,1,2, \ldots, 36\) ), consider the null hypothesis that the wheel is fair. a. For the null hypothesis, what is the probability for each pocket? b. For an experiment with 3,700 spins of the roulette wheel, find the expected number of times each pocket is selected. c. In the experiment, the 0 pocket occurred 110 times. Show the contribution to the \(X^{2}\) statistic of the results for this pocket. d. Comparing the observed and expected counts for all 37 pockets, we get \(X^{2}=34.4\) Specify the df value and indicate whether there is strong evidence that the roulette wheel is not balanced. (Hint: Recall that the \(d f\) value is the mean of the distribution.)

Exercise 11.32 showed the association between recreation and happiness. The table shown here gives the standardized residuals for those data in parentheses. a. Explain what a relatively small standardized residual such as -0.5 in the second cell represents. b. Identify the cells in which you would infer that the population has more cases than would occur if recreation and happiness were independent. Pick one of these cells and explain the association relative to independence.

Colon cancer and race The State Center for Health Statistics for the North Carolina Division of Public Health released a report in 2010 that indicates that there are racial disparities in colorectal cancer incidence and mortality rates. The report states, "African Americans are less likely to receive appropriate screenings that reduce the risk of developing or dying from colorectal cancer." During \(2002-2006,\) the rate of incidence for African Americans was 57.3 per 100,000 versus 46.5 per 100,000 for white residents (www.schs.state.nc.us). African Americans were \(19 \%\) more likely to have been diagnosed with colon cancer. Explain how to get this estimate.

Seat belt helps? The table refers to passengers in autos and light trucks involved in accidents in the state of Maine in a recent year. a. Use the difference of proportions to describe the strength of association. Interpret. b. Use the relative risk to describe the strength of association. Interpret. \begin{tabular}{lcc} \hline Maine Accident Data & \\ \hline & \multicolumn{2}{c} { Injury } \\ \cline { 2 - 3 } Seat Belt & Yes & No \\ \hline No & 3865 & 27,037 \\ Yes & 2409 & 35,383 \\ \hline \end{tabular} Source: Dr. Cristanna Cook, Medical Care Development, Augusta, Maine.
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