Question

Short answer

a.There is enough evidence tosupport the claim that nonsmokers exposed to tobacco smoke have a higher mean cotinine level than nonsmokers not exposed to tobacco smoke.

b. The 90% confidence interval is equal to (11.85 ng/mL, 92.61 ng/mL). Since the interval does not contain 0 and consists of only positive values, there is enough evidence tosupport the claim that nonsmokers exposed to tobacco smoke have a higher mean cotinine level than nonsmokers not exposed to tobacco smoke.

c. The effects of second-hand smoke are significant and serious.

Step-by-step solution

Given information

It is claimed that the mean cotinine level of nonsmokers exposed to tobacco smoke is higher than the mean cotinine level of nonsmokers not exposed to tobacco smoke.

The following values are obtained for the sample corresponding to the sample of nonsmokers exposed to tobacco smoke:

The sample size\(\left( {{n_1}} \right)\)is equal to 40.

The sample mean cotinine level\(\left( {{{\bar x}_1}} \right)\)is equal to 60.58 ng/mL

The sample standard deviation of the cotinine level\(\left( {{s_1}} \right)\)is equal to 138.08 ng/mL

The following values are obtained for the sample corresponding to the sample of nonsmokers not exposed to tobacco smoke:

The sample size\(\left( {{n_2}} \right)\)is equal to 40.

The sample mean cotinine level\(\left( {{{\bar x}_2}} \right)\)is equal to 16.35 ng/mL

The sample standard deviation of the cotinine level \(\left( {{s_2}} \right)\) is equal to 62.53 ng/mL

Hypotheses

Let \({\mu _1}\) be the population mean cotinine level of nonsmokers exposed to tobacco smoke.

Let \({\mu _2}\) be the population mean cotinine level of nonsmokers not exposed to tobacco smoke.

Null hypothesis: The population mean cotinine level of nonsmokers exposed to tobacco smoke is equal to the population mean cotinine level of nonsmokers not exposed to tobacco smoke.

\({H_{0\;}}:\;{\mu _1} = {\mu _2}\)

Alternative hypothesis: The population mean cotinine level of nonsmokers exposed to tobacco smoke is greater than the population mean cotinine level of nonsmokers not exposed to tobacco smoke.

\({H_1}\;:\;{\mu _1} > \;{\mu _2}\)

Test statistic

This is the problem of two independent samples t-test for the equality of means.

The test statistic value is computed below:

\(\begin{array}{c}t = \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }}\;\;\;\;\;\;\;\;{\rm{Under}}\;{H_0}\;\left( {{\mu _1} - {\mu _2}} \right) = 0\\ = \frac{{\left( {68.58 - 16.35} \right)}}{{\sqrt {\frac{{{{\left( {138.08} \right)}^2}}}{{40}} + \frac{{{{\left( {62.53} \right)}^2}}}{{40}}} }}\\ = 2.179\end{array}\)

Thus, t = 2.179.

Step 4:- Find the critical value and the p-value

The degrees of freedom is the smaller of the two values\(\left( {{n_1} - 1} \right)\)and\(\left( {{n_2} - 1} \right)\).

The values are computed below:

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

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

Thus, the value of the degrees of freedom is equal to 39.

The critical value can be obtained using the t distribution table with degrees of freedom equal to 39 and the significance level equal to 0.05 for a right-tailed test.

Thus, the critical value is equal to 1.6849.

The p-value for t equal to 2.179 is equal to 0.0177.

Conclusion of the test

a.

Since the test statistic value is greater than the critical value and the p-value is less than 0.05, the null hypothesis is rejected.

Thus, there is enough evidence to support the claim that that nonsmokers exposed to tobacco smoke have a higher mean cotinine level than nonsmokers not exposed to tobacco smoke.

Confidence Interval

b.

The formula for the confidence interval is given below:

\(\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - E < \left( {{\mu _1} - {\mu _2}} \right) < \left( {{{\bar x}_1} - {{\bar x}_2}} \right) + E\)

Here, \(\;E = {t_{\frac{\alpha }{2},df}} \times \sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \)

If the level of significance used in the one-tailed test is 0.05, the confidence level to construct the confidence interval is equal to 90%.

Thus, the corresponding significance level for the confidence interval \(\left( \alpha \right)\) is equal to 0.10.

E is the margin of error, and its value is computed below:

\(\begin{array}{c}E = {t_{\frac{\alpha }{2},df}} \times \sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \\ = {t_{0.05,39}} \times \sqrt {\frac{{{{138.08}^2}}}{{40}} + \frac{{{{62.53}^2}}}{{40}}} \\ = 1.6849 \times 23.96669\\ = 40.38147\end{array}\)

Substitute all the values in the formula, and find the confidence interval as follows:

\(\begin{array}{c}\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - E < \left( {{\mu _1} - {\mu _2}} \right) < \left( {{{\bar x}_1} - {{\bar x}_2}} \right) + E\\\left( {68.58 - 16.35} \right) - 40.38147 < \left( {{\mu _1} - {\mu _2}} \right) < \left( {68.58 - 16.35} \right) + 40.38147\\11.85 < \left( {{\mu _1} - {\mu _2}} \right) < 92.61\end{array}\)

Thus, the 90% confidence interval is equal to (11.85, 92.61).

Conclusion based on the confidence interval

Since the interval does not contain 0 and consists of only positive values, it can be said that the difference in the two mean values of the cotinine levels cannot be equal to 0.

Thus,there is enough evidence tosupport the claim that nonsmokers exposed to tobacco smoke have a higher mean cotinine level than nonsmokers not exposed to tobacco smoke.

Effect of second-hand smoking

c.

It can be seen that passive smokers also have considerable levels of cotinine (60.58 ng/mL) in their bodies, even if they do not actually smoke.

This implies that nicotine content in the bodies of passive smokers is quite high.

Thus, the effects of second-hand smoke are serious.