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Let \(X_{1}, X_{2}, \ldots, X_{25}\) denote a random sample of size 25 from a normal distribution \(N(\theta, 100) .\) Find a uniformly most powerful critical region of size \(\alpha=0.10\) for testing \(H_{0}: \theta=75\) against \(H_{1}: \theta>75\).

Short Answer

Expert verified
The uniformly most powerful critical region of size \(\alpha=0.10\) for testing the hypothesis \(H_{0}: \theta=75\) against \(H_{1}: \theta>75\) is \(\bar{X} > 77.56\), i.e. \(H_{0}\) should be rejected if the sample mean exceeds 77.56.

Step by step solution

01

Identify the Statistical Test

In this case, we are testing a hypothesis about the mean of a normally distributed population, where the population standard deviation is known. Therefore, we will use the Z test.
02

Formulate the Hypotheses

The null hypothesis \(H_{0}: \theta=75\) states that the population mean is 75. The alternative hypothesis \(H_{1}: \theta>75\) states that the population mean is greater than 75.
03

Calculate the Test Statistic

The formula for the Z test statistic is: \[ Z = \frac{\bar{X} - \theta_{0}}{\sigma/\sqrt{n}} \] where \(\bar{X}\) is the sample mean, \(\theta_{0}\) is the hypothesized population mean under the null hypothesis, \(\sigma\) is the population standard deviation, and \(n\) is the size of the sample. Here, the hypothesized value \(\theta_{0} = 75\), \(\sigma = 10\) (which is the square root of the variance 100), and \(n = 25\).
04

Determine the Rejection Region

Knowing that \(\alpha = 0.10\) and we reject the null hypothesis if the mean is greater than 75, we want to find the critical value \(Z_{0}\) such that the probability that a standard normal random variable is greater than \(Z_{0}\) is 0.10. From the standard normal tables, we find that \(Z_{0} \approx 1.28\). Thus, we reject the null hypothesis if our computed test statistic \(Z > Z_{0}\).
05

Express the Rejection Region in Terms of \(\bar{X}\)

This means we reject the null hypothesis if \[ \bar{X} > \theta_{0} + Z_{0} \cdot \sigma/\sqrt{n} \] Substituting in the known values, we get: \[ \bar{X}_{crit} = 75 + 1.28 \cdot (10/\sqrt{25}) = 75 + 1.28 \cdot 2 = 77.56 \] Therefore, the critical region, which is the set of all sample outcomes that lead us to reject the null hypothesis, in our case, means that we reject \(H_{0}: \theta=75\) if \(\bar{X} > 77.56\).

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a statistical process used to make decisions about a population parameter based on sample data. In the context of a uniformly most powerful test for a normal distribution, we aim to determine if there is significant evidence to reject a null hypothesis, which represents a default position or a claim of 'no effect'. The alternative hypothesis is a statement that contradicts the null hypothesis, suggesting that there is an effect or a difference.

The steps involve formulating two opposing hypotheses, calculating a test statistic from the sample, determining the critical region, and then making a decision. If the test statistic falls within the critical region, the null hypothesis is rejected in favor of the alternative. The level of significance, denoted by \(\alpha\), is predetermined and reflects the probability of rejecting the null hypothesis when it is actually true, known as a Type I error.
Normal Distribution
The normal distribution, often represented by the bell-shaped curve, is a continuous probability distribution that is symmetric around its mean. It's characterized by two parameters: the mean (\(\mu\)) and the variance (\(\sigma^2\)). In hypothesis testing, we frequently assume that the population from which we sample is normally distributed because of the Central Limit Theorem, which states that the sampling distribution of the sample mean will be approximately normal, regardless of the population distribution, as long as the sample size is sufficiently large.

