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It is recognized that cigarette smoking has a deleterious effect on lung function. In a study of the effect of cigarette smoking on the carbon monoxide diffusing capacity (DL) of the lung, researchers found that current smokers had DL readings significantly lower than those of either ex smokers or nonsmokers. The carbon monoxide diffusing capacities for a random sample of \(n=20\) current smokers are listed here: $$ \begin{array}{rrrrr} 103.768 & 88.602 & 73.003 & 123.086 & 91.052 \\\ 92.295 & 61.675 & 90.677 & 84.023 & 76.014 \\ 100.615 & 88.017 & 71210 & 82.115 & 89.222 \\ 102.754 & 108.579 & 73.154 & 106.755 & 90.479 \end{array} $$ a. Do these data indicate that the mean DL reading for current smokers is significantly lower than 100 DL, the average for nonsmokers? Use \(\alpha=.01\). b. Find a \(99 \%\) upper one-sided confidence bound for the mean DL reading for current smokers. Does this bound confirm your conclusions in part a?

Step by step solution

01

Perform a one-sample t-test

Firstly, let's state the null and alternative hypotheses: \(H_0: \mu = 100\) \(H_1: \mu < 100\) where \(\mu\) is the mean DL reading for current smokers. Now, let's perform a one-sample t-test. We need to calculate the following values: sample mean (\(\bar{x}\)), sample standard deviation (s), and the t-score. In this case, the sample mean and sample standard deviation: \(\bar{x} = \frac{\sum x_i}{n} = \frac{1861.759}{20} = 93.088\) \(s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} = 13.791\) Now, let's calculate the t-score: \(t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} = \frac{93.088 - 100}{13.791 / \sqrt{20}} = -2.274\)

02

Determine critical t-value

At a significance level of 0.01, we can find the critical t-value from a t-distribution table or by using software/calculators. For our dataset with 19 degrees of freedom (\(n-1=20-1=19\)), the critical t-value would be: \(t_{critical} = -2.539\)

03

Compare t-score and critical t-value

Now that we have both our t-score (-2.274) and critical t-value (-2.539), we can compare them: Since our t-score (-2.274) is not less than the critical t-value (-2.539), we fail to reject the null hypothesis (\(H_0\)) at a significance level of 0.01. Therefore, we cannot conclude that the mean DL reading for current smokers is significantly lower than 100 DL.

04

Calculate a 99% one-sided confidence interval

Now, let's calculate the upper bound of the 99% one-sided confidence interval for the mean DL reading for current smokers: Upper bound \(= \bar{x} - t_{critical} * \frac{s}{\sqrt{n}} = 93.088 - (-2.539) * \frac{13.791}{\sqrt{20}} = 103.397\) So, the 99% one-sided confidence interval for the mean DL reading for current smokers is \((-\infty, 103.397]\).

05

Interpret the confidence interval

Since the confidence interval for the mean DL reading for current smokers includes the value 100 DL (average for non-smokers), it agrees with our conclusion from part a. We cannot conclude, at a 99% confidence level, that the mean DL reading for current smokers is significantly lower than 100 DL.

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Key Concepts

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

One-Sample t-Test

A one-sample t-test is a statistical method used to determine whether the mean of a single sample differs significantly from a known or hypothesized population mean. In this case, researchers wanted to test if the mean carbon monoxide diffusing capacity (DL) for current smokers was significantly lower than 100, the average for nonsmokers.

To perform a one-sample t-test, you need:

• The sample mean (\(\bar{x}\)), which is the average of all observations in the sample.
• The sample standard deviation (s), which measures how spread out the numbers are in your data set.
• The t-score, calculated using the formula: \[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \] where \( \mu_0 \) is the hypothesized population mean, and \( n \) is the number of observations.

In this problem, after calculating a t-score of -2.274, it was compared to a critical t-value of -2.539 (from t-distribution tables). Because the t-score was not less than the critical value, the hypothesis that the smokers' mean DL is below 100 was not supported.

Confidence Interval

A confidence interval provides a range of values within which we expect the true population parameter, such as the mean, to lie. Here, a 99% one-sided confidence interval was used to assess the mean DL for smokers. It provides an upper bound because we're checking if the smokers' mean DL is significantly less than a specific value.

The one-sided confidence interval is calculated using:
\[ \text{Upper Bound} = \bar{x} - t_{\text{critical}} \times \frac{s}{\sqrt{n}} \]
Plugging in the values results in an upper bound of 103.397.

This means we are 99% confident that the true mean for smokers is less than 103.397. Since this interval includes 100, it confirms that we do not have evidence to claim the smokers' mean DL is lower than 100, reinforcing the t-test results.

Statistical Significance

Statistical significance helps determine if the observed effects or differences are likely genuine, or occurred by random chance. In hypothesis testing, a significance level \( \alpha \) is used; commonly set at 0.01 or 0.05. Here, \( \alpha \) was 0.01, indicating the researchers are looking for strong evidence to reject the null hypothesis.

When the t-score is compared to a critical t-value, and the t-score is more extreme (in the direction of the test) than the critical value, the result is statistically significant.

In this problem, the t-score was not extreme enough (\(-2.274\) vs \(-2.539\)), which means there isn't enough statistical evidence to claim that the mean DL reading for smokers is significantly different from 100. Thus, the study concludes there is no statistically significant difference at this confidence level. This understanding is crucial in assessing the reliability and implications of any research findings.