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Many people who quit smoking complain of weight gain. The results of an investigation of the relationship between smoking cessation and weight gain are given in the article "Does Smoking Cessation Lead to Weight Gain?" (American Journal of Public Health [1983]: \(1303-1305\) ). Three hundred twenty-two subjects, selected at random from those who successfully participated in a program to quit smoking, were weighed at the beginning of the program and again 1 year later. The mean change in weight was \(5.15 \mathrm{lb}\), and the standard deviation of the weight changes was \(11.45 \mathrm{lb}\). Is there sufficient evidence to conclude that the true mean change in weight is positive? Use \(\alpha=.05\).

Short Answer

Expert verified
Yes, there is sufficient evidence to conclude that the true mean change in weight is positive among former smokers.

Step by step solution

01

State the Hypotheses

The null hypothesis (\(H_0\)) is that the true mean weight change is not positive, i.e., \(H_0: \mu \leq 0\). The alternative hypothesis (\(H_a\)) is that the true mean weight change is positive, i.e., \(H_a: \mu > 0\). We are conducting a one-tailed t-test.
02

Compute the t score

The formula for the t score is \((\bar{X} - \mu)/(s/\sqrt{n})\), where \(\bar{X}\) is the sample mean (5.15 lb), \(\mu\) is the population mean (from \(H_0\), 0 lb), \(s\) is the sample standard deviation (11.45 lb), and \(n\) is the number of subjects (322). Plugging the values in, we get a t score of approximately 7.61.
03

Determine the Critical t score

Since alpha (\(\alpha\)) is 0.05 and the test is one-tailed, we need to find the t score that corresponds to the 95% of the data (100% - 5%). For the degrees of freedom (df = n-1 = 321), the critical t-score (t-critical) from a t-table is approximately 1.65.
04

Compare t score and t-critical

Our computed t score (7.61) is greater than the critical t score (1.65). Therefore, it falls in the rejection region.
05

Conclude the Test

Because our computed t score is in the rejection region, we reject the null hypothesis. This means we have sufficient evidence to conclude that the true mean weight change is positive. This suggests that the weight gain among former smokers is statistically significant.

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Key Concepts

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

t-test
A t-test is a statistical method used to determine if there is a significant difference between the means of two groups. In this exercise, we have a sample of people who quit smoking, and we wish to know if their average weight change is significantly positive. This is where the t-test comes in handy. It helps us to check if the observed mean weight change could have happened by chance. The formula for the t-test is: \[ t = \frac{\bar{X} - \mu}{s / \sqrt{n}} \]where:
  • \(\bar{X}\) is the sample mean,
  • \(\mu\) is the population mean under the null hypothesis,
  • \(s\) is the sample standard deviation,
  • \(n\) is the sample size.
This formula helps us to calculate the t-score, which can be compared to a critical value from the t-distribution to decide if the null hypothesis should be rejected.
statistical significance
Statistical significance plays a crucial role in hypothesis testing. It tells us whether our results from a study are likely to have occurred by chance. In this context, the **p-value** is used to determine the statistical significance. For a result to be statistically significant, the p-value must be less than the chosen significance level, \(\alpha\). In this exercise, \(\alpha\) is set at 0.05, which means there is a 5% risk of concluding that an effect exists when there is none. If the computed t-score leads to a p-value lower than 0.05, the results are considered statistically significant. This tells us that the observed mean weight change isn't likely due to random variations alone.
null hypothesis
The null hypothesis (\(H_0\)) is a statement used in hypothesis testing that assumes there is no effect or no difference. It serves as the default or starting assumption. In our exercise, the null hypothesis is that the true mean change in weight of people who quit smoking is not positive. Mathematically, this is represented as:\(H_0: \mu \leq 0\). We use statistical tests, like the t-test, to find evidence against it. Rejecting the null hypothesis suggests that there is enough evidence to support the idea that the true mean weight change is positive. If the null hypothesis is not rejected, the evidence is insufficient to suggest a difference from zero.
alternative hypothesis
An alternative hypothesis (\(H_a\)) is the statement that contradicts the null hypothesis. It proposes that there is an effect or a difference. In this context, it asserts that the true mean weight change is indeed positive after quitting smoking. Mathematically, it is:\(H_a: \mu > 0\).A successful hypothesis test will reject the null hypothesis, leading us to accept the alternative hypothesis as the more likely scenario.In the example from the exercise, by conducting our t-test and finding that the test statistic falls beyond the critical value, we conclude that there is significant statistical evidence in favor of the alternative hypothesis. This suggests that quitting smoking leads to a positive change in weight.

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Most popular questions from this chapter

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