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Chapter 7: Problem 56

Chronic exposure to asbestos fiber is a well-known health hazard. The article "The Acute Effects of Chrysotile Asbestos Exposure on Lung Function" (Erviron. Research, 1978: 360-372) reports results of a study based on a sample of construction workers who had been exposed to asbestos over a prolonged period. Among the data given in the article were the following (ordered) values of pulmonary compliance \(\left(\mathrm{cm}^{3} / \mathrm{cm} \mathrm{H}_{2} \mathrm{O}\right)\) for each of 16 subjects 8 months after the exposure period (pulmonary compliance is a measure of lung elasticity, or how effectively the lungs are able to inhale and exhale): \(\begin{array}{llllll}167.9 & 180.8 & 184.8 & 189.8 & 194.8 & 200.2 \\ 201.9 & 206.9 & 207.2 & 208.4 & 226.3 & 227.7 \\ 228.5 & 232.4 & 239.8 & 258.6 & & \end{array}\) a. Is it plausible that the population distribution is normal? b. Compute a \(95 \% \mathrm{CI}\) for the true average pulmonary compliance after such exposure. c. Calculate an interval that, with a confidence level of \(95 \%\), includes at least \(95 \%\) of the pulmonary compliance values in the population distribution.

Short Answer

Expert verified
a. Visually check normality using a Q-Q plot; it's plausible if linear. b. 95% CI: Calculate using sample mean, SD, and t-value. c. 95% prediction interval: Use mean, SD, and prediction formula.

Step by step solution

01

Check for Normality

To determine if the population distribution is normal, we can create a normal probability plot (Q-Q plot) using the given data. In a Q-Q plot, the data should approximately form a straight line if the distribution is normal. Alternatively, we could calculate skewness and kurtosis or perform a Shapiro-Wilk test, but for simplicity, we rely on a visual inspection of the Q-Q plot.
02

Compute Mean and Standard Deviation

First, calculate the mean of the data set: sum up all the measurements and divide by the number of observations (16 subjects).Mean \( \bar{x} = \frac{1}{16} (167.9 + 180.8 + 184.8 + \ldots + 258.6) \)Next, calculate the standard deviation using the formula:\[ s = \sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2} \]where \( x_i \) is each individual measurement, and \( n = 16 \).
03

Calculate the 95% Confidence Interval for the Mean

To find the 95% confidence interval for the true average pulmonary compliance, use the formula:\[ CI = \bar{x} \pm t_{\alpha/2, n-1} \left( \frac{s}{\sqrt{n}} \right) \]Where:- \( \bar{x} \) is the sample mean calculated in Step 2.- \( t_{\alpha/2, n-1} \) is the t-critical value for \( n-1 = 15 \) degrees of freedom from the t-table.- \( s \) is the sample standard deviation calculated in Step 2.- \( n = 16 \).
04

Calculate the 95% Prediction Interval

To calculate a prediction interval that includes at least 95% of pulmonary compliance values, use:\[ \text{Prediction Interval} = \bar{x} \pm t_{\alpha/2, n-1} \cdot s \sqrt{1 + \frac{1}{n}} \]This prediction interval accounts for individual points being predicted, unlike a confidence interval that estimates the population mean.

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Key Concepts

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

Normal Distribution
Understanding whether data follows a normal distribution is crucial in statistical analysis. In the case of pulmonary compliance values, to check for normality, a normal probability plot, also known as a Q-Q plot, can be used. If the plotted points form approximately a straight line, the data can be considered normally distributed. This method provides a visual confirmation without complicated statistics.

Besides graphical methods, skewness and kurtosis can numerically assess normality. Skewness measures asymmetry, while kurtosis indicates the "peakedness" of the distribution. However, graphical checks like Q-Q plots remain simple and effective, especially for small data sets as in this exercise with only 16 subjects.
Confidence Interval
A confidence interval (CI) provides a range within which we can be statistically certain the true population parameter lies. For the pulmonary compliance measurements after asbestos exposure, a 95% confidence interval can be calculated to estimate the true average lung compliance in the population.

To compute this interval, first determine the sample mean, which is the average of all given data. The formula to calculate a confidence interval is:
  • CI = \( \bar{x} \pm t_{\alpha/2, n-1} \left( \frac{s}{\sqrt{n}} \right) \)
Where \( \bar{x} \) is the sample mean, \( t_{\alpha/2, n-1} \) is the t-value for the desired confidence level, \( s \) is the sample standard deviation, and \( n \) is the number of observations. This CI tells us that we are 95% confident the true mean pulmonary compliance lies within the calculated range.
Prediction Interval
A prediction interval is different from a confidence interval in that it estimates the range where individual future observations will fall, rather than the mean of all observations. For the 95% prediction interval of our lung compliance data, we use almost the same steps as for the confidence interval, with an added component to account for variability in future measurements.