In the given exercise, the population variance is stated to be 100, which means the standard deviation \(\sigma\) is the square root of the variance, yielding a value of 10. Knowing the shape of the normal distribution is crucial as it helps us define the critical region for the Z test.
Z Test
A Z test is a type of hypothesis test that is used when the data is approximately normally distributed and the population variance is known. The Z statistic provides a means of determining how many standard deviations an element is from the mean. The formula for the Z test statistic is:
\[ Z = \frac{\bar{X} - \theta_{0}}{\sigma/\sqrt{n}} \]
where \(\bar{X}\) is the sample mean, \(\theta_{0}\) is the hypothesized population mean under the null hypothesis, \(\sigma\) is the population standard deviation, and \(n\) is the sample size. In our example, the test determines whether the true mean \(\theta\) is greater than 75 given the sample mean, assuming that sample data follows a normal distribution with known standard deviation.
Critical Region
The critical region, also known as the rejection region, is the range of values for which the null hypothesis is not considered plausible and is therefore rejected. It's determined using the significance level \(\alpha\), which is the probability of making a Type I error. In the context of a Z test, this region corresponds to the tails of the standard normal distribution beyond a certain Z value.

For a one-tailed test with \(\alpha = 0.10\), the critical value \(Z_{0}\) that marks the boundary of the critical region is found using statistical tables or software. Any test statistic that falls beyond this value indicates that the sample provides enough evidence to reject the null hypothesis. The critical region is straightforward to determine once we have \(\alpha\), the hypothesized mean, and the population standard deviation.

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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 \(c_{1}(n)\)

Verzani (2014) discusses a data set on healthy individuals, including their temperatures by gender. The data are in the file tempbygender. rda and the variables of interest are maletemp and femaletemp. Download this file from the site listed in the Preface. (a) Obtain comparison boxplots. Comment on the plots. Which, if any, gender seems to have lower temperatures? Based on the width of the boxplots, comment on the assumption of equal variances. (b) As discussed in Example 8.3.3, compute the two-sample, two-sided \(t\) -test that there is no difference in the true mean temperatures between genders. Obtain the \(p\) -value of the test and conclude in terms of the problem at the nominal \alpha-level of \(0.05 .\) (c) Obtain a \(95 \%\) confidence interval for the difference in means. What does it mean in terms of the problem?

Illustrative Example \(8.2 .1\) of this section dealt with a random sample of size \(n=2\) from a gamma distribution with \(\alpha=1, \beta=\theta .\) Thus the mgf of the distribution is \((1-\theta t)^{-1}, t<1 / \theta, \theta \geq 2 .\) Let \(Z=X_{1}+X_{2} .\) Show that \(Z\) has a gamma distribution with \(\alpha=2, \beta=\theta\). Express the power function \(\gamma(\theta)\) of Example 8.2.1 in terms of a single integral. Generalize this for a random sample of size \(n\).

Let \(X\) be \(N(0, \theta)\) and, in the notation of this section, let \(\theta^{\prime}=4, \theta^{\prime \prime}=9\), \(\alpha_{a}=0.05\), and \(\beta_{a}=0.10 .\) Show that the sequential probability ratio test can be based upon the statistic \(\sum_{1}^{n} X_{i}^{2} .\) Determine \(c_{0}(n)\) and \(c_{1}(n)\).

Let \(X\) and \(Y\) have a joint bivariate normal distribution. An observation \((x, y)\) arises from the joint distribution with parameters equal to either $$ \mu_{1}^{\prime}=\mu_{2}^{\prime}=0, \quad\left(\sigma_{1}^{2}\right)^{\prime}=\left(\sigma_{2}^{2}\right)^{\prime}=1, \quad \rho^{\prime}=\frac{1}{2} $$ Ior $$ \mu_{1}^{\prime \prime}=\mu_{2}^{\prime \prime}=1, \quad\left(\sigma_{1}^{2}\right)^{\prime \prime}=4, \quad\left(\sigma_{2}^{2}\right)^{\prime \prime}=9, \quad \rho^{\prime \prime}=\frac{1}{2} \text { . } $$ Show that the classification rule involves a second-degree polynomial in \(x\) and \(y\).

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