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Chapter 9: Problem 5

Assume that the sample \(\left(x_{1}, Y_{1}\right), \ldots,\left(x_{n}, Y_{n}\right)\) follows the linear model \((9.6 .1)\). Suppose \(Y_{0}\) is a future observation at \(x=x_{0}-\bar{x}\) and we want to determine a predictive interval for it. Assume that the model \((9.6 .1)\) holds for \(Y_{0}\); i.e., \(Y_{0}\) has a \(N\left(\alpha+\beta\left(x_{0}-\bar{x}\right), \sigma^{2}\right)\) distribution. We will use \(\hat{\eta}_{0}\) of Exercise \(9.6 .4\) as our prediction of \(Y_{0}\) (a) Obtain the distribution of \(Y_{0}-\hat{\eta}_{0}\). Use the fact that the future observation \(Y_{0}\) is independent of the sample \(\left(x_{1}, Y_{1}\right), \ldots,\left(x_{n}, Y_{n}\right)\) (b) Determine a \(t\) -statistic with numerator \(Y_{0}-\hat{\eta}_{0}\). (c) Now beginning with \(1-\alpha=P\left[-t_{\alpha / 2, n-2}

Short Answer

Expert verified
The predictive interval for \( Y_{0} \) is given by \( \hat{\eta}_{0}\pm t_{\alpha/2, n-2}s\sqrt{1+\frac{1}{n}+\frac{(x_{0}-\bar{x})^{2}}{\sum(x_{i}-\bar{x})^{2}}} \). This interval is generally larger than a confidence interval because it considers the variance of the new observation.

Step by step solution

01

Determine the distribution of \(Y_{0}-\hat{\eta}_{0}\)

First, note that the expected value of \(Y_{0}-\hat{\eta}_{0}\) is 0 and the variance is \(\sigma^{2}\left(1+\frac{1}{n}+\frac{\left(x_{0}-\bar{x}\right)^{2}}{\sum\left(x_{i}-\bar{x}\right)^{2}}\right)\). Hence, \(Y_{0}-\hat{\eta}_{0}\) follows a normal distribution with mean 0 and the given variance.
02

Determine the t-statistic

The t-statistic with numerator \(Y_{0}-\hat{\eta}_{0}\) is given by \(t=\frac{Y_{0}-\hat{\eta}_{0}}{s\sqrt{1+\frac{1}{n}+\frac{(x_{0}-\bar{x})^{2}}{\sum(x_{i}-\bar{x})^{2}}}}\) where \(s\) is the sample standard deviation.
03

Determine the predictive interval

A \( (1-\alpha)100\% \) predictive interval for \( Y_{0} \) is given by \( \hat{\eta}_{0}\pm t_{\alpha/2, n-2}s\sqrt{1+\frac{1}{n}+\frac{(x_{0}-\bar{x})^{2}}{\sum(x_{i}-\bar{x})^{2}}} \)
04

Compare with the confidence interval

A predictive interval is generally larger than a confidence interval as it takes into account the variance of the new observation, while the confidence interval only considers the variance of the estimated mean.

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Key Concepts

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

Linear Regression
Linear Regression is a statistical method used to model the relationship between a dependent variable and one or more independent variables. The goal is to find a linear equation, typically of the form \( Y = \alpha + \beta X + \varepsilon \) that best predicts the dependent variable \( Y \) from the independent variable(s) \( X \). In this expression, \( \alpha \) represents the y-intercept, \( \beta \) denotes the slope of the line, and \( \varepsilon \) is the error term that accounts for randomness in the data.

In the context of the exercise, a linear model is assumed where a future observation \( Y_0 \) depends linearly on a known value \( x_0 \) and the sample mean \( \bar{x} \). Predicting \( Y_0 \) involves estimating the \( \alpha \) and \( \beta \) parameters based on sample data and applying them to the known \( x_0 \). The solution involves calculating the predictive interval to gauge how much the prediction of \( Y_0 \) might vary from the actual value, which, unlike a simple point estimate, provides a range within which the future observation is likely to fall.
t-Statistic
The t-Statistic is a type of standardized statistic used when conducting hypothesis tests in statistics, especially when the population standard deviation is not known, and the sample size is small. It is calculated by dividing the difference between the sample mean and the population mean by the standard error of the sample mean. It follows a t-distribution with \( n-1 \) degrees of freedom, where \( n \) is the sample size.

