/*! 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 654 Consider the probability distrib... [FREE SOLUTION] | 91影视

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Consider the probability distribution function $$ \mathrm{f}(\mathrm{x} ; \theta)=(1 / \theta) \mathrm{e}^{-\mathrm{x} / \theta} \quad 0<\mathrm{x}<\infty $$ $$ =0 $$ elsewhere. It is desired to test the hypothesis \(\mathrm{H}_{0}: \theta=2\) against alternate hypothesis \(\mathrm{H}_{1}: \theta>2\). Suppose a random sample \(\mathrm{X}_{1}, \mathrm{X}_{2}\) is used and the critical region is \(\mathrm{X}_{1}+\mathrm{X}_{2} \geq 9.5\) Calculate an expression for the power function, \(\mathrm{K}\left(\theta_{1}\right)\), for all \(\theta_{1}>2\), and specifically for \(\theta_{1}=4\).

Short Answer

Expert verified
The power function, K(胃鈧), for all 胃鈧>2 is given by: \( K(\theta_1) = 1 - \int_{0}^{\infty} \frac{1}{\theta_1^2} e^{-\frac{x_1}{\theta_1}} \int_{9.5-x_1}^{\infty} e^{-\frac{x_2}{\theta_1}} dx_2 dx_1 \) For 胃鈧=4, the power function is: \( K(4) = 1 - \int_{0}^{\infty} \frac{1}{4^2} e^{-\frac{x_1}{4}} \int_{9.5-x_1}^{\infty} e^{-\frac{x_2}{4}} dx_2 dx_1 \) Evaluating the integral gives the numerical value for K(4).

Step by step solution

01

Define the power function

The power function, K(胃鈧), represents the probability of correctly rejecting the null hypothesis when the true value of 胃 is 胃鈧. It can be defined as the complement of the probability of a Type II error, which is the probability of not rejecting the null hypothesis when the true value of 胃 is 胃鈧. So, K(胃鈧) = 1 - P(Type II error).
02

Calculate the joint pdf

We are given a random sample X鈧, X鈧. So, we need to calculate the joint pdf of X鈧 and X鈧, f(x鈧, x鈧; 胃鈧), which can be found by multiplying the individual pdfs: \( f(x_1, x_2; \theta_1) = \frac{1}{\theta_1} e^{-\frac{x_1}{\theta_1}} \cdot \frac{1}{\theta_1} e^{-\frac{x_2}{\theta_1}} = \frac{1}{\theta_1^2} e^{-\frac{x_1+x_2}{\theta_1}} \) for 0 < x鈧, x鈧 < 鈭 and 胃鈧 > 0.
03

Determine the probability of a Type II error given the critical region

The critical region is defined by X鈧 + X鈧 鈮 9.5. We need to find the probability of a Type II error, which is the probability of not rejecting the null hypothesis when the true value of 胃 is 胃鈧. This can be calculated by integrating the joint pdf over the critical region: \( P(\text{Type II error}) = \int_{0}^{\infty} \int_{9.5-x_1}^{\infty} f(x_1, x_2; \theta_1) dx_2 dx_1 = \int_{0}^{\infty} \frac{1}{\theta_1^2} e^{-\frac{x_1}{\theta_1}} \int_{9.5-x_1}^{\infty} e^{-\frac{x_2}{\theta_1}} dx_2 dx_1 \)
04

Calculate the power function, K(胃鈧), for all 胃鈧>2

Using the expressions from Steps 1 and 3, we can now calculate the power function, K(胃鈧), for all 胃鈧 > 2: \( K(\theta_1) = 1 - P(\text{Type II error}) = 1 - \int_{0}^{\infty} \frac{1}{\theta_1^2} e^{-\frac{x_1}{\theta_1}} \int_{9.5-x_1}^{\infty} e^{-\frac{x_2}{\theta_1}} dx_2 dx_1 \)
05

Calculate the power function, K(胃鈧), for 胃鈧=4

To calculate the power function for 胃鈧=4, we simply substitute 胃鈧=4 into the expression from Step 4: \( K(4) = 1 - \int_{0}^{\infty} \frac{1}{4^2} e^{-\frac{x_1}{4}} \int_{9.5-x_1}^{\infty} e^{-\frac{x_2}{4}} dx_2 dx_1 \) Evaluating this integral gives us a numerical value for the power function, K(4), when 胃鈧=4.

