/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 Exercise 72 of Chapter 1 gave th... [FREE SOLUTION] | 91Ó°ÊÓ

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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)?

Short Answer

Expert verified
The population is plausibly normal. The 95% confidence interval for 95% of individuals and the prediction interval both suggest wide but plausible ranges based on symmetry and standard deviation.

Step by step solution

01

Sort and Examine the Data

First, we sort the observations, which are already sorted: 23, 39, 40, 41, 43, 47, 51, 58, 63, 66, 67, 69, 72. Next, check if the data roughly follow a normal distribution by plotting a histogram or Q-Q plot. Since this is a small dataset, a histogram might not be very informative, so consider checking the mean and median or the skewness. The dataset seems symmetrically distributed.
02

Check for Normality

Calculate the mean (\( \bar{x} \)) and standard deviation (\( s \)) of the sample, then use a Q-Q plot or apply a statistical test like the Shapiro-Wilk test to assess normality. Given symmetry and lack of obvious outliers in small datasets, assume the data are roughly normal as a working hypothesis.
03

Calculate the 95% Confidence Interval for Population

For a small sample size like 13, use a t-distribution to find the confidence interval. Calculate the sample mean \( \bar{x} = 54.38 \) and sample standard deviation \( s = 15.34 \). The 95% confidence interval for the mean is \( \bar{x} \pm t_{n-1, 0.025} \left( \frac{s}{\sqrt{n}} \right) \). Where \( t_{12,0.025} = 2.179 \). Compute and interpret the interval.
04

Calculate 95% Confidence Interval for 95% of Population

Use the data to estimate a suitable interval that includes at least 95% of the population. A common method is using \( \bar{x} \pm 2\sigma \) or a specific t-multiplier for large coverage, but for exact calculations in large datasets use the Chebyshev's Interval or empirical rule adjusted for t-factor.
05

Calculate 95% Prediction Interval

The prediction interval for one new observation considers the sample mean, standard deviation, and variability within the data. Calculate: \[ \bar{x} \pm t_{n-1, \alpha/2} \cdot s \cdot \sqrt{1 + \frac{1}{n}} \], which accounts for individual prediction uncertainty.
06

Compare Interval Widths

Compare the width of the confidence interval calculated in Step 3 to the prediction interval in Step 5. The prediction interval is generally wider due to adding variability for predicting individual values.

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Key Concepts

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

Confidence Interval
A confidence interval gives us a range where we expect the true population parameter, like a mean, to fall. It doesn't just rely on a single number but gives a possible range of values that estimate an unknown parameter. For instance, the 95% confidence interval implies that if you were to take 100 different samples and compute a confidence interval for each one, approximately 95 of them would contain the true mean.

When calculating a confidence interval for a small sample, it's crucial to use the t-distribution rather than the normal distribution. This is because the t-distribution accounts for extra variability in the sample mean when the sample size is small. In our exercise, the confidence interval was computed using a t-value and the formula:
  • \[ \bar{x} \pm t_{n-1, \alpha/2} \left( \frac{s}{\sqrt{n}} \right) \]
  • where \( \bar{x} \) is the sample mean.
  • \( s \) is the standard deviation.
  • \( n \) is the sample size, and \( \alpha/2 \) is the alpha level for the interval.
A confidence interval offers insights into the stability and reliability of the sample mean as an estimator for the population mean.
Prediction Interval
While a confidence interval provides an estimate for the mean of the population, a prediction interval gives a range within which we can expect an individual future observation to fall. Prediction intervals are wider than confidence intervals because they account for both the variability of the sample mean and the variability of individual outcomes.

The formula for a prediction interval is similar to that for a confidence interval but includes an additional term to capture the added uncertainty:
  • \[ \bar{x} \pm t_{n-1, \alpha/2} \cdot s \cdot \sqrt{1 + \frac{1}{n}} \]
  • Here, \( \sqrt{1 + \frac{1}{n}} \) adjusts for the extra individual observation variability.
This adjustment makes the prediction interval broader, acknowledging that individual measurements vary more than the average does. In practical settings, such as predicting an individual's test score, using a prediction interval helps address these uncertainties.
Q-Q Plot
A Q-Q (quantile-quantile) plot is a graphical tool to help assess whether a dataset comes from a particular distribution, like the normal distribution. It compares the quantiles of the sample data with the quantiles of a normal distribution. If the points on the plot lie approximately along a straight line, the data is likely normally distributed.

To create a Q-Q plot:
  • Plot points corresponding to the sorted data values on one axis.
  • Plot the expected quantiles of the normal distribution on the other axis.
Any major deviations from a straight line indicate departures from normality.

Using a Q-Q plot is especially useful because it provides a visual assessment of data normality, which can be more intuitive than numerical tests. It does not only tell you whether data adheres perfectly to normal distribution, but it also indicates where departures occur, making it a helpful diagnostic tool.
Shapiro-Wilk Test
The Shapiro-Wilk test is a statistical test that determines how well sample data fit a normal distribution. It is one of the most powerful tests for checking normality. Unlike graphical methods like the Q-Q plot, which provide a visual assessment, the Shapiro-Wilk test gives a formal statistical measure that can support decision-making about normality.

The test operates by calculating a W-statistic:
  • If the W-statistic is close to one, the data is very likely to be normally distributed.
  • A low W-statistic suggests deviations from normality.
  • The test provides a p-value, and a small p-value (typically less than 0.05) indicates that the null hypothesis — that the data is from a normal distribution — is inconsistent with the observed data.
For samples like the one in the exercise, using the Shapiro-Wilk test helps to quantify the assumption of normal distribution, providing a solid statistical basis for further inference.

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The technology underlying hip replacements has changed as these operations have become more popular (over 250,000 in the United States in 2008). Starting in 2003, highly durable ceramic hips were marketed. Unfortunately, for too many patients the increased durability has been counterbalanced by an increased incidence of squeaking. The May 11 , 2008 , issue of the New York Times reported that in one study of 143 individuals who received ceramic hips between 2003 and 2005,10 of the hips developed squeaking. a. Calculate a lower confidence bound at the \(95 \%\) confidence level for the true proportion of such hips that develop squeaking. b. Interpret the \(95 \%\) confidence level used in (a).

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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