The formula is:
  • Prediction Interval = \( \bar{x} \pm t_{\alpha/2, n-1} \cdot s \sqrt{1 + \frac{1}{n}} \)
This helps us understand the spread and potential outliers in individual observations, making it an essential tool when predicting how any one subject might react based on past data.
Standard Deviation
The standard deviation is a measure of the amount of variation or dispersion in a set of values. For our set of pulmonary compliance measurements, calculating the standard deviation helps us understand the variability between subjects after asbestos exposure.

The formula is:
  • \( s = \sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2} \)
Where \( x_i \) represents each individual measurement, \( \bar{x} \) is the sample mean, and \( n \) is the sample size. A high standard deviation indicates that values are spread out over a larger range, while a low standard deviation implies they are clustered close to the mean. This insight is vital for assessing the consistency of lung function among the studied workers.

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Most popular questions from this chapter

The article "Limited Yield Estimation for Visual Defect Sources" (IEEE Trans. on Semiconductor Manuf., 1997: 17-23) reported that, in a study of a particular wafer inspection process, 356 dies were examined by an inspection probe and 201 of these passed the probe. Assuming a stable process, calculate a \(95 \%\) (two-sided) confidence interval for the proportion of all dies that pass the probe.

Exercise 72 of Chapter 1 gave the following observations on a receptor binding measure (adjusted distribution volume) for a sample of 13 healthy individuals: \(23,39,40,41,43\), \(47,51,58,63,66,67,69,72\). a. Is it plausible that the population distribution from which this sample was selected is normal? b. Calculate an interval for which you can be \(95 \%\) confident that at least \(95 \%\) of all healthy individuals in the population have adjusted distribution volumes lying between the limits of the interval. c. Predict the adjusted distribution volume of a single healthy individual by calculating a \(95 \%\) prediction interval. How does this interval's width compare to the width of the interval calculated in part (b)?

A triathlon consisting of swimming, cycling, and running is one of the more strenuous amateur sporting events. The article "Cardiovascular and Thermal Response of Triathlon Performance" (Medicine and Science in Sports and Exercise, 1988: 385-389) reports on a research study involving nine male triathletes. Maximum heart rate (beats/min) was recorded during performance of each of the three events. For swimming, the sample mean and sample standard deviation were \(188.0\) and \(7.2\), respectively. Assuming that the heart-rate distribution is (approximately) normal, construct a \(98 \%\) CI for true mean heart rate of triathletes while swimming.

In a sample of 1000 randomly selected consumers who had opportunities to send in a rebate claim form after purchasing a product, 250 of these people said they never did so ("Rebates: Get What You Deserve," Consumer Reports, May 2009: 7). Reasons cited for their behavior included too many steps in the process, amount too small, missed deadline, fear of being placed on a mailing list, lost receipt, and doubts about receiving the money. Calculate an upper confidence bound at the \(95 \%\) confidence level for the true proportion of such consumers who never apply for a rebate. Based on this bound, is there compelling evidence that the true proportion of such consumers is smaller than \(1 / 3\) ? Explain your reasoning.

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a continuous probability distribution having median \(\tilde{\mu}\) (so that \(\left.P\left(X_{l} \leq \tilde{\mu}\right)=P\left(X_{l} \geq \tilde{\mu}\right)=.5\right)\). a. Show that $$ P\left(\min \left(X_{i}\right)<\tilde{\mu}<\max \left(X_{j}\right)\right)=1-\left(\frac{1}{2}\right)^{n-1} $$ so that \(\left(\min \left(x_{i}\right), \max \left(x_{i}\right)\right)\) is a \(100(1-\alpha) \%\) confidence interval for \(\tilde{\mu}\) with \(\alpha=\left(\frac{1}{2}\right)^{n-1}\). [Hint: The complement of the event \(\left\\{\min \left(X_{j}\right)<\widetilde{\mu}<\max \left(X_{j}\right)\right\\}\) is \(\left\\{\max \left(X_{j}\right) \leq\right.\) \(\tilde{\mu}\\} \cup\left\\{\min \left(X_{i}\right) \geq \tilde{\mu}\right\\}\). But \(\max \left(X_{i}\right) \leq \tilde{\mu}\) iff \(X_{i} \leq \tilde{\mu}\) for all \(i .]\) b. For each of six normal male infants, the amount of the amino acid alanine \((\mathrm{mg} / 100 \mathrm{~mL}\) ) was determined while the infants were on an isoleucine-free diet, resulting in the following data: \(\begin{array}{llllll}2.84 & 3.54 & 2.80 & 1.44 & 2.94 & 2.70\end{array}\) Compute a \(97 \%\) CI for the true median amount of alanine for infants on such a diet ("The Essential Amino Acid Requirements of Infants," Amer. \(J\). of Nutrition, 1964: 322-330). c. Let \(x_{(2)}\) denote the second smallest of the \(x_{i}\) 's and \(x_{(1 n-1)}\) denote the second largest of the \(x_{i}\) 's. What is the confidence level of the interval \(\left(x_{(2)}, x_{(n-1)}\right)\) for \(\tilde{\mu}\) ?
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