Question

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

The power function of the sign test is shown to be nonincreasing for the given hypotheses because as the actual parameter \(\theta\) decreases, implying an increase in the probability of negative differences \(p\), the power function - calculated as the cumulative binomial distribution function evaluated at given count \(\kappa\) - decreases due to its inherent property. Thus, the nonincreasing nature of the power function is established.

Step-by-step solution

Understanding the Problem

The problem requires an understanding of the sign test and its power function. The power function of a statistical test is the probability that it will reject a false null hypothesis. In this problem, the hypotheses stated are \(H_{0}: \theta=\theta_{0}\) and \(H_{1}: \theta<\theta_{0}\). The task is to show that this power function is nonincreasing for the above hypotheses, i.e., if the actual parameter \(\theta\) is less than \(\theta_{0}\), the power of the test does not increase. The sign test is a non-parametric test that uses the signs of the differences between paired samples.

Setting up the Power Function of the Sign Test

To calculate the power function of the sign test, we need to consider positive and negative differences between the paired observations. Let's denote the positive differences as \(X_{+}\) and the negative differences as \(X_{-}\). The number of observations is \(n\). The sign test then uses the count of the signs, \(\#(+)\) and \(\#(-)\) to make a decision. Suppose the count of negative signs (\(X_{-}\)) is smaller or equal to some value \(\kappa\), sign test state to reject the null hypothesis. The power function of the sign test is the probability of obtaining such a result, under the distribution characterized by \(H_{1}\). Under the alternative hypothesis, \(H_{1}\), the negative signs are likelier than the positive ones. The PDF of the binomial distribution is used here, as the signs of differences in each pair of observations represent independent Bernoulli trials with a success probability \(p\), where \(p\) denotes the probability of a negative difference. The power function of the sign test is then given by the cumulative distribution function of binomial distribution under \(H_{1}\), evaluated at \(\kappa\).

Proving Nonincreasing Nature

To prove the nonincreasing nature of the power function, we need to show that if the actual parameter \(\theta\) is less than \(\theta_{0}\), the power of the test does not increase. If \(\theta<\theta_{0}\), that implies the probability of negative differences, \(p\), increases; hence, the likelihood of rejecting \(H_{0}\) increases because rejection zone is when count of negative signs (\(X_{-}\)) is smaller or equal to some value \(\kappa\). However, because we are using the cumulative distribution function of the binomial distribution to obtain the power function, as \(p\) increases, the power function actually decreases. The CDF of the binomial distribution is a non-decreasing function and it’s reversed (1 - CDF) is nonincreasing. It’s important to note that the value of \(\kappa\) remains constant. Hence, the value of the power function is nonincreasing.