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Suppose \(X_{1}, \ldots, X_{n}\) is a random sample on \(X\) which has a \(N\left(\mu, \sigma_{0}^{2}\right)\) distribution, where \(\sigma_{0}^{2}\) is known. Consider the two-sided hypotheses $$ H_{0}: \mu=0 \text { versus } H_{1}: \mu \neq 0 $$ Show that the test based on the critical region \(C=\left\\{|\bar{X}|>\sqrt{\sigma_{0}^{2} / n} z_{\alpha / 2}\right\\}\) is an unbiased level \(\alpha\) test.

Short Answer

Expert verified
The test based on the critical region \(C=\left\{|\bar{X}|>\sqrt{\sigma_{0}^{2} / n} z_{\alpha / 2}\right\}\) is indeed an unbiased level \(\alpha\) test for the given hypotheses \(H_0: \mu=0 \text { versus } H_1: \mu \neq 0\).

Step by step solution

01

Understanding unbiased level \(\alpha\) test

An unbiased level \(\alpha\) test is a decision rule that does not favor the null or alternative hypothesis. The level \(\alpha\) represents the maximum probability that we reject the null hypothesis \(H_0\) when indeed it is true. This probability is also known as Type I error. Here, the null hypothesis is \(\mu = 0\) and alternative hypothesis is \(\mu \neq 0\)
02

Deriving the critical region

The critical region C is defined as \( C=\left\{|\bar{X}|>\sqrt{\sigma_{0}^{2} / n} z_{\alpha / 2}\right\} \). This is the area where we will reject the null hypothesis. The term \(\sqrt{\sigma_{0}^{2} / n} z_{\alpha / 2}\) is the threshold value at the \(\alpha/2\) level in the two tailed normal test. While \(\bar{X}\) is mean of the samples, \( n \) is the number of samples and \(\sigma_{0}^{2}\) is the variance.
03

Validating the test as unbiased level \(\alpha\) test

Now, let's validate this test. For the condition of unbiased level \(\alpha\) test, we have \( P(Reject \ H_{0}|H_{0} \ is \ true) = \alpha \). Given that \(H_0\) is true implies \(\mu = 0\), the distribution of \( \sqrt{n} \bar{X}/\sigma_0 \) follows \(N(0,1)\). Therefore the probability of rejection in case of null hypothesis becomes \( P\left( |\sqrt{n} \bar{X}/\sigma_0| > z_{\alpha / 2}\right) \) which equals to \(\alpha\). Hence, the given test is indeed an unbiased level \(\alpha\) test.

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Key Concepts

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

Critical Region
In hypothesis testing, the critical region is fundamental in deciding whether to reject the null hypothesis. It is basically a set of values for the test statistic that specifies where the null hypothesis would be considered unlikely. For example, suppose we have a sample mean, \( \bar{X} \) and we wish to test whether it significantly differs from a hypothesized population mean, \( \mu \). The critical region would be determined by a boundary or threshold. If the sample mean falls within that boundary, we conclude that there is sufficient evidence to reject the null hypothesis.

Using standard deviation, sample size, and a desired significance level \( \alpha \), we compute this threshold, often with the aid of the z-score from a normal distribution. In the given exercise, the critical region \( C \) is defined by \( |\bar{X}| > \sqrt{\sigma_{0}^{2} / n} z_{\alpha / 2} \) where \( z_{\alpha / 2} \) is the critical z-value that contains the central \( 1 - \alpha \) proportion of the distribution. It is crucial for students to understand that the critical region relates directly to the concept of significance level and determines when we should reject or fail to reject our null hypothesis based on the data.
Type I Error
The possibility of rejecting a true null hypothesis is denoted by a Type I error, often symbolized as \( \alpha \). It's essential to understand that a Type I error is a kind of statistical error that occurs when the test incorrectly indicates the presence of an effect or relationship that doesn't actually exist in the population – think of it as a 'false positive.'

A major goal in any statistical test, like the one in our exercise, is to control the probability of making a Type I error. We set a significance level \( \alpha \) to do this, which is the threshold for how much risk of error we are willing to accept. Typically, \( \alpha \) is set at 0.05 or 5%, meaning we would allow a 5% chance of rejecting the null hypothesis when it should actually be accepted. To prevent Type I errors as much as possible, researchers must choose the appropriate significance level and sample size before conducting the test.
Normal Distribution
The normal distribution is a continuous probability distribution that is symmetrical about its mean, visually represented by the characteristic 'bell curve.' It's a crucial tool in the toolbox of any student studying statistics because a large number of random variables are either naturally modeled by the normal distribution or are close enough that the distribution can provide a good approximation.