In our exercise, a t-statistic is used to help create a predictive interval for \( Y_0 \). The t-statistic is determined with the numerator \( Y_0 - \hat{\eta}_0 \) and the denominator incorporating the sample standard deviation and the square root of 1 plus additional terms considering the difference between \( x_0 \) and \( \bar{x} \) as well as the sum of squared differences from the sample. The methodology accounts for both the uncertainty in predicting \( Y_0 \) and the degree of freedom associated with the sample data.
Normal Distribution
The Normal Distribution, also known as the Gaussian distribution, is a continuous probability distribution that is symmetrical around its mean, showing that data near the mean are more frequent in occurrence than data far from the mean. The shape of the normal distribution is determined by the mean and the standard deviation. The density of the normal curve falls off rapidly on either side of the mean, resulting in the familiar bell-shaped curve.

In the case of the given exercise, it is assumed that the future observation \( Y_0 \) has a normal distribution centered around \( \alpha + \beta(x_0 - \bar{x}) \) with variance \( \sigma^2 \). The solution uses this assumption to determine the distribution of \( Y_0 - \hat{\eta}_0 \) by noting that it follows a normal distribution with its expected value being 0 and a specified variance. This normality plays a crucial role in calculating the predictive interval and utilizing the t-distribution for finding the t-statistic.
Confidence Interval
A Confidence Interval (CI) is a range of values, derived from statistical analysis, that is likely to contain the value of an unknown population parameter. The confidence level represents the frequency of possible confidence intervals that contain the actual value of the parameter. For example, a 95% confidence level indicates that 95 out of 100 samples will contain the actual mean.

In our exercise scenario, a CI is used to provide an interval estimate around an estimate of the mean value of the dependent variable for a given value of the independent variable. The predictive interval, in contrast, is broader as it accounts for the uncertainty in the predicted value itself. This uncertainty is why the predictive interval tends to be larger than the confidence interval and is useful when estimating the range where an actual future observation is expected to fall. This is critical for providing realistic expectations for predictions and accommodating the variability inherent in individual future observations.

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Most popular questions from this chapter

Let \(X_{1 j}, X_{2 j}, \ldots, X_{a_{f} j}\) represent independent random samples of sizes \(a_{j}\) from a normal distribution with means \(\mu_{j}\) and variances \(\sigma^{2}, j=1,2, \ldots, b\). Show that $$ \sum_{j=1}^{b} \sum_{i=1}^{a_{j}}\left(X_{i j}-\bar{X}_{. .}\right)^{2}=\sum_{j=1}^{b} \sum_{i=1}^{a_{j}}\left(X_{i j}-\bar{X}_{. j}\right)^{2}+\sum_{j=1}^{b} a_{j}\left(\bar{X}_{. j}-\bar{X}_{. .}\right)^{2} $$ or \(Q^{\prime}=Q_{3}^{\prime}+Q_{4}^{\prime} .\) Here \(\bar{X}_{. .}=\sum_{j=1}^{b} \sum_{i=1}^{a_{j}} X_{i j} / \sum_{j=1}^{b} a_{j}\) and \(\bar{X}_{. j}=\sum_{i=1}^{a_{j}} X_{i j} / a_{j} .\) If \(\mu_{1}=\mu_{2}=\) \(\cdots=\mu_{b}\), show that \(Q^{\prime} / \sigma^{2}\) and \(Q_{3}^{\prime} / \sigma^{2}\) have chi-square distributions. Prove that \(Q_{3}^{\prime}\) and \(Q_{4}^{\prime}\) are independent, and hence \(Q_{4}^{\prime} / \sigma^{2}\) also has a chi-square distribution. If the likelihood ratio \(\Lambda\) is used to test \(H_{0}: \mu_{1}=\mu_{2}=\cdots=\mu_{b}=\mu, \mu\) unspecified and \(\sigma^{2}\) unknown against all possible alternatives, show that \(\Lambda \leq \lambda_{0}\) is equivalent to the computed \(F \geq c\), where $$ F=\frac{\left(\sum_{j=1}^{b} a_{j}-b\right) Q_{4}^{\prime}}{(b-1) Q_{3}^{\prime}} $$ What is the distribution of \(F\) when \(H_{0}\) is true?

