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Chapter 6: Problem 4

Let \(X_{1}, X_{2}, \ldots, X_{n}\) and \(Y_{1}, Y_{2}, \ldots, Y_{m}\) be independent random samples from the two normal distributions \(N\left(0, \theta_{1}\right)\) and \(N\left(0, \theta_{2}\right)\). (a) Find the likelihood ratio \(\Lambda\) for testing the composite hypothesis \(H_{0}: \theta_{1}=\theta_{2}\) against the composite alternative \(H_{1}: \theta_{1} \neq \theta_{2}\). (b) This \(\Lambda\) is a function of what \(F\) -statistic that would actually be used in this test?

Short Answer

Expert verified
The likelihood ratio \(\Lambda\) for this hypothesis test is a function of the F-statistic, where the F-statistic is derived from the likelihood ratio using the formula \[F = \frac{n*m/2}{n+m-2} \times (1-\Lambda^{2/n})\].

Step by step solution

01

Setting Up the Likelihood Ratio

The likelihood ratio for the hypothesis test is defined as the maximized value of the likelihood function under the null hypothesis \(H_{0}: \theta_{1}=\theta_{2}\) divided by the maximized value of the likelihood function under the alternative hypothesis \(H_{1}: \theta_{1} \neq \theta_{2}\). In this case, the likelihood function for independent variables under a normal distribution is defined as the product of the individual probability density functions.
02

Express the Likelihood Ratio in terms of Sum of Squares

Transform the likelihood ratio by expressing it in terms of the sum of squares of the individual variables. This is achieved by replacing \(\theta_{1}\) and \(\theta_{2}\) in the formula by their estimates, which are the sum of squared deviations from the mean.
03

Deriving the F-statistic from the Likelihood Ratio

The relationship between the likelihood ratio (denoted by \(\Lambda\)) and the F-statistic is given by the formula -\[F = \frac{n*m/2}{n+m-2} \times (1-\Lambda^{2/n})\] Here, n and m are the sizes of the respective samples, \(\Lambda\) is the likelihood ratio.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from the distribution \(N\left(\theta_{1}, \theta_{2}\right)\). 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}

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be iid, each with the distribution having pdf \(f\left(x ; \theta_{1}, \theta_{2}\right)=\) \(\left(1 / \theta_{2}\right) e^{-\left(x-\theta_{1}\right) / \theta_{2}}, \theta_{1} \leq x<\infty,-\infty<\theta_{2}<\infty\), zero elsewhere. Find the maximum likelihood estimators of \(\theta_{1}\) and \(\theta_{2}\).

Rao (page 368,1973 ) considers a problem in the estimation of linkages in genetics. McLachlan and Krishnan (1997) also discuss this problem and we present their model. For our purposes it can be described as a multinomial model with the four categories \(C_{1}, C_{2}, C_{3}\) and \(C_{4}\). For a sample of size \(n\), let \(\mathbf{X}=\left(X_{1}, X_{2}, X_{3}, X_{4}\right)^{\prime}\) denote the observed frequencies of the four categories. Hence, \(n=\sum_{i=1}^{4} X_{i} .\) The probability model is $$\begin{array}{|c|c|c|c|}\hline C_{1} & C_{2} & C_{3} & C_{4} \\ \hline \frac{1}{2}+\frac{1}{4} \theta & \frac{1}{4}-\frac{1}{4} \theta & \frac{1}{4}-\frac{1}{4} \theta & \frac{1}{4} \theta \\\\\hline\end{array}$$ where the parameter \(\theta\) satisfies \(0 \leq \theta \leq 1 .\) In this exercise, we obtain the mle of \(\theta\). (a) Show that likelihood function is given by $$L(\theta \mid \mathbf{x})=\frac{n !}{x_{1} ! x_{2} ! x_{3} ! x_{4} !}\left[\frac{1}{2}+\frac{1}{4} \theta\right]^{x_{1}}\left[\frac{1}{4}-\frac{1}{4} \theta\right]^{x_{2}+x_{3}}\left[\frac{1}{4} \theta\right]^{x_{4}}$$ (b) Show that the log of the likelihood function can be expressed as a constant (not involving parameters) plus the term $$ x_{1} \log [2+\theta]+\left[x_{2}+x_{3}\right] \log [1-\theta]+x_{4} \log \theta $$ (c) Obtain the partial of the last expression, set the result to 0, and solve for the mle. (This will result in a quadratic equation which has one positive and one negative root.)

Let \(Y_{1}

A machine shop that manufactures toggle levers has both a day and a night shift. A toggle lever is defective if a standard nut cannot be screwed onto the threads. Let \(p_{1}\) and \(p_{2}\) be the proportion of defective levers among those manufactured by the day and night shifts, respectively. We shall test the null hypothesis, \(H_{0}: p_{1}=p_{2}\), against a two-sided alternative hypothesis based on two random samples, each of 1000 levers taken from the production of the respective shifts. Use the test statistic \(Z^{*}\) given in Example \(6.5 .3\). (a) Sketch a standard normal pdf illustrating the critical region having \(\alpha=0.05\). (b) If \(y_{1}=37\) and \(y_{2}=53\) defectives were observed for the day and night shifts, respectively, calculate the value of the test statistic and the approximate \(p-\) value (note that this is a two-sided test). Locate the calculated test statistic on your figure in Part (a) and state your conclusion. Obtain the approximate \(p\) -value of the test.
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