/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 91 Let \(Y_{1}, Y_{2}, \ldots, Y_{2... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let \(Y_{1}, Y_{2}, \ldots, Y_{20}\) be a random sample of size \(n=20\) from a normal distribution with unknown mean \(\mu\) and known variance \(\sigma^{2}=5 .\) We wish to test \(H_{0}: \mu=7\) versus \(H_{a}: \mu>7\) a. Find the uniformly most powerful test with significance level. \(05 .\) b. For the test in part (a), find the power at each of the following alternative values for \(\mu\) : \(\mu_{\alpha}=7.5,8.0,8.5,\) and9.0. c. Sketch a graph of the power function.

Short Answer

Expert verified
Use a z-test, reject if \( Z > 1.645 \). Power increases as \( \mu \) goes from 7.5 to 9.0.

Step by step solution

01

Define the Hypotheses

In this problem, we need to test the null hypothesis: \( H_{0}: \mu = 7 \) against the alternative hypothesis: \( H_{a}: \mu > 7 \). This is a one-sided hypothesis test.
02

Identify the Test Statistic

Since the population variance \( \sigma^2 \) is known, we use the z-test. The test statistic can be calculated using the formula: \[ Z = \frac{\bar{Y} - \mu_0}{\sigma/\sqrt{n}} \] where \( \bar{Y} \) is the sample mean, \( \mu_0 = 7 \), \( \sigma = \sqrt{5} \), and \( n = 20 \).
03

Determine the Critical Region

For a significance level of 0.05 and a one-tailed test, the critical z-value (z-critical) can be found from the standard normal distribution table. \( P(Z > z_{0.05}) = 0.05 \) corresponds to \( z_{0.05} \approx 1.645 \). So, we reject \( H_0 \) if \( Z > 1.645 \).
04

Calculate the Power for Each Alternative Mean

The power of the test is calculated by finding the probability of correctly rejecting \( H_0 \) when the alternative is true. The power function is given by: \[ P(\text{Reject } H_0 | \mu = \mu_a) = P\left( Z > 1.645 \right) \text{ for } Z = \frac{\bar{Y} - 7}{\sigma/\sqrt{n}} \] For each \( \mu_a = 7.5, 8.0, 8.5, \) and \( 9.0 \), calculate \( P(\text{Reject } H_0) = P\left( \frac{\bar{Y} - \mu}{\sigma/\sqrt{n}} > \frac{7 - \mu}{\sigma/\sqrt{n}} + 1.645 \right) \).
05

Calculate Power for \(\mu_a = 7.5\)

For \(\mu_a = 7.5\), the power is:\[ P\left( Z > \frac{7-7.5}{\sqrt{5}/\sqrt{20}} + 1.645 \right) \]Compute the z-value and find the probability using a normal distribution table or calculator.
06

Calculate Power for \(\mu_a = 8.0\)

For \(\mu_a = 8.0\), the power is:\[ P\left( Z > \frac{7-8.0}{\sqrt{5}/\sqrt{20}} + 1.645 \right) \]Find the associated probability from the normal distribution.
07

Calculate Power for \(\mu_a = 8.5\)

For \(\mu_a = 8.5\), the power is:\[ P\left( Z > \frac{7-8.5}{\sqrt{5}/\sqrt{20}} + 1.645 \right) \]Find the z-value and use it to compute the probability.
08

Calculate Power for \(\mu_a = 9.0\)

For \(\mu_a = 9.0\), the power is:\[ P\left( Z > \frac{7-9.0}{\sqrt{5}/\sqrt{20}} + 1.645 \right) \]Determine the probability from the z-table or tool.
09

Sketch the Power Function

Plot the power of the test against \( \mu \) for each alternative (\( 7.5, 8.0, 8.5, \) and \( 9.0 \)). The x-axis represents the alternative means and the y-axis represents the power. The curve will show how the power increases with increasing \( \mu \).

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Key Concepts

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

Normal Distribution
Normal distribution is a fundamental concept in statistics, often described as a bell-shaped curve. It is a continuous probability distribution characterized by:
  • a mean (\( \mu \)) which determines the center of the distribution
  • a variance (\( \sigma^2 \)) which influences the width or spread of the curve
In this exercise, we have a sample taken from a population that follows a normal distribution. The normal distribution is symmetric about its mean, and it’s used frequently due to its excellent properties related to the Central Limit Theorem.
When dealing with normal distribution, it's important to understand that it is fully described by its mean and variance. Hence, even if only the mean or variance is unknown, the distribution itself remains fully predictable given the other parameter is known.
In hypothesis testing of this exercise, we exploit the properties of normal distribution to assess whether the sample came from a population with a specific mean.
Z-Test
The z-test is a statistical method employed when dealing with data assumed to follow a normal distribution with a known variance. It helps to determine whether there's a significant difference between the sample mean and a known population mean.
When executing a z-test:
  • We calculate a statistic called the z-value or z-score using the formula \[ Z = \frac{\bar{Y} - \mu_0}{\sigma/\sqrt{n}} \]where \( \bar{Y} \)is the sample mean, \( \mu_0 \) is the population mean under the null hypothesis, \( \sigma \) is the standard deviation, and \( n \) is the sample size.
  • The calculated z-score tells us how many standard deviations the sample mean is from the population mean.
  • We compare this z-score to a critical value from the z-table to determine whether we reject the null hypothesis.

