91Ó°ÊÓ

Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. (Note: Answers in Appendix D include technology answers based on Formula 9 - 1 along with "Table" answers based on Table \(A\) - 3 with df equal to the smaller of \(\boldsymbol{n}_{I}-\boldsymbol{I}\) and \(\boldsymbol{n}_{2}-\boldsymbol{I} .\) ) Second-Hand Smoke Data Set 12 "Passive and Active Smoke" in Appendix B includes cotinine levels measured in a group of nonsmokers exposed to tobacco smoke \((n=40,\) \(\bar{x}=60.58 \mathrm{ng} / \mathrm{mL}, s=138.08 \mathrm{ng} / \mathrm{mL})\) 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 nonsmokers exposed to tobacco smoke have a higher mean cotinine level than nonsmokers not exposed to tobacco smoke. b. Construct the confidence interval appropriate for the hypothesis test in part (a). c. What do you conclude about the effects of second-hand smoke?

Step by step solution

01

State the hypotheses

Let \(\bar{x}_1\) and \(\bar{x}_2\) be the sample means for nonsmokers exposed to tobacco smoke and nonsmokers not exposed, respectively. Test the claim \(H_0: \mu_1 \leq \mu_2\) against the alternative \(H_a: \mu_1 > \mu_2\).

02

Determine the test statistic

Since the population standard deviations are not assumed equal, we use the t-test for unequal variances. The test statistic is given by: \[ t = \frac{(\bar{x}_1 - \bar{x}_2)}{\text{sqrt} \bigg( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \bigg)} \] where \( \bar{x}_1 = 60.58 \), \( s_1 = 138.08 \), \( n_1 = 40 \), \( \bar{x}_2 = 16.35 \), \( s_2 = 62.53 \), and \( n_2 = 40 \).

03

Calculate the degrees of freedom

The degrees of freedom (df) are given by the formula: \[ df = \frac{\bigg( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \bigg)^2}{\frac{\bigg(\frac{s_1^2}{n_1}\bigg)^2}{n_1 - 1} + \frac{\bigg(\frac{s_2^2}{n_2}\bigg)^2}{n_2 - 1}} \]. Calculate using the provided values.

04

Find the critical value and make a decision

For a one-tailed test at \( \alpha = 0.05 \) and the calculated df, find the critical t-value from the t-distribution table. Compare the test statistic to the critical value.

05

Construct the confidence interval

The confidence interval for the difference in means is given by: \[ (\bar{x}_1 - \bar{x}_2) \pm t_{crit} \times \text{sqrt}\bigg(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\bigg) \]. Use the t-value from the t-distribution table for 95% confidence level.

06

Conclusion about the effects of second-hand smoke

Based on the hypothesis test results and the confidence interval, conclude whether the mean cotinine level for nonsmokers exposed to tobacco smoke is significantly higher than for nonsmokers not exposed.

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Key Concepts

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

t-test for unequal variances

In hypothesis testing, a t-test checks if the means of two groups are significantly different from each other. When we do not assume equal variances for the two populations, we use the t-test for unequal variances, also known as Welch's t-test. This is particularly useful when you have two independent samples, like in our given problem with cotinine levels.
The test statistic for the t-test for unequal variances is calculated by:
\[ t = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{ \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }} \]
Here:

• \( \bar{x}_1 \) and \( \bar{x}_2 \) are the sample means
• \( s_1 \) and \( s_2 \) are the sample standard deviations
• \( n_1 \) and \( n_2 \) are the sample sizes

By applying this t-test, we can decide if the difference in means could have happened by random chance or if it’s statistically significant.

significance level

The significance level, denoted by \( \alpha \), is the threshold we use to determine whether a result is statistically significant. In many studies, a common choice for \( \alpha \) is 0.05, which means there is a 5% chance of concluding that a difference exists when there is no actual difference.
In our exercise, we use a 0.05 significance level to test the hypothesis that nonsmokers exposed to tobacco smoke have higher cotinine levels than those not exposed. If the p-value calculated from our t-test is less than 0.05, we reject the null hypothesis and accept the alternative hypothesis.

confidence interval

A confidence interval gives a range of values within which we expect the true population parameter to lie, given a certain level of confidence (commonly 95%). For our study on cotinine levels, the confidence interval helps us understand the difference in means between the two groups.
To construct the confidence interval for the difference in means, we calculate:
\[ (\bar{x}_1 - \bar{x}_2) \pm t_{crit} \times \sqrt{ \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} } \]
Here, \( t_{crit} \) is the critical value from the t-distribution table for our confidence level, and the other terms are the same as defined previously. This interval helps us determine if the mean difference is statistically significant and not just due to random variability.

degrees of freedom

The degrees of freedom (df) are crucial in determining the critical value from the t-distribution table. In the context of the t-test for unequal variances, the formula for calculating df is given by:
\[ df = \frac{ \bigg( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \bigg)^2 }{ \frac{ \bigg( \frac{s_1^2}{n_1} \bigg)^2 }{n_1 - 1} + \frac{ \bigg( \frac{s_2^2}{n_2} \bigg)^2 }{n_2 - 1} } \]
This formula accounts for the variability in both samples and adjusts our critical t-value accordingly. Accurately calculating the degrees of freedom ensures that our t-test results are valid and reliable.

second-hand smoke effects

Second-hand smoke, also known as passive smoke, can have significant health effects on nonsmokers. In our study, we measured cotinine levels, a metabolite of nicotine, to compare the exposure of nonsmokers to tobacco smoke.
By analyzing the difference in mean cotinine levels between nonsmokers exposed to second-hand smoke and those who are not, we can make conclusions about the health impact. Higher cotinine levels in the exposed group indicate that second-hand smoke can significantly increase a nonsmoker's nicotine intake, leading to potential health risks.
This type of analysis can help in forming public health policies aimed at protecting nonsmokers from the harmful effects of second-hand smoke.