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Chapter 8: Problem 7

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be iid \(N\left(\theta_{1}, \theta_{2}\right) .\) Show that the likelihood ratio principle for testing \(H_{0}: \theta_{2}=\theta_{2}^{\prime}\) specified, and \(\theta_{1}\) unspecified, against \(H_{1}: \theta_{2} \neq \theta_{2}^{\prime}, \theta_{1}\) unspecified, leads to a test that rejects when \(\sum_{1}^{n}\left(x_{i}-\bar{x}\right)^{2} \leq c_{1}\) or \(\sum_{1}^{n}\left(x_{i}-\bar{x}\right)^{2} \geq c_{2}\) where \(c_{1}

Short Answer

Expert verified
A likelihood ratio test for the given hypotheses leads to a test that rejects \(H_{0}\) when the sum of the squared deviations from the mean is either less than a certain value \(c_{1}\) or greater than another value \(c_{2}\), with \(c_{1}<c_{2}\).

Step by step solution

01

Statement of Hypotheses

The exercise is testing the null hypothesis \(H_{0}: \theta_{2}=\theta_{2}^{\prime}\) against the alternative hypothesis \(H_{1}: \theta_{2} \neq \theta_{2}^{\prime}\). In this context, \(H_{0}\) represents the variance of a normal distribution being a specified value and \(H_{1}\) denotes the variance as an unspecified value different from \(H_{0}\).
02

Derivation of the Likelihood Ratio

The likelihood ratio for the hypotheses can be represented as \(\lambda = \frac{max_{\theta_{1}, \theta_{2}=\theta_{2}^{\prime}}L(\theta_{1}, \theta_{2}=\theta_{2}^{\prime})}{max_{\theta_{1}, \theta_{2}}L(\theta_{1}, \theta_{2})}\), where \(L(\theta_{1}, \theta_{2})\) is the likelihood function. We maximize the likelihood functions in the numerator and the denominator for the given restrictions. In the numerator we take \(\theta_{2}\) as \(\theta_{2}^{\prime}\), whereas in the denominator both \(\theta_{1}\) and \(\theta_{2}\) can vary. This ratio will tell us the degree to which the evidence supports \(H_{0}\) vs \(H_{1}\).
03

Formulation of the Test Rule

The likelihood ratio test rule is typically set up to reject \(H_{0}\) if \(\lambda\) is too small i.e., if the likelihood of the data under the null hypothesis is significantly less than the likelihood under the alternative. This can be reformulated into a criteria that in this case leads us to reject when \(\sum_{1}^{n}\left(x_{i}-\bar{x}\right)^{2} \leq c_{1}\) or \(\sum_{1}^{n}\left(x_{i}-\bar{x}\right)^{2} \geq c_{2}\) where \(c_{1}<c_{2}\) are selected to control the Type I and Type II error risks of the test.
04

Implementation of the Test

Finally, to implement the test, you would evaluate your sample data and see whether it falls into the region defined by these inequality constraints. If it does, you would reject the null hypothesis. If not, you would fail to reject the null hypothesis.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots, X_{n}\) denote a random sample from a normal distribution \(N(\theta, 16)\). Find the sample size \(n\) and a uniformly most powerful test of \(H_{0}: \theta=25\) against \(H_{1}: \theta<25\) with power function \(\gamma(\theta)\) so that approximately \(\gamma(25)=0.10\) and \(\gamma(23)=0.90\).

Show that the likelihood ratio principle leads to the same test when testing a simple hypothesis \(H_{0}\) against an alternative simple hypothesis \(H_{1}\), as that given by the Neyman-Pearson theorem. Note that there are only two points in \(\Omega\).

Illustrative Example \(8.2 .1\) of this section dealt with a random sample of size \(n=2\) from a gamma distribution with \(\alpha=1, \beta=\theta\). Thus the mgf of the distribution is \((1-\theta t)^{-1}, t<1 / \theta, \theta \geq 2 .\) Let \(Z=X_{1}+X_{2} .\) Show that \(Z\) has a gamma distribution with \(\alpha=2, \beta=\theta\). Express the power function \(\gamma(\theta)\) of Example 8.2.1 in terms of a single integral. Generalize this for a random sample of size \(n\).

Let \(X\) be a random variable with pdf \(f_{X}(x)=\left(2 b_{X}\right)^{-1} \exp \left\\{-|x| / b_{X}\right\\}\), for \(-\infty0\). First, show that the variance of \(X\) is \(\sigma_{X}^{2}=2 b_{X}^{2} .\) Next, let \(Y\), independent of \(X\), have pdf \(f_{Y}(y)=\left(2 b_{Y}\right)^{-1} \exp \left\\{-|y| / b_{Y}\right\\}\), for \(-\infty0\). Consider the hypotheses $$ H_{0}: \sigma_{X}^{2}=\sigma_{Y}^{2} \text { versus } H_{1}: \sigma_{X}^{2}>\sigma_{Y}^{2} $$ To illustrate Remark \(8.3 .2\) for testing these hypotheses, consider the following data set. Sample 1 represents the values of a sample drawn on \(X\) with \(b_{X}=1\), while Sample 2 represents the values of a sample drawn on \(Y\) with \(b_{Y}=1\). Hence, in this case \(H_{0}\) is true. (a) Obtain comparison boxplots of these two samples. Comparison boxplots consist of boxplots of both samples drawn on the same scale. Based on these plots, in particular the interquartile ranges, what do you conclude about \(H_{0}\) ? (b) Obtain the \(F\) -test (for a one-sided hypothesis) as discussed in Remark 8.3.2 at level \(\alpha=0.10\). What is your conclusion? (c) The test in part (b) is not exact. Why?

Suppose that a manufacturing process makes about \(3 \%\) defective items, which is considered satisfactory for this particular product. The managers would like to decrease this to about \(1 \%\) and clearly want to guard against a substantial increase, say to \(5 \%\). To monitor the process, periodically \(n=100\) items are taken and the number \(X\) of defectives counted. Assume that \(X\) is \(b(n=100, p=\theta)\). Based on a sequence \(X_{1}, X_{2}, \ldots, X_{m}, \ldots\), determine a sequential probability ratio test that tests \(H_{0}: \theta=0.01\) against \(H_{1}: \theta=0.05 .\) (Note that \(\theta=0.03\), the present level, is in between these two values.) Write this test in the form $$ h_{0}>\sum_{i=1}^{m}\left(x_{i}-n d\right)>h_{1} $$ and determine \(d, h_{0}\), and \(h_{1}\) if \(\alpha_{a}=\beta_{a}=0.02\).
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