Second-Hand Smoke Samples from Data Set 15 “Passive and Active...

Question

Second-Hand Smoke Samples from Data Set 15 “Passive and Active Smoke” include cotinine levels measured in a group of smokers ( n = 40, x_bar = 172.48 ng/mL, 119.50 ng/mL ) and a group of nonsmokers not exposed to tobacco smoke ( n = 40, x_bar = 16.35 ng/mL, 62.53 ng/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.

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.

Accepted Answer

Hello everyone. Let's take a look at this question together. Researchers are studying blood lead levels in two groups children living near a highway with a sample size of 35, a sample mean of 8.2 mcg per deciliter, and a standard deviation of 3.9 mcg per deciliter, as well as children living in rural areas with a sample size of 35, a sample mean of 2.7 mcg per deciliter, and a standard deviation of 1. 6 mcg per deciliter at the 0.05 significance level, test the claim that the variation in blood lead levels is greater for children living near the highway than for those in rural areas. So in order to solve this question, we must first do a requirement check where we know that the two populations are independent as the data collected from one group does not influence or relate to the data from the other group. As well as we assume that the two samples are simple random samples and that the researchers carefully plan the design for the study so that the selection process follows a random procedure, and lastly, we assume that each of the two populations is assumed to be normally distributed. And so now we can state our hypotheses where we let population. 1 be the population with the larger sample variants, and the population 2 be the population with the smaller sample variants. Thus, population 1, which is the population with the larger sample variants, is the children living near the highway, and population 2, which is the population with the smaller sample variants, is the children living in rural areas. So then our hypotheses include our null hypothesis, where the sample variance of population 1 is less than or equal to the sample variants of population 2 compared to our alternative hypothesis, where the sample variance of population 1 is greater than the sample variance of population 2. And then we can calculate the test statistic F using the formula for the test statistic, which is F is equal to the sample variance of population 1 divided by the sample variance of population 2, where we know that the sample variance of 1 is the variance of the group with the larger. Sample standard deviation and the sample variance of 2 is the variance of the group with the smaller sample standard deviation. And since 3.9 is greater than 1.6, the formula we use for the test statistic F is equal to 3.9 squared divided by 1.6 squared. Which is approximately 5.941, so our test statistic F is approximately 5.941. And then we determine our degrees of freedom. Where our first degree of freedom is N1 minus 1, so 35 minus 1, which equals 34. And our second degree of freedom is N2 minus 1, so 35 minus 1 again, which is equal to 34. And then we need to find the critical value using the table for the F distribution with an alpha equaling 0.05 in the right tail. However, since our first degree of freedom and second degree of freedom do not have a corresponding specific column and row in the table, We can simply take the possible range of the critical F value instead of interpolating twice, and based on our F table, we note that our degree of freedom 1, which equals 34, and our degree of freedom 2, which also equals 34, are both between 30 and 40. And we also consider the minimum and maximum F values between 30 and 40 inclusive. Which are 1.6928 and 1.8409 respectively. Therefore, we are certain that our F critical value is greater than 1.6928, but less than 1.8409. And so we are ready to make our decision, which is that we reject the null hypothesis if our FS statistic is greater than our F critical value. And since 5.941 is greater than 1.8409, which is greater than 1.6928, we reject the null hypothesis. Therefore, at the 0.05 significance level, there is sufficient evidence to support the claim that the variation in blood lead levels is greater for children living near the highway than those in rural areas. So answer choice D is the correct answer, and all other answer choices are incorrect. I hope you found this video to be helpful. Thank you and goodbye.