Let \(\mu_{1}, \mu_{2}, \mu_{3}\) be, respectively, the means of three normal distributions with a common but unknown variance \(\sigma^{2}\). In order to test, at the \(\alpha=5\) percent significance level, the hypothesis \(H_{0}: \mu_{1}=\mu_{2}=\mu_{3}\) against all possible alternative hypotheses, we take an independent random sample of size 4 from each of these distributions. Determine whether we accept or reject \(H_{0}\) if the observed values from these three distributions are, respectively, $$ \begin{array}{lrrrr} X_{1}: & 5 & 9 & 6 & 8 \\ X_{2}: & 11 & 13 & 10 & 12 \\ X_{3}: & 10 & 6 & 9 & 9 \end{array} $$

Let \(\mathbf{A}=\left[a_{i j}\right]\) be a real symmetric matrix. Prove that \(\sum_{i} \sum_{j} a_{i j}^{2}\) is equal to the sum of the squares of the eigenvalues of \(\mathbf{A}\). Hint: If \(\boldsymbol{\Gamma}\) is an orthogonal matrix, show that \(\sum_{j} \sum_{i} a_{i j}^{2}=\operatorname{tr}\left(\mathbf{A}^{2}\right)=\operatorname{tr}\left(\mathbf{\Gamma}^{\prime} \mathbf{A}^{2} \mathbf{\Gamma}\right)=\) \(\operatorname{tr}\left[\left(\mathbf{\Gamma}^{\prime} \mathbf{A} \mathbf{\Gamma}\right)\left(\mathbf{\Gamma}^{\prime} \mathbf{A} \boldsymbol{\Gamma}\right)\right]\)

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a normal distribution \(N\left(\mu, \sigma^{2}\right)\). Show that $$ \sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2}=\sum_{i=2}^{n}\left(X_{i}-\bar{X}^{\prime}\right)^{2}+\frac{n-1}{n}\left(X_{1}-\bar{X}^{\prime}\right)^{2}, $$ where \(\bar{X}=\sum_{i=1}^{n} X_{i} / n\) and \(\bar{X}^{\prime}=\sum_{i=2}^{n} X_{i} /(n-1)\). Hint: \(\quad\) Replace \(X_{i}-\bar{X}\) by \(\left(X_{i}-\bar{X}^{\prime}\right)-\left(X_{1}-\bar{X}^{\prime}\right) / n\). Show that \(\sum_{i=2}^{n}\left(X_{i}-\bar{X}^{\prime}\right)^{2} / \sigma^{2}\) has a chi-square distribution with \(n-2\) degrees of freedom. Prove that the two terms in the right-hand member are independent. What then is the distribution of $$ \frac{[(n-1) / n]\left(X_{1}-\bar{X}^{\prime}\right)^{2}}{\sigma^{2}} ? $$

(Bonferroni Multiple Comparison Procedure). In the notation of this section, let \(\left(k_{i 1}, k_{i 2}, \ldots, k_{i b}\right), i=1,2, \ldots, m\), represent a finite number of \(b\) -tuples. The problem is to find simultaneous confidence intervals for \(\sum_{j=1}^{b} k_{i j} \mu_{j}, i=1,2, \ldots, m\), by a method different from that of Scheffé. Define the random variable \(T_{i}\) by $$ \left(\sum_{j=1}^{b} k_{i j} \bar{X}_{. j}-\sum_{j=1}^{b} k_{i j} \mu_{j}\right) / \sqrt{\left(\sum_{j=1}^{b} k_{i j}^{2}\right) V / a}, \quad i=1,2, \ldots, m $$ (a) Let the event \(A_{i}^{c}\) be given by \(-c_{i} \leq T_{i} \leq c_{i}, i=1,2, \ldots, m\). Find the random variables \(U_{i}\) and \(W_{i}\) such that \(U_{i} \leq \sum_{1}^{b} k_{i j} \mu_{j} \leq W_{j}\) is equivalent to \(A_{i}^{c}\) (b) Select \(c_{i}\) such that \(P\left(A_{i}^{c}\right)=1-\alpha / m ;\) that is, \(P\left(A_{i}\right)=\alpha / m .\) Use Exercise 9.4.1 to determine a lower bound on the probability that simultaneously the random intervals \(\left(U_{1}, W_{1}\right), \ldots,\left(U_{m}, W_{m}\right)\) include \(\sum_{j=1}^{b} k_{1 j} \mu_{j}, \ldots, \sum_{j=1}^{b} k_{m j} \mu_{j}\) respectively. (c) Let \(a=3, b=6\), and \(\alpha=0.05\). Consider the linear functions \(\mu_{1}-\mu_{2}, \mu_{2}-\mu_{3}\), \(\mu_{3}-\mu_{4}, \mu_{4}-\left(\mu_{5}+\mu_{6}\right) / 2\), and \(\left(\mu_{1}+\mu_{2}+\cdots+\mu_{6}\right) / 6 .\) Here \(m=5 .\) Show that the lengths of the confidence intervals given by the results of Part (b) are shorter than the corresponding ones given by the method of Scheffé as described in the text. If \(m\) becomes sufficiently large, however, this is not the case.
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