Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a Bernoulli \(b(1, \theta)\) distribution, where \(0<\theta<1\) (a) Show that the likelihood ratio test of \(H_{0}: \theta=\theta_{0}\) versus \(H_{1}: \theta \neq \theta_{0}\) is based upon the statistic \(Y=\sum_{i=1}^{n} X_{i} .\) Obtain the null distribution of \(Y\). (b) For \(n=100\) and \(\theta_{0}=1 / 2\), find \(c_{1}\) so that the test rejects \(H_{0}\) when \(Y \leq c_{1}\) or \(Y \geq c_{2}=100-c_{1}\) has the approximate significance level of \(\alpha=0.05 .\) Hint: Use the Central Limit Theorem.

Short Answer

Expert verified
To answer part (a), the likelihood ratio test statistic \(\lambda=\left(\frac{\theta_{0}^{Y}(1-\theta_{0})^{n-Y}}{(\bar{X})^{Y}(1-\bar{X})^{n-Y}}\right)\) and the master \(Y\) follows a \(binomial(n, \theta_{0})\) distribution under the null hypothesis. In part (b), to find the critical values \(c_{1}\) and \(c_{2}\) for the test, one can use the Central Limit Theorem to approximate them by finding the z-score for the provided significance level, and then derive the approximate boundary values as \(c_{1}= n \theta_0 - z_{0.025}(n\theta_0 (1-\theta_0))^{0.5}\) and \(c_{2}= n\theta_{0} + z_{0.025}(n\theta_0 (1-\theta_0))^{0.5}\).

Step by step solution

01

Derive the Likelihood Ratio Test Statistic

From the definition of the likelihood function for Bernoulli distribution, for \(n\) samples, the likelihood function is: \(L(\theta;X)=\theta^{\sum_{i=1}^{n} X_{i}}(1-\theta)^{n - \sum_{i=1}^{n} X_{i}}.\) The likelihood ratio test statistic is the ratio of the maximum likelihood under \(H_{0}\) to the maximum likelihood under \(H_{1}\), i.e. \(\lambda=\frac{L(\theta_{0};X)}{L(\hat{\theta};X)}\), where \(\hat{\theta}\) is MLE of \(\theta\) which equals to \(\bar{X}=Y/n\). Therefore, the likelihood ratio test statistic \(\lambda\) takes the form: \(\lambda=\left(\frac{\theta_{0}^{Y}(1-\theta_{0})^{n-Y}}{(\bar{X})^{Y}(1-\bar{X})^{n-Y}}\right)\).
02

Obtain the Null Distribution of Y

Under the null hypothesis \(H_{0}: \theta=\theta_{0}\), each \(X_{i}\) follows a Bernoulli(\(\theta_{0}\)) distribution, which means \(Y=\sum_{i=1}^{n} X_{i}\) follows a \(binomial(n, \theta_{0})\) distribution.
03

Find the Critical Values Using the Central Limit Theorem

To reject the null hypothesis at significance level \(\alpha=0.05\), you need to find the values \(c_{1}\) and \(c_{2}=100-c_{1}\) such that \(P(Y \leq c_{1} or Y \geq c_{2})=0.05\). You can use the Central Limit Theorem to approximate this probability. The Central Limit Theorem states that the variable \(Y\) follows a distribution that is roughly \(normal(n\theta_{0}, n\theta_{0}(1-\theta_{0}))\) when \(n\) is large. Therefore when \(n=100\) and \(\theta_{0}=0.5\), the normal distribution is \(normal(50, 25)\), and you can use the standard normal z-table to find \(z_{0.025}\) so that \(P(Z > z_{0.025}) = 0.025\). Then, use the approximation \(c_{1}= n \theta_0 - z_{0.025} \sqrt{n\theta_0 (1-\theta_0)}\) and \(c_{2}= n\theta_{0} + z_{0.025} \sqrt{n\theta_0 (1-\theta_0)}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(\left(X_{1}, Y_{1}\right),\left(X_{2}, Y_{2}\right), \ldots,\left(X_{n}, Y_{n}\right)\) be a random sample from a bivariate normal distribution with \(\mu_{1}, \mu_{2}, \sigma_{1}^{2}=\sigma_{2}^{2}=\sigma^{2}, \rho=\frac{1}{2}\), where \(\mu_{1}, \mu_{2}\), and \(\sigma^{2}>0\) are unknown real numbers. Find the likelihood ratio \(\Lambda\) for testing \(H_{0}: \mu_{1}=\mu_{2}=0, \sigma^{2}\) unknown against all alternatives. The likelihood ratio \(\Lambda\) is a function of what statistic that has a well- known distribution?

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}

If \(X_{1}, X_{2}, \ldots, X_{n}\) is a random sample from a distribution with pdf $$f(x ; \theta)=\left\\{\begin{array}{ll}\frac{3 \theta^{3}}{(x+\theta)^{2}} & 0

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a Bernoulli distribution with parameter \(p .\) If \(p\) is restricted so that we know that \(\frac{1}{2} \leq p \leq 1\), find the mle of this parameter.

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a distribution with pdf \(f(x ; \theta)=\theta \exp \left\\{-|x|^{\theta}\right\\} / 2 \Gamma(1 / \theta),-\infty0 .\) Suppose \(\Omega=\) \(\\{\theta: \theta=1,2\\}\). Consider the hypotheses \(H_{0}: \theta=2\) (a normal distribution) versus \(H_{1}: \theta=1\) (a double exponential distribution). Show that the likelihood ratio test can be based on the statistic \(W=\sum_{i=1}^{n}\left(X_{i}^{2}-\left|X_{i}\right|\right)\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free