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Incentives for Quitting Smoking: Do They Work? Exercise 5.31 describes a study examining incentives to quit smoking. With no incentives, the proportion of smokers trying to quit who are still abstaining six months later is about 0.06 . Participants in the study were randomly assigned to one of four different incentives, and the proportion successful was measured six months later. Of the 498 participants in the group with the least success, 47 were still abstaining from smoking six months later. We wish to test to see if this provides evidence that even the smallest incentive works better than the proportion of 0.06 with no incentive at all. (a) State the null and alternative hypotheses, and give the notation and value of the sample statistic. (b) Use a randomization distribution and the observed sample statistic to find the p-value. (c) Give the mean and standard error of the normal distribution that most closely matches the randomization distribution, and then use this normal distribution with the observed sample statistic to find the p-value. (d) Use the standard error found from the randomization distribution in part (b) to find the standardized test statistic, and then use that test statistic to find the p-value using a standard normal distribution. (e) Compare the p-values from parts (b), (c), and (d). Use any of these p-values to give the conclusion of the test.

Step by step solution

01

Formulation of Hypotheses

Firstly, we identify the null hypothesis (H0) and the alternative hypothesis (Ha). \n\nH0: Proportion of success = 0.06 i.e., incentives do not improve the success rate.\nHa: Proportion of success > 0.06 i.e., incentives do improve the success rate.

02

Notation and Value of Sample Statistic

Then, we calculate the success proportion from given value. It's \( p \), with: \n\n\( p = \frac{{number \, of \, successful \, attempts}}{{total \, number \, of \, attempts}} = \frac{47}{498} = 0.094 \)

03

P-Value Calculation from Sample Statistic

Using a randomization distribution or simulation, in an ideal situation, we can assess the number of scenarios where the outcome is equal to or more extreme than our observed sample statistic (p = 0.094). Ratio of these events to total number of events is the p-value. As it's hard to calculate the exact p-value in real-time, we generally use normal approximation techniques.

04

Calculation of Mean and Standard Error

The mean of the normal distribution is the value from null hypothesis \(p_0\), which is 0.06. The standard error (se) can be calculated from the formula \n\n\(se = \sqrt{ \frac{{p_0(1 - p_0)}}{n}} = \sqrt{\frac{{0.06*0.94}}{498}} = 0.011 \)

05

P-Value Calculation Using Normal Distribution

Then, we use normal distribution with observed sample statistic to calculate the p-value. The p-value is the probability of getting a value more extreme than observed statistic under null hypothesis. It is gotten by finding the z-score from the formula \n\n\( Z = \frac{{observed \, statistic - null \, value}}{se} = \frac{{0.094 - 0.06}}{0.011} = 3.09 \)\n\nUsing statistical tables or calculator, the one sided p-value corresponding to z-score of 3.09 is less than 0.001.

06

P-Value Comparison and Conclusion

Comparing the p-values from steps 3 and 5, both are very small (less than 0.05), thus we have strong evidence to reject the null hypothesis. So, we can conclude that the smallest incentive works better than no incentive at all.

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Key Concepts

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

Null Hypothesis

A null hypothesis is an initial claim that assumes no effect or no difference. It is denoted as \( H_0 \). In hypothesis testing stages, this acts as the standard or baseline that we test against. In our smoking incentive study, the null hypothesis states that the incentive does not improve the success rate beyond the standard no-incentive rate. Mathematically, it can be expressed as \( H_0: p = 0.06 \), where \( p \) is the true proportion of individuals still abstaining. Here, 0.06 represents the typical success rate observed in individuals without any smoking cessation incentives. In hypothesis testing, rejecting \( H_0 \) implies that the observed data is significantly different from what \( H_0 \) states under certain confidence levels.
Setting the null hypothesis is a critical step, as it provides the criteria for decision-making in the API (Alternative Hypothesis Interaction, more below). This ensures that conclusions are not made based merely on chance or one-offs.

Alternative Hypothesis

The alternative hypothesis, denoted by \( H_a \), challenges the null hypothesis by proposing that there is a real effect or difference. It serves as the hypothesis that researchers typically aim to support. In the smoking incentive exercise, the alternative hypothesis is that incentives increase the likelihood of success in quitting smoking. Formally, it is expressed as \( H_a: p > 0.06 \). This suggests that the incentive group achieves a greater proportion of success compared to the standard rate of 0.06, indicating a potential benefit provided by even the smallest incentives.
The alternative hypothesis positions itself as the foil to the null hypothesis, and testing results reveal whether there is enough evidence to support this claim. If the data shows a significant difference, we infer that the null hypothesis is incorrect, thus validating the alternative as plausible.

P-Value

A p-value is a fundamental concept used in hypothesis testing. It measures the probability of observing results as extreme as the observed data, assuming the null hypothesis is true. If this p-value is sufficiently low, typically less than a threshold of 0.05, it provides strong evidence against the null hypothesis and suggests that an alternative hypothesis might be more appropriate.
In our smoking cessation study, the p-value helps to decide whether the observed success rate with incentives (\( p = 0.094 \)) is statistically significant. By comparing observed values against a randomization distribution—an arrangement constructed from multiple unbiased scenario replications—we gauge the rarity of our outcome. Finding a p-value of less than 0.05 suggests the observed effect (due to incentives) is unlikely due to chance, thus favoring the alternative hypothesis over the null.

Randomization Distribution

A randomization distribution is a tool used to assess the probability of observing a test statistic as extreme as, or more extreme than, the observed one under the null hypothesis. In hypothesis testing, it serves as a visual and numerical reference to determine how data behaves under random assignment of observations.
In the context of the smoking incentive scenario, a randomization distribution would entail simulating what could happen if there were actually no effect due to incentives—all using extensive data mimicking the population studied. By assembling many random samples of what success rates might look like under the assumption that the null hypothesis is true, statisticians can discern how unusual the observed sample statistic really is.
Implementing the randomization distribution helps focus on whether observed data deviate enough from expectations under the null hypothesis, thereby informing the calculation and understanding of the p-value.