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(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.

Short Answer

Expert verified
In this Bonferroni procedure, the random variables \(U_i\) and \(W_i\) are derived first from the condition of the given event \(A_i^c\). With a set value for \(c_i\), the probability of the intervals \(U_i\) and \(W_i\) is then determined applying Exercise 9.4.1. Lastly, we compare the results with the Scheffé method to determine which one yields shorter confidence intervals.

Step by step solution

01

Finding Random Variables \(U_i\) and \(W_i\)

To find \(U_i\) and \(W_i\) we use the given event \(A_i^c\), which states \(-c_i \leq T_i = \leq c_i\). We can rewrite \(T_i\), and use reverse algebra to solve for \(\sum_{j=1}^{b} k_{ij} \mu_j\). Thus, our equivalent condition \(U_i \leq \sum_{j=1}^{b} k_{ij} \mu_j \leq W_i\) can be written implying \(U_i\) and \(W_i\) in terms of the given variables.
02

Selecting \(c_i\) and Applying Exercise 9.4.1

In this part, \(c_i\) is selected such that the probability \(P(A_i^c) = 1 - \alpha / m\). From this, we can calculate the probability \(P(A_i)\). With these set, we then use Exercise 9.4.1 to determine a lower bound on the probability that the random intervals \((U_1, W_1), ...,(U_m, W_m)\) include \(\sum_{j=1}^{b} k_{ij} \mu_j, ..., \sum_{j=1}^{b} k_{mj} \mu_j\) respectively.
03

Comparing with Scheffé's Method

Given three variables \(a\), \(b\) and \(\alpha\), we use these in our linear functions. After calculating the required values, we compare the lengths of the confidence intervals given by the results of Part (b) with the corresponding ones given by the Scheffé method and determine which yields shorter intervals.

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Key Concepts

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

Simultaneous Confidence Intervals
Simultaneous confidence intervals are a powerful statistical tool used to estimate multiple parameters at once. Imagine you want to estimate several population means simultaneously. In doing this, you want to ensure confidence that your estimates fall within a certain range, all at the same time. This is where simultaneous confidence intervals come into play.

To construct these intervals, especially in a setup involving multiple comparisons, it's crucial to broaden the confidence level to account for the multiplicity of tests. If you construct individual intervals for each mean with, say, a 95% confidence level, the combined confidence across all means would drift below 95%. This is where specialized methods like the Bonferroni adjustment aid by adjusting each test to maintain an overall confidence level. The key is to maintain the chance of correctly capturing parameters across all tests, while controlling inaccuracies from performing many tests simultaneously.
Scheffé Method
The Scheffé method is a statistical technique specifically designed for making multiple comparisons among group means in a dataset. It is known for being quite robust, as it allows for a wide range of comparison types. This method forms a simultaneous confidence region, which is applicable to any number of linear contrasts of group means.

Unlike other methods, Scheffé's method does not require specific contrasts to be defined before data collection. It is flexible, offering a way to test hypotheses post-data collection. However, the main tradeoff is that the intervals formed via Scheffé's method tend to be wider compared with other procedures like Bonferroni. Though this guarantees coverage, it may result in less precision, especially when handling a moderate number of contrasts. It's crucial to weigh the benefits of flexibility against the downside of larger confidence intervals.
Linear Functions in Statistical Analysis
Linear functions play a central role in statistical analysis, manifesting in various algorithms and methods, including those in hypothesis testing and regression models.

In the context of the exercise, linear functions are combinations of population parameters \( \mu_j \) weighted by some coefficients \( k_{ij} \). For instance, forms like \( \mu_1 - \mu_2 \) or averages like \( (\mu_1 + \mu_2 + \ldots + \mu_6)/6 \) are examples of linear contrasts. These contrasts help dissect relationships between groups or variables, providing deeper insights into tendencies or trends.

By leveraging linear functions, statisticians can refine their models, comprehend underlying patterns, and make educated predictions about the population under study. Such functions simplify complex data into interpretable components, ensuring clearer interpretations and effective strategic decisions.

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Most popular questions from this chapter

Show that \(\sum_{i=1}^{n}\left[Y_{i}-\alpha-\beta\left(x_{i}-\bar{x}\right)\right]^{2}=n(\hat{\alpha}-\alpha)^{2}+(\hat{\beta}-\beta)^{2} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}+\sum_{i=1}^{n}\left[Y_{i}-\hat{\alpha}-\hat{\beta}\left(x_{i}-\bar{x}\right)\right]^{2} .\)

Given the following observations associated with a two-way classification with \(a=3\) and \(b=4\), compute the \(F\) -statistic used to test the equality of the column means \(\left(\beta_{1}=\beta_{2}=\beta_{3}=\beta_{4}=0\right)\) and the equality of the row means \(\left(\alpha_{1}=\alpha_{2}=\alpha_{3}=0\right)\), respectively. $$ \begin{array}{ccccc} \hline \text { Row/Column } & 1 & 2 & 3 & 4 \\ \hline 1 & 3.1 & 4.2 & 2.7 & 4.9 \\ 2 & 2.7 & 2.9 & 1.8 & 3.0 \\ 3 & 4.0 & 4.6 & 3.0 & 3.9 \\ \hline \end{array} $$

The following are observations associated with independent random samples from three normal distributions having equal variances and respective means \(\mu_{1}, \mu_{2}, \mu_{3}\) $$ \begin{array}{rrr} \hline \text { I } & \text { II } & \text { III } \\ \hline 0.5 & 2.1 & 3.0 \\ 1.3 & 3.3 & 5.1 \\ -1.0 & 0.0 & 1.9 \\ 1.8 & 2.3 & 2.4 \\ & 2.5 & 4.2 \\ & & 4.1 \\ \hline \end{array} $$ Compute the \(F\) -statistic that is used to test \(H_{0}: \mu_{1}=\mu_{2}=\mu_{3} .\)

Let \(Y_{1}, Y_{2}, \ldots, Y_{n}\) be \(n\) independent normal variables with common unknown variance \(\sigma^{2}\). Let \(Y_{i}\) have mean \(\beta x_{i}, i=1,2, \ldots, n\), where \(x_{1}, x_{2}, \ldots, x_{n}\) are known but not all the same and \(\beta\) is an unknown constant. Find the likelihood ratio test for \(H_{0}: \beta=0\) against all alternatives. Show that this likelihood ratio test can be based on a statistic that has a well-known distribution.

Show that \(R=\frac{\sum_{1}^{n}\left(X_{i}-\bar{X}\right)\left(Y_{i}-\bar{Y}\right)}{\sqrt{\sum_{1}^{n}\left(X_{i}-\bar{X}\right)^{2} \sum_{1}^{n}\left(Y_{i}-Y\right)^{2}}}=\frac{\sum_{1}^{n} X_{i} Y_{i}-n \overline{X Y}}{\sqrt{\left(\sum_{1}^{n} X_{i}^{2}-n \bar{X}^{2}\right)\left(\sum_{1}^{n} Y_{i}^{2}-n \bar{Y}^{2}\right)}}\)

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