Question

(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

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

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.

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.

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.