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Let \(X_{1}, X_{2}, \ldots, X_{m}\) denote a random sample from the exponential density with mean \(\theta_{1}\) and let \(Y_{1}, Y_{2}, \ldots, Y_{n}\) denote an independent random sample from an exponential density with \(\theta_{2}\). a. Find the likelihood ratio criterion for testing \(H_{0}: \theta_{1}=\theta_{2}\) versus \(H_{a}: \theta_{1} \neq \theta_{2}\). b. Show that the test in part (a) is equivalent to an exact \(F\) test [Hint: Transform \(\sum X_{i}\) and \(\sum Y_{j}\) to \(\left.\chi^{2} \text { random variables. }\right]\)

Short Answer

Expert verified
Use an F-test comparing \(\sum X_i/m\) and \(\sum Y_j/n\).

Step by step solution

01

Set Up the Problem

We are given two independent samples: \(X_1, X_2, \ldots, X_m\) from the exponential distribution with mean \(\theta_1\), and \(Y_1, Y_2, \ldots, Y_n\) from another exponential distribution with mean \(\theta_2\). We need to test the null hypothesis \(H_0: \theta_1 = \theta_2\) against the alternative hypothesis \(H_a: \theta_1 eq \theta_2\).
02

Write the Likelihood Functions

The likelihood functions for exponential distributions are \(L_X(\theta_1) = \prod_{i=1}^{m} \frac{1}{\theta_1} e^{-X_i/\theta_1}\) for the \(X\)-sample, and \(L_Y(\theta_2) = \prod_{j=1}^{n} \frac{1}{\theta_2} e^{-Y_j/\theta_2}\) for the \(Y\)-sample.
03

Combine the Likelihoods

The joint likelihood function under the hypothesis \(H_0: \theta_1 = \theta_2 = \theta\) becomes \(L(\theta) = \left( \frac{1}{\theta} \right)^{m+n} e^{-(\sum X_i + \sum Y_j)/\theta}\).
04

Construct the Likelihood Ratio

Under \(H_0\), the likelihood is \(L(\theta)\). Under \(H_a\), we have the separate likelihoods \(L_X(\theta_1)\) and \(L_Y(\theta_2)\). The likelihood ratio is then \(\Lambda = \frac{L(\theta)}{L_X(\theta_1) L_Y(\theta_2)}\). Substitute the likelihood functions and simplify.
05

Express in Terms of Chi-Squared Variables

Recognize that \(\sum X_i \) follows a gamma distribution, which is related to the chi-squared distribution since \(2\sum X_i / \theta_1\) follows a chi-squared distribution with \(2m\) degrees of freedom, and \(2\sum Y_j / \theta_2\) follows a chi-squared distribution with \(2n\) degrees of freedom.
06

Relate to an F-test

The ratio \(\frac{\sum X_i / m}{\sum Y_j / n}\) follows an \(F\)-distribution under \(H_0\) with \(2m\) and \(2n\) degrees of freedom. Therefore, testing the equality \(\theta_1 = \theta_2\) can be done by performing an exact F-test on these chi-squared transformed variables.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Distribution
The exponential distribution is a continuous probability distribution often used to model the time until an event occurs. It is defined by one parameter, the mean, \(\theta\). Let's say we have a random variable \(X\) that follows this distribution, the probability density function is given by:
  • \( f(x|\theta) = \frac{1}{\theta} e^{-x/\theta} \) for \(x \geq 0\)
This function basically tells us how likely different intervals of time are to include the event.
The distribution is memoryless, meaning the probability of an event occurring in the next time moment is independent of how much time has already passed.
This characteristic makes it applicable in various fields such as reliability analysis and queueing theory. In terms of mean and variance, both are \(\theta\), which can also represent the scale or expected time until the event.
F-Test
The F-Test is a statistical method used to compare variances. This test checks if the variances of two populations are equal. It's useful in the context of the problem where we're comparing exponential distributions. The key here is that the variance of the exponential distribution approaches the mean, so testing variance can help assert equality of means.
  • The test statistic is formed as: \( F = \frac{s_1^2}{s_2^2} \), where \( s_1^2 \) and \( s_2^2 \) are the sample variances.
In the exercise's context, an F-Test supports the hypothesis testing section by transforming sums from gamma variables to chi-squared distribution variables. This transformation assesses if the populations have the same mean, which directly relates to our original hypothesis that \( \theta_1 = \theta_2 \). Under the null hypothesis, the ratio of these sums should follow an F-distribution.
Chi-Squared Distribution
The Chi-Squared distribution is a vital tool in statistics, especially in hypothesis testing and confidence interval estimation. It's generally used for variance comparison, like in the F-test, and in our problem's context, it allows transformation and testing of exponential distribution variables.
  • Defined primarily for non-negative real numbers, a Chi-Squared distribution with \(k\) degrees of freedom is the sum of the squares of \(k\) independent standard normal random variables.
In our exercise, a key application was to use the Chi-Squared distribution to connect sums of exponential variables to more familiar test statistics.
This is important because the exponential's related Gamma distribution makes it possible to reframe exponential data variables as Chi-Squared, allowing straightforward variance tests or comparisons like the F-Test under the hypothesized mean equality. This calculation supports determining whether the samples are from populations with equal parameters.
Hypothesis Testing
Hypothesis Testing is a statistical method that helps determine if there is enough evidence in a sample to infer a particular condition is true for the entire population. The core components include null and alternative hypotheses.
  • Null Hypothesis (H_0): It's the statement being tested, typically expressing no effect or no difference. In this exercise, \(\theta_1 = \theta_2\).
  • Alternative Hypothesis (H_a): The statement considered if the null is rejected, suggesting a difference exists. Here, it was \(\theta_1 eq \theta_2\).
The goal is to use the data to determine if there is enough evidence to reject \(H_0\) in favor of \(H_a\).
We proceed through a series of steps including stating hypotheses, selecting a significance level, calculating a test statistic, and making a decision to reject or not reject the null hypothesis. In structured approaches, like applying a Likelihood Ratio Test or transforming data to perform an F-Test as seen in our example, the hypothesis test becomes a formal pathway to inferential statistics, affirming or refuting parameter equality.

