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Let \(X_{1}, X_{2}, \ldots, X_{n}\) and \(Y_{1}, Y_{2}, \ldots, Y_{n}\) be independent random samples from two normal distributions \(N\left(\mu_{1}, \sigma^{2}\right)\) and \(N\left(\mu_{2}, \sigma^{2}\right)\), respectively, where \(\sigma^{2}\) is the common but unknown variance. (a) Find the likelihood ratio \(\Lambda\) for testing \(H_{0}: \mu_{1}=\mu_{2}=0\) against all alternatives. (b) Rewrite \(\Lambda\) so that it is a function of a statistic \(Z\) which has a well-known distribution. (c) Give the distribution of \(Z\) under both null and alternative hypotheses.

Short Answer

Expert verified
The likelihood ratio \(\Lambda\) for testing \(H_{0}: \mu_{1}=\mu_{2}=0\) against all alternatives is \(\Lambda = \exp(n)\). After rewriting \(Λ\), it becomes a function of a statistic \(Z\), specifically \(Λ = \exp(\frac{Z^2}{2})\). The distribution of \(Z\) under the null hypothesis is a standard normal distribution \(N(0,1)\), and under the alternative hypothesis, it is not precisely defined without additional information since it depends on the difference between the means \(\mu_{1}\) and \(\mu_{2}\).

Step by step solution

01

Calculate Likelihood Ratio

The likelihood ratio \(Λ\) is given by \[Λ = \frac{L(\mu_{1}=0, \mu_{2}=0)}{L(\hat{\mu}_{1}, \hat{\mu}_{2})}\] Where \(L(\mu_{1}=0, \mu_{2}=0)\) is the likelihood of the null hypothesis and \(L(\hat{\mu}_{1}, \hat{\mu}_{2}\) is the likelihood of the alternate hypothesis. After inserting the given distributions and doing the needed calculations, this ratio simplifies to \(Λ = \exp(n)\.
02

Rewrite the Ratio

The next step is to rewrite the likelihood ratio \(Λ\) so that it is a function of the statistic \(Z\). The \(Z\) statistic is defined as \(Z = \frac{\bar{X} - \bar{Y}}{s}\) where \(\bar{X}\) and \(\bar{Y}\) are the means, and \(s\) is the standard deviation. After substitution, the new expression for \(Λ\) will become \(Λ = \exp(\frac{Z^2}{2})\).
03

Provide the Distribution

The final step is to provide the distribution of \(Z\) under both null and alternative hypotheses. Under \(H_{0}\) the \(Z\) score follows a standard normal distribution, typically denoted as \(N(0,1)\). Contrarily, under any other alternative hypothesis \(H_{1}\), since \(Z\) is a random variable that varies according to the difference of the means \(\mu_{1}\) and \(\mu_{2}\), the distribution cannot be precisely determined without further information.

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Most popular questions from this chapter

The power function for the one-sample \(t\) -test is discussed. (a) Plot the power function for the following setup: \(X\) has a \(N\left(\mu, \sigma^{2}\right)\) distribution; \(H_{0}: \mu=50\) versus \(H_{1}: \mu \neq 50 ; \alpha=0.05 ; n=25 ;\) and \(\sigma=10\). (b) Overlay the power curve in (a) with that for \(\alpha=0.01\). Comment. (c) Overlay the power curve in (a) with that for \(n=35\). Comment. (d) Determine the smallest value of \(n\) so the power exceeds \(0.80\) to detect \(\mu=53\). Hint: Modify the R function tpowerg.R so it returns the power for a specified alternative.

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 \(\Omega\).

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from the normal distribution \(N(\theta, 1)\). Show that the likelihood ratio principle for testing \(H_{0}: \theta=\theta^{\prime}\), where \(\theta^{\prime}\) is specified, against \(H_{1}: \theta \neq \theta^{\prime}\) leads to the inequality \(\left|\bar{x}-\theta^{\prime}\right| \geq c\). (a) Is this a uniformly most powerful test of \(H_{0}\) against \(H_{1} ?\) (b) Is this a uniformly most powerful unbiased test of \(H_{0}\) against \(H_{1}\) ?

Let \(X\) and \(Y\) have a joint bivariate normal distribution. An observation \((x, y)\) arises from the joint distribution with parameters equal to either $$ \mu_{1}^{\prime}=\mu_{2}^{\prime}=0, \quad\left(\sigma_{1}^{2}\right)^{\prime}=\left(\sigma_{2}^{2}\right)^{\prime}=1, \quad \rho^{\prime}=\frac{1}{2} $$ Ior $$ \mu_{1}^{\prime \prime}=\mu_{2}^{\prime \prime}=1, \quad\left(\sigma_{1}^{2}\right)^{\prime \prime}=4, \quad\left(\sigma_{2}^{2}\right)^{\prime \prime}=9, \quad \rho^{\prime \prime}=\frac{1}{2} \text { . } $$ Show that the classification rule involves a second-degree polynomial in \(x\) and \(y\).

Consider a random sample \(X_{1}, X_{2}, \ldots, X_{n}\) from a distribution with pdf \(f(x ; \theta)=\theta(1-x)^{\theta-1}, 00 .\) (a) Find the form of the uniformly most powerful test of \(H_{0}: \theta=1\) against \(H_{1}: \theta>1\) (b) What is the likelihood ratio \(\Lambda\) for testing \(H_{0}: \theta=1\) against \(H_{1}: \theta \neq 1 ?\)

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