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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.)

Short Answer

Expert verified
The exercise demonstrates the relation of the expectation and variance of random variables, and statistical testing of hypotheses. The test statistic \(T\) under \(H_{0}: \theta_{1}=\theta_{2}\) follows a standard normal (N(0,1)) distribution. The large sample test (decision rule) of \(H_{0}\) versus \(H_{1}\) is developed based on this information, where the null hypothesis is rejected if the absolute value of \(T\) is sufficiently large.

Step by step solution

01

E(Xi) Calculation

We know that \(E(X_i)\) is equal to \(\theta_1 + E(Z_i)\). Since assumed that \(E(Zi) = 0\), we can infer that \(E(X_i) = \theta_1\).
02

E(Yi) Calculation

Similar to the calculation of \(E(X_i)\), we can perform for \(E(Y_i)\) as well. Since \(E(Yi) = \theta_2 + E(Z_i)\) and \(E(Zi) = 0\), we get \(E(Y_i) = \theta_2\).
03

Variance Calculation

Since variances are not affected by the change of location, we are given that \(\operatorname{Var}\left(Z_{i}\right)=\theta_{3}<\infty\). Therefore, \(\operatorname{Var}\left(X_{i}\right)=\operatorname{Var}\left(Y_{i}\right)=\theta_{3}\).
04

Statistics and Hypotheses

Under the null hypothesis \(H_{0}\), \(\theta_{1}=\theta_{2}\). Given this, the test statistic \(T\) simplifies to a value that follows a standard normal (N(0,1)) distribution. This is due to the Central Limit Theorem as the sample size goes to infinity.
05

Large Sample Test (Decision Rule)

It follows from part (b) that we reject the null hypothesis \(H_{0}: \theta_{1}=\theta_{2}\) if the absolute value of \(T\) is sufficiently large. This is because under the null hypothesis, \(T\) should follow a standard normal distribution and hence should not take on extreme values. This gives us a decision rule for testing the hypothesis based on the given location model.

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Key Concepts

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

Location Model
A location model is a statistical model where observations are expressed as the sum of a parameter of interest (known as the location parameter) and a disturbance term. In our case, with random variables \(X_i\) and \(Y_i\), we represent them as \(X_i = \theta_1 + Z_i\) and \(Y_i = \theta_2 + Z_{n+i}\), where \(Z_i\) are independent random variables.

The location parameters \(\theta_1\) and \(\theta_2\) serve as shifts in the data. These shifts can be seen literally as "locations" of the data on the number line. The role of these parameters is crucial because they signify the central tendency or average value of the data being considered.

In this model, it’s important to determine the mean and variance of the variables:\[E(X_i) = \theta_1\]and\[E(Y_i) = \theta_2\].
This shows how the location parameters appear directly as means, assuming the expected value of disturbances \(E(Z_i)\) is zero.

The variances are \(\operatorname{Var}(X_i) = \operatorname{Var}(Y_i) = \theta_3\), indicating how spread out the values are around the mean, unaffected by these shifts.
Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental principle in statistics. It states that, given a sufficiently large sample size, the distribution of the sample mean will be approximately normally distributed, regardless of the original distribution.

In the context of our hypothesis test, this theorem is used to justify why the test statistic \(T\), which compares the differences in means of our two samples, follows a standard normal distribution under the null hypothesis. The sample means comprise many independent random disturbances \(Z_i\), each with finite variance \(\theta_3\).

This aggregation causes the distribution of the test statistic to approach a normal distribution as the sample size increases. Therefore, despite potentially non-normal characteristics of individual data points, CLT guarantees the reliability of using normal distribution assumptions for large samples.
  • This makes hypothesis tests such as these quite robust and efficient.
  • It also means that we can use the \(N(0,1)\) distribution because it provides a meaningful threshold for determining if the observed data significantly deviate from what the null hypothesis predicts.
Probability Density Function
A probability density function (pdf), which is denoted by \(f(z)\), describes the likelihood of a random variable to take on a particular value. For continuous random variables, the pdf gives us the relative likelihood for this random variable to obtain some specific value.

In a location model like the one described, \(Z_i\) is assumed to have a common pdf \(f(z)\) with mean zero and finite variance \(\theta_3\). This pdf helps in specifying the distribution shape of \(Z_i\) and thus plays an essential role in establishing the properties of \(X_i\) and \(Y_i\).

A key attribute of a pdf is that the area under the curve across its range is equal to 1. This property is fundamental, as it represents the total probability, ensuring that outcomes spanning all possible values are accounted for. Additionally, the pdf is employed in various statistical methods to derive not only means and variances but also to conduct transformative operations necessary for inferential statistics, like calculating the probability of events in a hypothesis test.
  • The pdf provides the foundation for computing expectations and variances.
  • Understanding the pdf of \(Z_i\) is crucial to maintain consistency in assumptions about \(X_i\) and \(Y_i\).

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Most popular questions from this chapter

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 \(\left(X_{1}, Y_{1}\right),\left(X_{2}, Y_{2}\right), \ldots,\left(X_{n}, Y_{n}\right)\) 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?

The effect that a certain drug (Drug A) has on increasing blood pressure is a major concern. It is thought that a modification of the drug (Drug B) will lessen the increase in blood pressure. Let \(\mu_{A}\) and \(\mu_{B}\) be the true mean increases in blood pressure due to Drug \(\mathrm{A}\) and \(\mathrm{B}\), respectively. The hypotheses of interest are \(H_{0}: \mu_{A}=\mu_{B}=0\) versus \(H_{1}: \mu_{A}>\mu_{B}=0 .\) The two-sample \(t\) -test statistic discussed in Example \(8.3 .3\) is to be used to conduct the analysis. The nominal level is set at \(\alpha=0.05\) For the experimental design assume that the sample sizes are the same; i.e., \(m=n .\) Also, based on data from Drug \(A, \sigma=30\) seems to be a reasonable selection for the common standard deviation. Determine the common sample size, so that the difference in means \(\mu_{A}-\mu_{B}=12\) has an \(80 \%\) detection rate. Suppose when the experiment is over, due to patients dropping out, the sample sizes for Drugs \(A\) and \(B\) are respectively \(n=72\) and \(m=68 .\) What was the actual power of the experiment to detect the difference of \(12 ?\)

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 \(X_{1}, \ldots, X_{n}\) denote a random sample from a gamma-type distribution with \(\alpha=2\) and \(\beta=\theta\). Let \(H_{0}: \theta=1\) and \(H_{1}: \theta>1\). (a) Show that there exists a uniformly most powerful test for \(H_{0}\) against \(H_{1}\), determine the statistic \(Y\) upon which the test may be based, and indicate the nature of the best critical region. (b) Find the pdf of the statistic \(Y\) in part (a). If we want a significance level of \(0.05\), write an equation that can be used to determine the critical region. Let \(\gamma(\theta), \theta \geq 1\), be the power function of the test. Express the power function as an integral.

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