/*! 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 7 Show that the power function of ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 7

Show that the power function of the sign test is nonincreasing for the hypotheses $$ H_{0}: \theta=\theta_{0} \text { versus } H_{1}: \theta<\theta_{0} $$

Short Answer

Expert verified
The non-increasing nature of a power function in a sign test shows that as the alternate hypothesis parameter decreases, the test's power function does not increase. This, in essence, honors the 'distribution-free' feature of the sign test.

Step by step solution

01

Define Power function

The power function of a statistical test is denoted as \( 尾(\theta) \) is defined as the probability of rejecting the null hypothesis \( H_{0}: \theta=\theta_{0} \) when the alternative hypothesis \( H_{1}: \theta<\theta_{0} \) is true, that is \( 尾(\theta) = P(Test \ Rejects \ H_{0}|H_{1} \ is \ True) \)
02

Apply the Sign Test

In the context of a sign test, we are considering a test statistic based on order or rank rather than exact values. Essentially, in a sign test, we do not value how different each observation is from the hypothesized median \( 胃_{0} \); we only count an observation as a 鈥榮uccess鈥� or a 鈥榝ailure鈥�.
03

Prove non-increasing nature of the power function

We need to prove that the power function of the sign test is non increasing, or in other words, if \( 胃_{1}<胃_{2} \) then \( 尾(\theta_{1}) \geq 尾(\theta_{2}) \). The goal is to show that, as the alternate hypothesis parameter \( 胃 \) decreases, the power function does not increase. This means that as the underlying distribution changes in favor of the alternative hypothesis (i.e., decreases), the probability of rejecting the null hypothesis does not increase, which matches with the basic intuition of a sign test being 'distribution-free'.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(x_{1}, x_{2}, \ldots, x_{n}\) be a realization of a random sample. Consider the Hodges-Lehmann estimate of location given in expression (10.9.4). Show that the breakdown point of this estimate is \(0.29\). Hint: Suppose we corrupt \(m\) data points. We need to determine the value of \(m\) which results in corruption of one half of the Walsh averages. Show that the corruption of \(m\) data points leads to $$ p(m)=m+\left(\begin{array}{c} m \\ 2 \end{array}\right)+m(n-m) $$ corrupted Walsh averages. Hence the finite sample breakdown point is the "correct" solution of the quadratic equation \(p(m)=n(n+1) / 4\).

(a) For \(n=3\), expand the mgf (10.3.6) to show that the distribution of the signed-rank Wilcoxon is given by $$ \begin{array}{|l|ccccccc|} \hline j & -6 & -4 & -2 & 0 & 2 & 4 & 6 \\ \hline P(T=j) & \frac{1}{8} & \frac{1}{8} & \frac{1}{8} & \frac{2}{8} & \frac{1}{8} & \frac{1}{8} & \frac{1}{8} \\ \hline \end{array} $$$$ \text { (b) Obtain the distribution of the signed-rank Wilcoxon for } n=4 \text { . } $$

Suppose the random variable \(e\) has \(\operatorname{cdf} F(t) .\) Let \(\varphi(u)=\sqrt{12}[u-(1 / 2)]\), \(0

_{j}\left\\{R\left(Y_{j}\right)>\frac鈥� # Consider the sign scores test procedure discussed in Example \(10.5 .4\). (a) Show that \(W_{S}=2 W_{S}^{*}-n_{2}\), where \(W_{S}^{*}=\\#_{j}\left\\{R\left(Y_{j}\right)>\frac{n+1}{2}\right\\} .\) Hence \(W_{S}^{*}\) is an equivalent test statistic. Find the null mean and variance of \(W_{S}\). (b) Show that \(W_{S}^{*}=\\#_{j}\left\\{Y_{j}>\theta^{*}\right\\}\), where \(\theta^{*}\) is the combined sample median. (c) Suppose \(n\) is even. Letting \(W_{X S}^{*}=\\#_{i}\left\\{X_{i}>\theta^{*}\right\\}\), show that we can table \(W_{S}^{*}\) in the following \(2 \times 2\) contingency table with all margins fixed: $$ \begin{array}{|c|c|c|c|} \hline & Y & X & \\ \hline \text { No. items }>\theta^{*} & W_{S}^{*} & W_{X S}^{*} & \frac{n}{2} \\\ \hline \text { No. items }<\theta^{*} & n_{2}-W_{S}^{*} & n_{1}-W_{X S}^{*} & \frac{n}{2} \\ \hline & n_{2} & n_{1} & n \\ \hline \end{array} $$ Show that the usual \(\chi^{2}\) goodness-of-fit is the same as \(Z_{S}^{2}\), where \(Z_{S}\) is the standardized \(z\) -test based on \(W_{S}\). This is often called Mood's median test; see Example \(10.5 .4 .\)

Let \(\widehat{F}_{n}(x)\) denote the empirical cdf of the sample \(X_{1}, X_{2}, \ldots, X_{n} .\) The distribution of \(\widehat{F}_{n}(x)\) puts mass \(1 / n\) at each sample item \(X_{i} .\) Show that its mean is \(\bar{X}\). If \(T(F)=F^{-1}(1 / 2)\) is the median, show that \(T\left(\widehat{F}_{n}\right)=Q_{2}\), the sample median.
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.