91Ó°ÊÓ

Test the given claim. Data Set 12 "Passive and Active Smoke" includes cotinine levels measured in a group of smokers \((n=40, \bar{x}=172.48 \mathrm{ng} / \mathrm{mL}, s=119.50 \mathrm{ng} / \mathrm{mL})\) and a group of nonsmokers not exposed to tobacco smoke \((n=40, \bar{x}=16.35 \mathrm{ng} / \mathrm{mL}, s=62.53 \mathrm{ng} / \mathrm{mL})\) Cotinine is a metabolite of nicotine, meaning that when nicotine is absorbed by the body, cotinine is produced. a. Use a 0.05 significance level to test the claim that the variation of cotinine in smokers is greater than the variation of cotinine in nonsmokers not exposed to tobacco smoke. b. The 40 cotinine measurements from the nonsmoking group consist of these values (all in ng/mL): \(1,1,90,244,309,\) and 35 other values that are all \(0 .\) Does this sample appear to be from a normally distributed population? If not, how are the results from part (a) affected?

Step by step solution

01

State the Hypotheses

Identify the null and alternative hypotheses. The null hypothesis (H_0) is that the variance of cotinine levels in smokers (σ^2_smokers) is equal to the variance of cotinine levels in nonsmokers (σ^2_nonsmokers). The alternative hypothesis (H_1) is that the variance in smokers is greater than the variance in nonsmokers. Symbolically, H_0: σ^2_smokers = σ^2_nonsmokers or H_0: σ^2_smokers / σ^2_nonsmokers = 1 H_1: σ^2_smokers > σ^2_nonsmokers or H_1: σ^2_smokers / σ^2_nonsmokers > 1

02

Choose the Significance Level

The problem states to use a significance level of α = 0.05.

03

Compute the Test Statistic

To compare variances, use the F-test: F = s^2_smokers / s^2_nonsmokers Using the given data: s^2_smokers = (119.50)^2 = 14280.25 s^2_nonsmokers = (62.53)^2 = 3908.80 F = 14280.25 / 3908.80 ≈ 3.655

04

Determine the Critical Value

With α = 0.05 and df1 = 39 (for the numerator, smokers) and df2 = 39 (for the denominator, nonsmokers), use the F-distribution table to find the critical value F_crit. For α = 0.05, df1 = 39, df2 = 39: F_crit ≈ 1.86.

05

Compare and Make a Decision

Compare the calculated F-statistic with the critical value. F = 3.655 F_crit = 1.86 Since F > F_crit , reject the null hypothesis. Conclusion: There is evidence to support the claim that the variation of cotinine levels in smokers is greater than in nonsmokers.

06

Checking Normal Distribution in Nonsmokers

Examine the nonsmokers' data: [1, 1, 90, 244, 309,and 35 values of 0]. Most values are 0, with a few large values, suggesting a non-normal distribution. Since normality of the data is an assumption for the F-test, the results from part (a) might be affected if the assumption is violated.

07

Discuss the Effect of Non-Normality

Because the nonsmoker data does not appear to come from a normally distributed population, the F-test results may be unreliable. Variance tests like the F-test assume normality, and non-normal data can affect the validity of the test. Alternative non-parametric tests or transformation of data may be considered.

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Key Concepts

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

F-test

When we want to compare the variances of two populations, we use an F-test. The F-test evaluates if one population's variance is significantly greater than the other's. Variance measures how spread out the values are in a data set. In our case, we have cotinine levels for a group of smokers and nonsmokers. The F-test formula is simple: \[F = \frac{s^2_{\text{group 1}}}{s^2_{\text{group 2}}} \] where \(s^2\) represents the sample variance.
For this exercise, smokers are considered group 1 and nonsmokers are group 2. We can compute the sample variances and then the F value. If this computed F value is greater than the predetermined critical value from the F-distribution table, we reject the null hypothesis. This suggests that the variances are statistically different and, in our specific hypothesis, that the variance in smokers is greater.

Significance Level

The significance level, denoted as \( \alpha \), is the threshold we set to decide whether to reject the null hypothesis. It represents the probability of rejecting the null hypothesis when it is actually true. Commonly used significance levels are 0.05, 0.01, and 0.10. In this problem, we use a 0.05 significance level. This means there's a 5% risk of concluding that there is a difference in variance when there is none.
Setting \( \alpha = 0.05 \) helps balance between being too lenient and too strict in our test criteria. We then compare our calculated test statistic with the critical value corresponding to our chosen \( \alpha \) to make our decision.

Normal Distribution

Normal distribution is a key assumption in many statistical methods, including the F-test. It's a bell-shaped curve where most data points cluster around the mean, and the probabilities for values taper off symmetrically as they move further from the mean. This exercise's part (b) examines whether the nonsmokers' data appears to come from a normally distributed population.
Given the data [1, 1, 90, 244, 309, and 35 values of 0], it looks skewed rather than normally distributed. This can affect the validity of our F-test because the test assumes normality. If a dataset significantly deviates from normal distribution, the reliability of the F-test results declines. In such cases, alternative methods like non-parametric tests might be necessary.

Statistical Variance

Statistical variance measures the dispersion of data points in a data set relative to the mean. It is the average of the squared differences from the mean, providing insight into how ‘spread out’ the data is. The formula for variance in a sample is: \[ s^2 = \frac{\sum (x_i - \overline{x})^2}{n-1} \] where \( x_i \) are the data points, \( \overline{x} \) is the sample mean, and \( n \) is the number of data points.
In our exercise, we compare the variances of cotinine levels in smokers and nonsmokers. Higher variance indicates greater spread. For instance, the variance for smokers and nonsmokers came out to be quite different, contributing to the F-value calculation. This variance comparison helps determine if the spread of cotinine levels is significantly different between the two groups.