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Chapter 8: Problem 64

In a 2018 study reported in The Lancet, Mercie et al. investigated the efficacy and safety of varenicline for smoking cessation in people living with HIV. The study was a randomized, double-blind, placebo-controlled trial. Of the 123 subjects treated with varenicline, 18 abstained from smoking for the entire 48-week study period. Of the 124 subjects assigned to the placebo group, 8 abstained from smoking for the entire study period. a. Find the sample percentage of subjects in each group who abstained from smoking for the entire study period. b. Determine whether varenicline is effective in reducing smoking among HIV patients. Note that this means we should test if the proportion of the varenicline group who abstained from smoking for the entire study period is significantly greater than that of the placebo group. Use a significance level of \(0.05\)

Short Answer

Expert verified
Based on the data: a) The sample percentage who successfully abstained from smoking for the entire study period is 14.63% for the varenicline group and 6.45% for the placebo group. b) Given that the Z-score (2.61) is higher than the critical Z-value (1.645), it can be concluded that varenicline is indeed effective in decreasing smoking in HIV patients with a significance level of 0.05.

Step by step solution

01

Calculate the percentages

First, it is necessary to calculate the sample percentage of subjects who abstained from smoking for the entire study period for each group. The formula used is:\[ Percentage = \frac{Number of successes}{Total number in the group} * 100 \] For the varenicline group:\[ Percentage = \frac{18}{123} * 100 = 14.63\% \] For the placebo group:\[ Percentage = \frac{8}{124} * 100 = 6.45\% \]
02

Set up hypotheses

To determine if the varenicline is effective in reducing smoking among HIV patients, a statistical hypothesis test for proportions is used. The null hypothesis (\(H_0\)) is that there is no difference between the proportions of subjects who abstained from smoking in the two groups. The alternative hypothesis (\(H_1\)) is that the proportion who abstained in the varenicline group is greater than in the placebo group.So,\[ H_0: p_1 = p_2\]\[ H_1: p_1 > p_2\]where \(p_1\) is the proportion of subjects who abstained in the varenicline group and \(p_2\) is the proportion in the placebo group.
03

Test statistic

Given the sample size is large enough, the test statistic follows a standard normal distribution. Here, the z-test is used, given by the formula:\[ z = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{p(1 - p)(\frac{1}{n_1} + \frac{1}{n_2})}} \]where\(\hat{p}_1\) = observed proportion in the varenicline group = 0.1463,\(\hat{p}_2\) = observed proportion in the placebo group = 0.0645,\(n_1\) = size of the varenicline group = 123,\(n_2\) = size of the placebo group = 124,and \(p\) is the pooled proportion.First, calculate the pooled proportion:\[ p = \frac{n_1\hat{p}_1 + n_2\hat{p}_2}{n_1 + n_2} = \frac{123*0.1463 + 124*0.0645}{123 + 124} = 0.1054\]Now, the test statistic:\[ z = \frac{0.1463 - 0.0645}{\sqrt{0.1054(1 - 0.1054)(\frac{1}{123} + \frac{1}{124})}} = 2.61\]
04

Step 4:Decision

Now, this z-value has to be compared with critical z-value. Using a level of significance of 0.05 and the fact that the test is one-tailed, acquire the critical z-value from the z-table, which is 1.645. If the calculated z-value is greater than the critical z-value, reject the null hypothesis.Here, since our calculated z-value, 2.61, is greater than the critical value of 1.645, we reject the null hypothesis.
05

Step 5:Conclusion

After rejecting the null hypothesis, this indicates that there is significant evidence to suggest that the proportion of patients who abstained from smoking is higher in the varenicline group than in the placebo group, and therefore that varenicline is effective in reducing smoking among HIV patients.

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Key Concepts

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

Randomized Controlled Trial
One of the gold standards of clinical research is the randomized controlled trial (RCT). This method involves randomly assigning participants to either the treatment group or the control group, thus minimizing potential biases and ensuring that the only expected difference between groups is the intervention being tested. In our example, patients living with HIV were randomly assigned to receive either the study medication, varenicline, or a placebo.

