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Chapter 8: Problem 8

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 \(\gamma(23)=0.90\).

Short Answer

Expert verified
The sample size required for the 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 \(\gamma(23)=0.90\) is 62.

Step by step solution

01

Define the Hypotheses and Test Statistic

We have two hypotheses: \(H_{0}: \theta=25\), the null and \(H_{1}: \theta<25\), the alternative. Because we're dealing with a normal distribution, we will use a z-score as our test statistic. Where \(Z = \frac{{\bar{X} - \theta}}{{\sigma/\sqrt{n}}}\) with \(\bar{X}\) as the sample mean, \(\theta\) as the population mean, \(\sigma\) as the population standard deviation (which is 4 here, because \(\sigma^{2}=16\)) and \(n\) as the sample size.
02

Find the Critical Region and Rejection Criterion

For a lower-tailed test (\(\theta<25\)), the critical region is \(\{Z < Z_{\alpha}\}\), where \(Z_{\alpha}\) is the lower \(\alpha\) percentile of the standard normal distribution. As \(\gamma(25) = 0.10\), we have \(\alpha = 0.10\). Thus, \(Z_{\alpha} = Z_{0.10} = -1.28\). Therefore, the rejection criterion becomes \(\bar{X} < \theta + Z_{\alpha} \cdot (\sigma/\sqrt{n})\).
03

Determine the Sample Size

We know the rejection criterion for the test at \(\theta = 23\) should be \(\gamma(23) = 0.90\). Hence, \(\theta + Z_{\alpha} \cdot (\sigma/\sqrt{n}) = 23\). Solving for \(n\), we get \(n = \left(\frac{{\sigma \cdot (25 - 23)}}{{Z_{\alpha}}}\right)^2 \approx 61.9\). Since the number of trials must be an integer, choose \(n = 62\).

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Most popular questions from this chapter

Let \(Y_{1}

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a distribution with \(\operatorname{pdf} f(x ; \theta)=\) \(\theta x^{\theta-1}, 0

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

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 \(H_{1} ?\)

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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