The normal distribution is determined by two parameters: mean \( \mu \) and variance \( \sigma^{2} \). The standard normal distribution in particular, which has a mean of 0 and a variance of 1, is especially important for hypothesis testing, as it allows for straightforward calculation of p-values and critical regions. This distribution is used as the basis for the z-score in the exercise, where the test statistic, after standardization, follows a normal distribution. Understanding the properties of the normal distribution is key to comprehending how we make inferences about populations based on sample data.
Unbiased Test
An unbiased test is a statistical test that is equally likely to reject the null hypothesis when it is false as it is to retain it when it is true. In the context of hypothesis testing, 'bias' refers to the tendency of a test to err in one direction – either in favor of rejecting the null or not. An unbiased test ensures that this propensity is neutralized, meaning that the test does not favor the null or alternative hypothesis unfairly.

In the exercise we have, demonstrating that a test is unbiased involves showing that the probability of rejecting a true null hypothesis (Type I error) is exactly \( \alpha \) – no more, no less. This characteristic is desirable because it means that the test maintains the pre-set probability of making a Type I error across all possible alternative hypotheses. It's a critical concept for students to grasp because this fairness is fundamental to the validity of any statistical conclusions drawn from hypothesis tests.

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Most popular questions from this chapter

Let the independent random variables \(Y\) and \(Z\) be \(N\left(\mu_{1}, 1\right)\) and \(N\left(\mu_{2}, 1\right)\), respectively. Let \(\theta=\mu_{1}-\mu_{2} .\) Let us observe independent observations from each distribution, say \(Y_{1}, Y_{2}, \ldots\) and \(Z_{1}, Z_{2}, \ldots .\) To test sequentially the hypothesis \(H_{0}: \theta=0\) against \(H_{1}: \theta=\frac{1}{2}\), use the sequence \(X_{i}=Y_{i}-Z_{i}, i=1,2, \ldots .\) If \(\alpha_{a}=\beta_{a}=0.05\), show that the test can be based upon \(\bar{X}=\bar{Y}-\bar{Z} .\) Find \(c_{0}(n)\) and \(c_{1}(n)\)

Let \(X_{1}, X_{2}, \ldots, X_{20}\) be a random sample of size 20 from a distribution that is \(N(\theta, 5)\). Let \(L(\theta)\) represent the joint pdf of \(X_{1}, X_{2}, \ldots, X_{20}\). The problem is to test. \(H_{0}: \theta=1\) against \(H_{1}: \theta=0 .\) Thus \(\Omega=\\{\theta: \theta=0,1\\}\). (a) Show that \(L(1) / L(0) \leq k\) is equivalent to \(\bar{x} \leq c\). (b) Find \(c\) so that the significance level is \(\alpha=0.05 .\) Compute the power of this test if \(H_{1}\) is true. (c) If the loss function is such that \(\mathcal{L}(1,1)=\mathcal{L}(0,0)=0\) and \(\mathcal{L}(1,0)=\mathcal{L}(0,1)>0\), find the minimax test. Evaluate the power function of this test at the points \(\theta=1\) and \(\theta=0 .\)

Let \(X_{1}, \ldots, X_{n}\) and \(Y_{1}, \ldots, Y_{m}\) follow the location model $$ \begin{aligned} X_{i} &=\theta_{1}+Z_{i}, \quad i=1, \ldots, n \\ Y_{i} &=\theta_{2}+Z_{n+i}, \quad i=1, \ldots, m, \end{aligned} $$ where \(Z_{1}, \ldots, Z_{n+m}\) are iid random variables with common pdf \(f(z) .\) Assume that \(E\left(Z_{i}\right)=0\) and \(\operatorname{Var}\left(Z_{i}\right)=\theta_{3}<\infty\) (a) Show that \(E\left(X_{i}\right)=\theta_{1}, E\left(Y_{i}\right)=\theta_{2}\), and \(\operatorname{Var}\left(X_{i}\right)=\operatorname{Var}\left(Y_{i}\right)=\theta_{3}\). (b) Consider the hypotheses of Example 8.3.1, i.e., $$ H_{0}: \theta_{1}=\theta_{2} \text { versus } H_{1}: \theta_{1} \neq \theta_{2} \text { . } $$ Show that under \(H_{0}\), the test statistic \(T\) given in expression \((8.3 .4)\) has a limiting \(N(0,1)\) distribution. (c) Using part (b), determine the corresponding large sample test (decision rule) of \(H_{0}\) versus \(H_{1}\). (This shows that the test in Example \(8.3 .1\) is asymptotically correct.)

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 Theorem 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 \(\gamma\left(\frac{1}{10}\right)=0.90\)

Let \(X\) have a Poisson distribution with mean \(\theta\). Find the sequential probability ratio test for testing \(H_{0}: \theta=0.02\) against. \(H_{1}: \theta=0.07\). Show that this test can be based upon the statistic \(\sum_{1}^{n} X_{i}\). If \(\alpha_{a}=0.20\) and \(\beta_{a}=0.10\), find \(c_{0}(n)\) and \(c_{1}(n)\)

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