The importance of RCTs cannot be overstated; they help establish the cause-and-effect relationship between an intervention and its outcomes. Participants do not choose which group to be in, and this randomization process is crucial to reduce selection bias. Moreover, to avoid influencing results through expectations, trials are often blinded, meaning that neither the participants nor the researchers know who is receiving the treatment, which was the case in the study by Mercie et al.
Proportion Test
The proportion test is used to compare the proportion of outcomes in two different groups. In our smoking cessation study, researchers used a proportion test to determine if a greater percentage of participants in the varenicline group abstained from smoking compared to the placebo group. To perform this test, you need to calculate the sample proportions and then find the test statistic for comparison.

In simple terms, the proportion test is effectively asking if the difference between two percentages is large enough to be considered statistically significant or if it might just be due to chance. This test can be imagined as a way to measure the signal (true effect) against the noise (random variations) in the data. As seen in the step-by-step solution, researchers calculated that the varenicline group had a higher percentage of non-smokers, which led to the next step: assessing the significance of this finding.
Significance Level
The significance level is a critical concept in hypothesis testing, representing the threshold at which you're willing to reject the null hypothesis - the default assumption that there is no effect or no difference. It's denoted as \(\alpha\) and commonly set at \(0.05\), or 5%, indicating a 5% risk of concluding that a difference exists when there actually is none (Type I error).

When the calculated test statistic value is greater than the critical value from the z-table at the chosen significance level, we have sufficient evidence to reject the null hypothesis. In the provided study, the calculated z-value of 2.61 exceeded the critical z-value of 1.645 for a one-tailed test at an \(\alpha\) of \(0.05\). This result suggests that there is less than a 5% probability that the results are due to random chance, implying that varenicline indeed has a statistically significant effect on smoking cessation in HIV patients.

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Most popular questions from this chapter

In a Northeastern University/Gallup poll of 461 young Americans aged 18 to 35,152 reported they would be comfortable riding in a self-driving car. Suppose we are testing the hypothesis that more than \(30 \%\) of Americans in this age group would be comfortable riding in a self-driving car, using a significance level of \(0.05 .\) Which of the following figures correctly matches the alternative hypothesis \(p>0.30 .\) Report and interpret the correct p-value.

A true/false test has 50 questions. Suppose a passing grade is 35 or more correct answers. Test the claim that a student knows more than half of the answers and is not just guessing. Assume the student gets 35 answers correct out of \(50 .\) Use a significance level of \(0.05 .\) Steps 1 and 2 of a hypothesis test procedure are given. Show steps 3 and 4, and be sure to write a clear conclusion. Step $$\text { 1: } \begin{aligned}&\mathrm{H}_{0}: p=0.50 \\\&\mathrm{H}_{\mathrm{a}}: p>0.50\end{aligned}$$ Step 2: Choose the one-proportion \(z\) -test. Sample size is large enough, because \(n p_{0}\) is \(50(0.5)=25\) and \(n\left(1-p_{0}\right)=50(0.50)=25\), and both are more than \(10 .\) Assume the sample is random and \(\alpha=0.05\).

Votes for Independents Judging on the basis of experience, a politician claims that \(50 \%\) of voters in Pennsylvania have voted for an independent candidate in past elections. Suppose you surveyed 20 randomly selected people in Pennsylvania, and 12 of them reported having voted for an independent candidate. The null hypothesis is that the overall proportion of voters in Pennsylvania that have voted for an independent candidate is \(50 \%\). What value of the test statistic should you report?

A hospital readmission is an episode when a patient who has been discharged from a hospital is readmitted again within a certain period. Nationally the readmission rate for patients with pneumonia is \(17 \% .\) A hospital was interested in knowing whether their readmission rate for pneumonia was less than the national percentage. They found 11 patients out of 70 treated for pneumonia in a two-month period were readmitted. a. What is \(\hat{p}\), the sample proportion of readmission? b. Write the null and alternative hypotheses. c. Find the value of the test statistic and explain it in context. d. The p-value associated with this test statistic is \(0.39 .\) Explain the meaning of the \(\mathrm{p}\) -value in this context. Based on this result, does the \(\mathrm{p}\) -value indicate the null hypothesis should be doubted?

A 2003 study of dreaming published in the journal Perceptual and Motor Skills found that out of a random sample of 113 people, 92 reported dreaming in color. However, the proportion of people who reported dreaming in color that was established in the 1940 s was \(0.29\) (Schwitzgebel 2003). Check to see whether the conditions for using a one-proportion z-test are met assuming the researcher wanted to see whether the proportion dreaming in color had changed since the \(1940 \mathrm{~s}\).
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