In Exercises 5–16, test the given claim.
Second-Hand Smoke Data Set 12 “Passive and Active Smoke” includes cotinine levelsmeasured in a group of smokers (n = 40, \(\bar x\) = 172.48 ng/mL, s = 119.50 ng/mL) and a groupof nonsmokers not exposed to tobacco smoke (n = 40, \(\bar x\) = 16.35 ng/mL, s = 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.

b. The 40 cotinine measurements from the nonsmoking group consist of these values (all inng/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?

In a sample of 40 people who are smokers, the mean cotinine level is equal to 172.48 ng/mL, and the standard deviation of the cotinine level is equal to 119.50 ng/mL. In another sample of 40 people who are nonsmokers, the mean cotinine level is equal to 16.35 ng/mLand the standard deviation is equal to 62.53 ng/mL

It is claimed that the variation of the cotinine level in smokersis more than the variation of cotinine level in nonsmokers.

Let\({\sigma _1}\)and\({\sigma _2}\)be the populationstandard deviations of the cotinine level in the group of smokers and the group of nonsmokers, respectively.

Null Hypothesis:The populationstandard deviationof the cotinine level for the group of smokers is equal to the populationstandard deviationof the cotinine level for the group of nonsmokers.

Symbolically,

\({H_0}:{\sigma _1} = {\sigma _2}\)

Alternative Hypothesis:The populationstandard deviationof the cotinine level for the group of smokers is greater than the populationstandard deviationof the cotinine level for the group of nonsmokers.

Symbolically,

\({H_1}:{\sigma _1} > {\sigma _2}\)

Since two independent samples involve a claim about the population standard deviation, apply an F-test.

Consider the larger sample variance to be\(s_1^2\)and the corresponding sample size to be\({n_1}\).

The following values are obtained:

\({\left( {119.50} \right)^2} = 14280.25\)

\({\left( {62.53} \right)^2} = 3910.001\)

Here,\(s_1^2\)is the sample variance corresponding to the group of smokers and has a value equal to 14280.25.

\(s_2^2\)is the sample variance corresponding to the group of nonsmokersand has a value equal to 3910.001.

Substitute the respective values to calculate the F statistic:

\(\begin{array}{c}F = \frac{{s_1^2}}{{s_2^2}}\\ = \frac{{{{\left( {119.50} \right)}^2}}}{{{{\left( {62.53} \right)}^2}}}\\ = 3.652\end{array}\)

The value of the numerator degrees of freedom is equal to:

\(\begin{array}{c}{n_1} - 1 = 40 - 1\\ = 39\end{array}\)

The value of the denominator degrees of freedom is equal to:

\(\begin{array}{c}{n_2} - 1 = 40 - 1\\ = 39\end{array}\)

For the F test, the critical value corresponding to the right-tail is considered.

The critical value can be obtained using the F-distribution table with numerator degrees of freedom equal to 39 and denominator degrees of freedom equal to 39 for a right-tailed test.

The level of significance is equal to 0.05.

Thus, the critical value is equal to 1.7045.

The right-tailed p-value for F equal to 3.652 is equal to 0.0001.

a.

Since the test statistic value is greaterthan the critical value and the p-value is lessthan 0.05, the null hypothesis is rejected.

Thus, there is enough evidence to supportthe claimthat the variation of cotinine in smokers is greater than the variation of cotinine in nonsmokers not exposed to tobacco smoke.

b.

Follow the given steps to construct a normal quantile plot for the cotinine levels of the nonsmoker group.

Arrange the given values in ascending order as shown:

Compute the cumulative areas to the left for each sample value in the following manner:

The corresponding z-scores of the areas computed above are obtained as shown:

Now, plot the original data values on the x-axis and the corresponding z-scores on the y-axis. The final table containing the 40 observations is shown below:

• Mark the values 0, 50, 100, ……., 250 on the horizontal scale. Label the axis as “Sample Value”.
• Mark the values -2.500, -2.000, -1.500, …….., 2.500 on the vertical axis. Label the axis as “z-score”.
• Place a dot for the values of the z-scores corresponding to the sample values on the x-axis.

The following normal quantile plot is obtained:

It can be seen that the points on the plot do not follow a straight-line pattern.

Therefore, the given sample of cotinine levels for the nonsmoker group does not appear to come from a normally distributed population.

The F distribution (used for conducting the hypothesis test) is not robust against departures from normality.This means that if the population of the sample does not appear to be normally distributed, irrespective of the sample size, the test will not give accurate results and should not be used for concluding the claim.

Thus, the non-normality of the cotinine levels of the nonsmokerssignificantly affects the results regarding the claim.