/*! 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 37 How to quit smoking It's hard fo... [FREE SOLUTION] | 91Ó°ÊÓ

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How to quit smoking It's hard for smokers to quit. Perhaps prescribing a drug to fight depression will work as well as the usual nicotine patch. Perhaps combining the patch and the drug will work better than either treatment alone. Here are data from a randomized, double-blind trial that compared four treatments. \({ }^{19} \mathrm{~A}\) "success" means that the subject did not smoke for a year following the beginning of the study. $$ \begin{array}{llcc} \hline \text { Group } & \text { Treatment } & \text { Subjects } & \text { Successes } \\ 1 & \text { Nicotine patch } & 244 & 40 \\ 2 & \text { Drug } & 244 & 74 \\ 3 & \text { Patch plus drug } & 245 & 87 \\ 4 & \text { Placebo } & 160 & 25 \\ \hline \end{array} $$ (a) Summarize these data in a two-way table. Then compare the success rates for the four treatments. (b) Explain in words what the null hypothesis \(H_{0}: p_{1}=\) \(p_{2}=p_{3}=p_{4}\) says about subjects' smoking habits. (c) Do the data provide convincing evidence of a difference in the effectiveness of the four treatments at the \(\alpha=0.05\) significance level?

Short Answer

Expert verified
Group 3 has the highest success rate, and the data suggest significant differences among treatments.

Step by step solution

01

Summarize Data in a Two-Way Table

First, we create a two-way table showing the number of successes and failures for each treatment group. This will help us compare the effectiveness of different treatments. The summary table will look like this:\[\begin{array}{lccc}\hline \text{Group} & \text{Subjects} & \text{Successes} & \text{Failures} \\hline1 & 244 & 40 & 204 \2 & 244 & 74 & 170 \3 & 245 & 87 & 158 \4 & 160 & 25 & 135 \\hline\end{array}\]
02

Calculate Success Rates

For each treatment group, calculate the success rate as the ratio of successes to the number of subjects. The success rates for the groups are:- Group 1 (Nicotine patch): \( \frac{40}{244} \approx 0.164 \, (16.4\%) \)- Group 2 (Drug): \( \frac{74}{244} \approx 0.303 \, (30.3\%) \)- Group 3 (Patch plus drug): \( \frac{87}{245} \approx 0.355 \, (35.5\%) \)- Group 4 (Placebo): \( \frac{25}{160} \approx 0.156 \, (15.6\%) \)
03

Interpret Null Hypothesis

The null hypothesis \(H_0 : p_1 = p_2 = p_3 = p_4\) states that there is no difference in the success rates of not smoking among the four groups. In other words, all treatments (including placebo) are equally effective at helping subjects quit smoking.
04

Conduct Chi-square Test for Independence

To determine if there are significant differences in success rates among the groups, use a chi-square test for independence. Calculate the chi-square statistic based on the two-way table and compare it to the critical value for the \( \alpha = 0.05 \) significance level. Without doing manual calculations here, you'd look up or compute the smallest critical chi-square value for \( (4-1)(2-1) = 3 \) degrees of freedom, which is 7.815. If the computed chi-square statistic is greater than 7.815, reject the null hypothesis, indicating there is a significant difference among the treatments.
05

Conclusion Based on Testing

Assuming the chi-square test indicated a significant difference (since the solution asks about convincing evidence at \(\alpha = 0.05\)), we conclude that at least one treatment is significantly different in effectiveness from the others. In practical terms, the combination of patch and drug seems to yield the highest success rate at 35.5% compared to others.