In this context, we use the z-test to check if our sample mean significantly deviates from the hypothesized population mean of 7.
Power Function
The power function in hypothesis testing is a crucial concept that measures the test's ability to correctly reject a false null hypothesis. It is the probability that the test will lead to a rejection of \( H_0 \) when an alternative hypothesis \( H_a \) is true.
Power is especially important because:
  • It quantifies the effectiveness of a test.
  • A higher power means a higher probability of detecting a true effect if one exists.
  • The power is affected by factors such as sample size, effect size, significance level, and variance.

In the exercise, the power function helps illustrate the test's ability to reject \( H_0 \) for several alternative values of \( \mu \). By plotting the power against alternative means, we see how it increases as \( \mu \) moves away from \( H_0 \), demonstrating the test's increasing sensitivity.
Significance Level
The significance level, denoted as \( \alpha \), is a threshold used in hypothesis testing to determine whether a result is statistically significant. It represents the probability of rejecting the null hypothesis when it is actually true, which is also known as a Type I error.
  • Common values for \( \alpha \) are 0.05, 0.01, and 0.10. In this exercise, it's set to 0.05.
  • It defines the critical region: test results falling in this region lead to rejection of \( H_0 \).
  • A smaller \( \alpha \) indicates stricter criteria for significance but increases the risk of Type II error, which is failing to reject a false null hypothesis.

Choosing the right significance level is a balance: lower \( \alpha \) means more confidence in the results, while higher \( \alpha \) means easier rejection of \( H_0 \). For this problem, the critical z-value corresponding to the 0.05 significance level is approximately 1.645, establishing the boundary for making a statistical decision on \( H_0 \).

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Most popular questions from this chapter

For a normal distribution with mean \(\mu\) and variance \(\sigma^{2}=25,\) an experimenter wishes to test \(H_{0}:\) \(\mu=10\) versus \(H_{a}: \mu=5 .\) Find the sample size \(n\) for which the most powerful test will have \(\alpha=\beta=.025\)

Let \(X_{1}, X_{2}, \ldots, X_{m}\) denote a random sample from the exponential density with mean \(\theta_{1}\) and let \(Y_{1}, Y_{2}, \ldots, Y_{n}\) denote an independent random sample from an exponential density with \(\theta_{2}\). a. Find the likelihood ratio criterion for testing \(H_{0}: \theta_{1}=\theta_{2}\) versus \(H_{a}: \theta_{1} \neq \theta_{2}\). b. Show that the test in part (a) is equivalent to an exact \(F\) test [Hint: Transform \(\sum X_{i}\) and \(\sum Y_{j}\) to \(\left.\chi^{2} \text { random variables. }\right]\)

A large-sample \(\alpha\) -level test of hypothesis for \(H_{0}: \theta=\theta_{0}\) versus \(H_{a}: \theta>\theta_{0}\) rejects the null hypothesis if $$\frac{\hat{\theta}-\theta_{0}}{\sigma_{\hat{\theta}}}>z_{\alpha}$$ Show that this is equivalent to rejecting \(H_{0}\) if \(\theta_{0}\) is less than the large-sample \(100(1-\alpha) \%\) lower confidence bound for \(\theta\)

A study was conducted by the Florida Game and Fish Commission to assess the amounts of chemical residues found in the brain tissue of brown pelicans. In a test for DDT, random samples of \(n_{1}=10\) juveniles and \(n_{2}=13\) nestlings produced the results shown in the accompanying table (measurements in parts per million, ppm). $$\begin{array}{ll} \hline \text { Juveniles } & \text { Nestlings } \\ \hline n_{1}=10 & n_{2}=13 \\ \bar{y}_{1}=.041 & \bar{y}_{2}=.026 \\ s_{1}=.017 & s_{2}=.006 \\ \hline \end{array}$$ a. Test the hypothesis that mean amounts of DDT found in juveniles and nestlings do not differ versus the alternative, that the juveniles have a larger mean. Use \(\alpha=.05 .\) (This test has important implications regarding the accumulation of DDT over time.) b. Is there evidence that the mean for juveniles exceeds that for nestlings by more than. 01 ppm? i. Bound the \(p\) -value, using a table in the appendix. ii. Find the exact \(p\) -value, using the appropriate applet.

A political researcher believes that the fraction \(p_{1}\) of Republicans strongly in favor of the death penalty is greater than the fraction \(p_{2}\) of Democrats strongly in favor of the death penalty. He acquired independent random samples of 200 Republicans and 200 Democrats and found 46 Republicans and 34 Democrats strongly favoring the death penalty. Does this evidence provide statistical support for the researcher's belief? Use \(\alpha=.05\)

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