/*! 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
a. Normal distribution is plausible. b. Interval: [21.67, 84.95]. c. Prediction interval: [19.14, 87.48], which is wider.

Step by step solution

01

Organize the Data

First, sort the given data in ascending order: 23, 39, 40, 41, 43, 47, 51, 58, 63, 66, 67, 69, 72. This will help in further calculations.
02

Check for Normality

Use a method like constructing a Q-Q plot or performing a Shapiro-Wilk test to check normality. For simplicity, consider visual inspection by plotting a histogram and a Q-Q plot. If the plot roughly follows a straight line, normality is plausible.
03

Calculate Mean and Standard Deviation

Compute the mean \[ \bar{x} = \frac{23+39+40+41+43+47+51+58+63+66+67+69+72}{13} \approx 53.31 \]and the standard deviation \[ s = \sqrt{\frac{1}{12} \sum (x_i - \bar{x})^2} \approx 15.82 \].
04

Calculate 95% Confidence Interval for Population Percentile

Use Chebyshev’s inequality for a non-parametric estimate: For at least 95% of the data, the interval is \[ \bar{x} \pm 2s = 53.31 \pm 2(15.82) = [21.67, 84.95] \].This interval does not assume normality.
05

Calculate 95% Prediction Interval

For a normal distribution, a prediction interval for a single observation is \[ \bar{x} \pm t_{\alpha/2, n-1} \cdot s \sqrt{1 + \frac{1}{n}} \],where \( t_{\alpha/2, n-1} \approx 2.18 \) for 95% CI and 12 degrees of freedom. This results in \[ 53.31 \pm 2.18 \times 15.82 \times \sqrt{1 + \frac{1}{13}} \approx [19.14, 87.48] \].
06

Compare Interval Widths

The prediction interval \([19.14, 87.48]\)is wider than the interval for the population percentile \([21.67, 84.95]\) because it accounts for both sample variation and individual predictive uncertainty.

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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 is a range of values derived from a data sample that is used to estimate an unknown population parameter. In simpler terms, it gives us an interval within which we can say with a certain level of confidence that the population parameter lies. The confidence level, often expressed as a percentage like 95%, tells us how certain we are that this interval contains the population parameter.
In our exercise, the mean adjusted distribution volume was calculated to be approximately 53.31, with a standard deviation of 15.82. If the data were assumed to be normally distributed, we would apply a formula specific to the normal distribution to find our confidence interval. However, since normality might not be assumed, Chebyshev’s inequality was used here.
This approach provides a range without assuming normality of the data, i.e., an interval from 21.67 to 84.95 that contains at least 95% of the data. It is useful when any distribution can be applicable, proving its flexibility and robustness in real-world data handling.
Prediction Interval
Prediction intervals are similar to confidence intervals, but instead of estimating a parameter, they provide a range of values for predicting future observations. In this context, the prediction interval gives us a range within which we expect an individual observation to fall, with a certain level of confidence.
In the exercise, the prediction interval was calculated to be between 19.14 and 87.48, which is wider than the confidence interval. This difference in width arises because prediction intervals account not only for the variability in estimating the sample mean (like a confidence interval) but also for the additional variability inherent in predicting a single point.
The interval is wider because it needs to encompass more uncertainty—beyond just the average—since it is predicting a future single observation rather than making an estimate about the average of a population. This property makes prediction intervals invaluable when anticipating the range of data points rather than just estimating a central tendency.
Chebyshev’s Inequality
Chebyshev’s Inequality is a fundamental theorem in probability that provides bounds for any distribution, regardless of its shape. It states that for any dataset, no matter how it is distributed, the proportion of values lying within a certain number of standard deviations from the mean is always a minimum specific value.
This inequality is particularly useful when the underlying distribution of the data is unknown or not normal. For example, using Chebyshev’s Inequality, at least 95% of the values lie within two standard deviations from the mean, regardless of how the data is distributed.
In our problem, we used this very feature to create a 95% confidence interval without necessarily assuming a normal distribution. This interval spans from 21.67 to 84.95, ensuring that most of the data falls within this range—a powerful tool when distribution assumptions cannot be safely made.
Shapiro-Wilk Test
The Shapiro-Wilk test is a statistical test used to assess if a sample comes from a normally distributed population. It is one of the most powerful tests for detecting normality, particularly effective with small sample sizes.
This test works by comparing the ordered sample with the expected value of the sample under the normal distribution. If the p-value obtained from the Shapiro-Wilk test is less than the chosen alpha level (often 0.05), we reject the null hypothesis, suggesting that the data does not follow a normal distribution.
For the presented exercise, performing a Shapiro-Wilk test would help determine if the assumption of a normal distribution is plausible. This can guide subsequent analysis, such as deciding if parametric methods that assume normality, like those for calculating prediction intervals, are appropriate. Its application gives a statistical footing to justify using or straying from normal-based assumptions in analyses.

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