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Most popular questions from this chapter

A large-sample \(\alpha\) -level test of hypothesis for \(H_{0}: \theta=\theta_{0}\) versus \(H_{a}: \theta>\theta_{0}\) rejects the null hypothesis if $$\frac{\hat{\theta}-\theta_{0}}{\sigma_{\hat{\theta}}}>z_{\alpha}$$ Show that this is equivalent to rejecting \(H_{0}\) if \(\theta_{0}\) is less than the large-sample \(100(1-\alpha) \%\) lower confidence bound for \(\theta\)

Suppose that we are interested in testing the simple null hypothesis \(H_{0}: \theta=\theta_{0}\) versus the simple alternative hypothesis \(H_{a}: \theta=\theta_{a} .\) According to the Neyman-Pearson lemma, the test that maximizes the power at \(\theta_{a}\) has a rejection region determined by $$\frac{L\left(\theta_{0}\right)}{L\left(\theta_{a}\right)}

A pharmaceutical manufacturer purchases a particular material from two different suppliers. The mean level of impurities in the raw material is approximately the same for both suppliers, but the manufacturer is concerned about the variability of the impurities from shipment to shipment. If the level of impurities tends to vary excessively for one source of supply, it could affect the quality of the pharmaceutical product. To compare the variation in percentage impurities for the two suppliers, the manufacturer selects ten shipments from each of the two suppliers and measures the percentage of impurities in the raw material for each shipment. The sample means and variances are shown in the accompanying table. $$\begin{array}{ll} \hline \text { Supplier A } & \text { Supplier B } \\ \hline \bar{y}_{1}=1.89 & \bar{y}_{2}=1.85 \\ s_{1}^{2}=.273 & s_{2}^{2}=.094 \\ n_{1}=10 & n_{2}=10 \\ \hline \end{array}$$ a. Do the data provide sufficient evidence to indicate a difference in the variability of the shipment impurity levels for the two suppliers? Test using \(\alpha=.10 .\) Based on the results of your test, what recommendation would you make to the pharmaceutical manufacturer? b. Find a \(90 \%\) confidence interval for \(\sigma_{\mathrm{B}}^{2}\) and interpret your results.

What conditions must be met for the \(Z\) test to be used to test a hypothesis concerning a population mean \(\mu ?\)

Refer to Exercise \(10.10 .\) Click the button "Clear Summary" to delete the results of any previous simulations. Change the sample size for each simulation to \(n=30\) and set up the applet to simulate testing \(H_{0}: p=.4\) versus \(H_{a}: p>.4\) at the .05 level of significance. a. Click the button "Clear Summary" to erase the results or any previous simulations. Set the real value of \(p\) to .4 and implement at least 200 simulations. What is the percentage simulated tests that result in rejecting the null hypothesis? Does the test work as you expected? b. Leave all settings as they were in part (a) but change the real value of \(p\) to \(.5 .\) Simulate at least 200 tests. Repeat when the real value of \(p\) is .6 and .7 . Click the button "Show Summary." What do you observe about the rejection rate as the true value of \(p\) gets further from .4 and closer to \(1 ?\) Does the pattern that you observe match your impression of how a good test should perform?

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