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Key Concepts

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

Randomized Controlled Trial
A randomized controlled trial (RCT) is a type of scientific experiment that is designed to reduce bias when testing the effectiveness of new treatments or interventions. In the context of quitting smoking, the trial includes four different treatment groups: a nicotine patch, a drug, a combination of both, and a placebo. Each participant is randomly assigned to one of these groups. This random assignment helps ensure that the groups are comparable in all important factors, thus allowing any differences in outcomes to be attributed to the treatment itself, rather than external factors.
A key feature of RCTs is the "double-blind" aspect, where neither the participants nor the researchers know which treatment the participants are receiving during the trial. This further prevents bias in the results, as neither expectations of the patients nor the care provided by researchers are influenced by knowledge of the treatment. By utilizing these methods, RCTs aim to provide the strongest possible evidence.
  • Random assignment to groups reduces selection bias.
  • Placebo groups help measure treatment effectiveness against no treatment.
  • Double-blinding enhances objectivity in results.
Null Hypothesis
The null hypothesis is a foundational concept in statistical testing. It represents a statement of no effect or no difference. In the case of the smoking cessation trial, the null hypothesis (\( H_0 : p_1 = p_2 = p_3 = p_4 \)) asserts that there is no difference in the success rates across the four treatment groups. This means that, theoretically, the nicotine patch, the drug, their combination, and the placebo all have the same probability of helping a person quit smoking.
The null hypothesis acts as a starting assumption that researchers aim to test. When conducting experiments, researchers typically look for evidence to reject the null hypothesis in favor of an alternative hypothesis, which suggests there is a difference worth noting.
  • No effect or difference is the default assumption.
  • Testing aims to provide evidence against the null hypothesis.
  • Rejection of the null hypothesis indicates a significant finding.
Chi-Square Test
The chi-square test is a statistical method used to determine if there is a significant association between categorical variables. In this experiment, it examines whether observed differences in smoking cessation success rates between treatment groups are statistically significant.
To conduct a chi-square test, you would compare the observed frequencies (the actual number of successes and failures) with the expected frequencies that you would expect if the null hypothesis were true. The chi-square statistic is calculated using the formula:\[\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\]where \( O_i \) is the observed frequency and \( E_i \) is the expected frequency. If this statistic exceeds the critical value from the chi-square distribution table, particularly at the chosen significance level (like 0.05), the null hypothesis is rejected.
  • Evaluates association between variables.
  • Compares observed and expected frequencies.
  • Critical value helps decide hypothesis rejection.
Success Rate Calculation
Success rate calculation is a vital step when analyzing the effectiveness of treatments. It's calculated as the proportion of "successes" (in this case, individuals who did not smoke for a year) to the total number of subjects in each treatment group. This provides a percentage that reflects the efficacy of each treatment.
For example:
  • Group 1 (Nicotine patch): \( \frac{40}{244} \approx 0.164 \), or 16.4%
  • Group 2 (Drug): \( \frac{74}{244} \approx 0.303 \), or 30.3%
  • Group 3 (Patch plus drug): \( \frac{87}{245} \approx 0.355 \), or 35.5%
  • Group 4 (Placebo): \( \frac{25}{160} \approx 0.156 \), or 15.6%
These calculations help us easily compare which treatment is the most effective in helping participants quit smoking. Observing these percentages, you can see that the combination of patch and drug yields the highest success, offering insights into treatment efficiency.

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Most popular questions from this chapter

Students and catalog shopping What is the most important reason that students buy from catalogs? The answer may differ for different groups of students. Here are results for separate random samples of American and Asian students at a large midwestern university: \(^{26}\) $$ \begin{array}{lcc} \hline & \text { American } & \text { Asian } \\ \text { Save time } & 29 & 10 \\ \text { Easy } & 28 & 11 \\ \text { Low price } & 17 & 34 \\ \text { Live far from stores } & 11 & 4 \\ \text { No pressure to buy } & 10 & 3 \\ \hline \end{array} $$ (a) Should we use a chi-square test for homogeneity or a chi-square test for independence in this setting? Justify your answer. (b) State appropriate hypotheses for performing the type of test you chose in part (a). (c) Check that the conditions for carrying out the test are met. (d) Interpret the \(P\) -value in context. What conclusion would you draw?

Roulette Casinos are required to verify that their games operate as advertised. American roulette wheels have 38 slots -18 red, 18 black, and 2 green. In one casino, managers record data from a random sample of 200 spins of one of their American roulette wheels. The one-way table below displays the results. $$ \begin{array}{lccc} \hline \text { Color: } & \text { Red } & \text { Black } & \text { Green } \\\ \text { Count: } & 85 & 99 & 16 \\ \hline \end{array} $$ (a) State appropriate hypotheses for testing whether these data give convincing evidence that the distribution of outcomes on this wheel is not what it should be. (b) Calculate the expected counts for each color. Show your work.

Sorry, no chi-square How do U.S. residents who travel overseas for leisure differ from those who travel for business? The following is the breakdown by occupation: \({ }^{17}\) $$ \begin{array}{lcc} \hline \text { 0ccupation } & \text { Leisure } & \text { Business } \\ \text { Professional/technical } & \text { travelers (\%) } & \text { travelers (\%) } \\ \text { Manager/executive } & 36 & 39 \\ \text { Retired } & 23 & 48 \\ \text { Student } & 14 & 3 \\ \text { 0ther } & 7 & 3 \\ \text { Total } & 20 & 7 \\ \hline \end{array} $$ Explain why we can't use a chi-square test to learn whether these two distributions differ significantly.

Going Nuts The UR Nuts Company sells Deluxe and Premium nut mixes, both of which contain only cashews, brazil nuts, almonds, and peanuts. The Premium nuts are much more expensive than the Deluxe nuts. A consumer group suspects that the two nut mixes are really the same. To find out, the group took separate random samples of 20 pounds of each nut mix and recorded the weights of each type of nut in the sample. Here are the data: \({ }^{18}\) $$ \begin{array}{lcc} {\text { Type of mix }} \\ \text { Type of nut } & \text { Premium } & \text { Deluxe } \\ \text { Cashew } & 6 \mathrm{lb} & 5 \mathrm{lb} \\ \text { Brazil nut } & 3 \mathrm{lb} & 4 \mathrm{lb} \\ \text { Almond } & 5 \mathrm{lb} & 6 \mathrm{lb} \\ \text { Peanut } & 6 \mathrm{lb} & 5 \mathrm{lb} \end{array} $$ Explain why we can't use a chi-square test to determine whether these two distributions differ significantly.

When analyzing survey results from a two-way table, the main distinction between a test for independence and a test for homogeneity is (a) how the degrees of freedom are calculated. (b) how the expected counts are calculated. (c) the number of samples obtained. (d) the number of rows in the two-way table. (e) the number of columns in the two-way table.

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