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Let \(X\) have the pdf \(f(x ; \theta)=\theta^{x}(1-\theta)^{1-x}, x=0,1\), zero elsewhere. We test \(H_{0}: \theta=\frac{1}{2}\) against \(H_{1}: \theta<\frac{1}{2}\) by taking a random sample \(X_{1}, X_{2}, \ldots, X_{5}\) of size \(n=5\) and rejecting \(H_{0}\) if \(Y=\sum_{1}^{n} X_{i}\) is observed to be less than or equal to a constant \(c\). (a) Show that this is a uniformly most powerful test. (b) Find the significance level when \(c=1\). (c) Find the significance level when \(c=0\). (d) By using a randomized test, as discussed in Example \(5.6 .4\), modify the tests given in Parts (b) and (c) to find a test with significance level \(\alpha=\frac{2}{32}\).

Short Answer

Expert verified
The testing procedure described is indeed a uniformly most powerful test for the given null and alternative hypotheses. The significance levels for \(c=1\) and \(c=0\) are calculated as probabilities under the binomial distribution assumption. By using a randomized test and solving the equation, we obtain the value of \(p\) that helps us to adjust the tests to achieve a specific significance level \(\alpha = \frac{2}{32}\).

Step by step solution

01

Verifying Uniformly Most Powerful Test

We apply the Neyman Pearson Lemma to show that this is a uniformly most powerful test. Given the null hypothesis \(H_0: \theta = \frac{1}{2}\) and alternative hypothesis \(H_1: \theta < \frac{1}{2}\), we need to find the test that maximizes the power function \(\beta(\theta)\) under the constraint \(P_{\theta}(Y \leq c) \leq \alpha\), where \(\alpha\) is the significance level.
02

Find the significance level for c=1

By choosing \(c=1\), the significance level is \(P_{\frac{1}{2}}(Y \leq 1) = \sum_{i=0}^{1} {5 \choose i}\left(\frac{1}{2}\right)^i \left(1-\frac{1}{2}\right)^{5-i}\). This corresponds to \(Y=0\) or \(1\); and using the binomial theorem, we calculate these probabilities and add them up to get the significance level.
03

Find the significance level for c=0

With \(c=0\), the significance level is \(P_{\frac{1}{2}}(Y = 0) = \left(\frac{1}{2}\right)^0 \left(1-\frac{1}{2}\right)^{5}\), corresponding to the case when \(Y=0\). Using the expression for the binomial probability, we find this probability.
04

Modify the Tests to Achieve Specific Significance Level

We use a randomized test to find a test with significance level \(\alpha = \frac{2}{32}\), the method involves solving the equation \(p P_{\frac{1}{2}}(Y=1)+(1-p)P_{\frac{1}{2}}(Y=0)= \frac{2}{32}\) to find the value of \(p\). Here, \(p\) is the probability of accepting \(H_0\) when \(Y=1\) in the randomized test. This equation is a result of combining and modifying the tests in steps 2 and 3.

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Key Concepts

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

Neyman Pearson Lemma
The Neyman-Pearson Lemma is a fundamental result in the field of hypothesis testing in statistics. It provides a method to construct the most powerful test for a given size when testing a simple null hypothesis against a simple alternative hypothesis. According to this lemma, the test that maximizes the probability of correctly rejecting the null hypothesis (power of the test) for a fixed probability of incorrectly rejecting it (type I error) involves a likelihood ratio test.

The likelihood ratio is the ratio of the probabilities of observing the given sample under the two hypotheses. In hypothesis testing, we use the likelihood ratio to decide whether to reject the null hypothesis, based on how likely the observed sample would be under the alternative hypothesis compared to the null hypothesis. In our exercise, we apply the Neyman-Pearson Lemma to show that a test on the sum of a random sample is uniformly most powerful for testing a binomial distribution parameter.
Significance Level
The significance level, denoted by \( \alpha \), is a threshold that we set for deciding whether to reject the null hypothesis in hypothesis testing. It represents the probability of committing a type I error, which is rejecting the null hypothesis when it is actually true. The typical values for \( \alpha \) are 0.05, 0.01, or 0.10, and it is determined before conducting the test.

In our exercise, the significance level depends on the value we choose for the constant \( c \). For instance, with \( c=1 \), we find the probability of observing a test statistic less than or equal to 1 when the null hypothesis is true. This probability is calculated using the binomial distribution, as the random variables in question follow a binomial pattern due to being indicators of success or failure.
Randomized Test
A randomized test is a statistical technique that involves randomization in the decision of rejecting or not rejecting the null hypothesis. In certain scenarios, especially when dealing with discrete distributions, it is not possible to achieve the desired significance level by using a simple test. In such cases, a randomized test allows us to obtain an exact significance level.

In the exercise, a randomized test is used to modify the existing tests to achieve a specific significance level, \( \alpha = \frac{2}{32} \). The probability of accepting the null hypothesis when the test statistic is on the boundary (in this case, when \( Y=1 \) and \( Y=0 \) for our two tests) is adjusted through a calculated probability \( p \), which effectively randomizes the test decision at the boundary point, ensuring that the overall test has the desired significance level.
Binomial Theorem
The binomial theorem provides a powerful way to expand expressions that are raised to a power. Specifically, it tells us how to expand expressions of the form \( (x + y)^n \) where \( n \) is a non-negative integer. According to the theorem, you can expand this expression into a sum involving terms of the form \( {n \choose k} x^k y^{n-k} \) for \( k \) from 0 to \( n \), where \( {n \choose k} \) are binomial coefficients.

In the context of our exercise, the binomial theorem is used to calculate the probabilities that comprise the significance levels for different values of \( c \). For a binomial distribution with parameters \( n \) and \( p \) (the probability of success on a single trial), the theorem is used to find the probability of \( k \) successes in \( n \) trials.
Power Function
The power function of a hypothesis test, denoted by \( \beta(\theta) \), represents the probability of correctly rejecting the null hypothesis for a given value of the parameter \( \theta \). It measures the strength of a test in detecting an effect or a difference when one truly exists, thereby reflecting the test's ability to avoid a type II error (failing to reject the null hypothesis when the alternative hypothesis is true).

The ultimate goal in hypothesis testing is to maximize the power function while maintaining a predefined significance level. In the exercise, we discuss finding a test that is uniformly most powerful, which means it has the highest power function among all tests with the same significance level. In simpler terms, it's the best test for detecting a real effect without exceeding the acceptable risk of a false positive, as defined by the significance level.

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Most popular questions from this chapter

If \(X_{1}, X_{2}, \ldots, X_{n}\) is a random sample from a distribution having pdf of the form \(f(x ; \theta)=\theta x^{\theta-1}, 0

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.

Consider a normal distribution of the form \(N(\theta, 4)\). The simple hypothesis \(H_{0}: \theta=0\) is rejected, and the alternative composite hypothesis \(H_{1}: \theta>0\) is accepted if and only if the observed mean \(\bar{x}\) of a random sample of size 25 is greater than or equal to \(\frac{3}{5}\). Find the power function \(\gamma(\theta), 0 \leq \theta\), of this test.

Suppose that a manufacturing process makes about 3 percent defective items, which is considered satisfactory for this particular product. The managers would like to decrease this to about 1 percent and clearly want to guard against a substantial increase, say to 5 percent. To monitor the process, periodically \(n=100\) items are taken and the number \(X\) of defectives counted. Assume that \(X\) is \(b(n=100, p=\theta)\). Based on a sequence \(X_{1}, X_{2}, \ldots, X_{m}, \ldots\), determine a sequential probability ratio test that tests \(H_{0}: \theta=0.01\) against \(H_{1}: \theta=0.05 .\) (Note that \(\theta=0.03\), the present level, is in between these two values.) Write this test in the form $$h_{0}>\sum_{i=1}^{m}\left(x_{i}-n d\right)>h_{1}$$ and determine \(d, h_{0}\), and \(h_{1}\) if \(\alpha_{a}=\beta_{a}=0.02\).

Consider a distribution having a pmf of the form \(f(x ; \theta)=\theta^{x}(1-\theta)^{1-x}, x=\) 0,1, zero elsewhere. Let \(H_{0}: \theta=\frac{1}{20}\) and \(H_{1}: \theta>\frac{1}{20} .\) Use the central limit theorcu? to determine the sample size \(n\) of a random sample so that a uniformly most powerful test of \(H_{0}\) against \(H_{1}\) has a power function \(\gamma(\theta)\), with approximately \(\gamma\left(\frac{1}{20}\right)=0.05\) and \(\gamma\left(\frac{1}{10}\right)=0.90\).

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