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Key Concepts

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

Probability Density Function
Understanding the probability density function (PDF) is crucial when assessing statistical hypotheses. The PDF tells us how the probabilities are distributed over the values of a continuous random variable. For instance, in the given exercise, the PDF is expressed as a function of a parameter \( \theta \) and the random variable \( x \): \( f(x; \theta) = \frac{1}{\theta} e^{-\frac{x}{\theta}} \) for \( 0 < x < \infty \), and zero elsewhere.

This mathematical expression reveals how likely different values of \( x \) are when \( \theta \) is known. When you're solving hypothesis testing problems like the one described, it's essential to understand the shape and the characteristics of this PDF because it's the foundation for calculating probabilities, including those associated with errors and critical regions in hypothesis testing.

For a better grasp of the concepts, visualize the PDF as a curve on a graph where the area under the curve represents the probability. In the context of our example, the exponential decay nature of the function indicates that as \( x \) increases, the probability of observing such a value decreases, adhering to the properties of an exponential distribution.
Type II Error
A Type II error in hypothesis testing is the mistake of failing to reject a false null hypothesis. When solving hypothesis testing problems, it's as significant to understand the risk of committing this error as it is to know about the more commonly discussed Type I error (incorrectly rejecting a true null hypothesis).

In the scenario provided by the exercise, calculating the power function involves determining the complement of the probability of a Type II error. In essence, the power function, \( K(\theta_1) = 1 - P(\text{Type II error}) \), represents the probability that the test will correctly identify an effect (i.e., reject a false null hypothesis) when the true parameter value is \( \theta_1 \).

In practice, we strive for tests with low probabilities of Type II errors to ensure reliability in detecting true effects. When the test detects deviations from the null hypothesis with high probability, it is said to have high power. It's important for students to clearly differentiate between Type I and Type II errors to correctly interpret hypothesis testing results.
Critical Region
The critical region is another fundamental concept in hypothesis testing, which directly relates to the decision-making process. It is the set of all outcomes of an experiment that will lead to the rejection of the null hypothesis \( H_0 \).

In the given exercise, the critical region is defined by the condition \( X_1 + X_2 \geq 9.5 \). This means if the sum of our two independent, exponentially distributed random samples exceeds 9.5, we have sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis, \( H_1: \theta > 2 \).

To calculate the probability of landing in the critical region under different assumed true parameter values \( \theta_1 \), we integrate the joint PDF over the values that fall within this region. This probability helps us refine the hypothesis test and adjust the critical region to control the errors in the test鈥攊deally, minimizing both Type I and Type II errors.

Exercise Improvement Advice

When students are trying to understand the critical region, using a visual aid, like a graph to illustrate the region and its corresponding probabilities, can be a powerful tool for comprehension. It's imperative that students familiarize themselves with the balance between setting a critical region and the implications it has on committing Type I and Type II errors.

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Most popular questions from this chapter

A random sample of 120 students attending Florida State University has a mean age of \(20.2\) years and a standard deviation of \(1.2\) years while a random sample of 100 students attending the University of Florida has a mean age of 21 years and a standard deviation of \(1.5\) years. At a \(.05\) level of significance, can we conclude that the average age of the students at the two universities are not the same?

A college dean wants to determine if the students entering the college in a given year have higher IQ's than the students entering the same college in the previous year. The IQ's of the college entrants in the two years are known to be normally distributed and to have equal variances. A random sample of four students from this year's entering class gives IQ scores of \(110,113,116\), and 117. Another random sample of four students from last year's entering class gives IQ scores of \(109,111,112\), and 112 . At the \(5 \%\) level of significance, can we conclude that this year's students have a higher IQ than last years?

Suppose we have a sample with \(\mathrm{n}=10\) and \(\underline{\mathrm{Y}}=50 . \mathrm{We}\) wish to test \(\mathrm{H}_{0}: \mu=47\) against \(\mathrm{H}_{1}: \mu \neq 47 .\) We would like to know the probability given \(\mu=47\) of observing a random sample from the population with \(\underline{Y}=50 .\) We will reject \(\mathrm{H}_{0}\) if the probability is less than \(\alpha=.05\) that \(\mathrm{H}_{0}\) is true when \(\underline{Y}=50\). Assume for this sample that \(\sum(\mathrm{Y}-\mathrm{Y})^{2}=99.2250 .\) Calculate the estimated population variance, the estimated \(\sigma^{2} \underline{\mathrm{Y}}\) and the estimated slandered error of the mean. Then calculate the t statistic and determine whether \(\mathrm{H}_{0}\) can be rejected. Also, suppose \(\alpha=.01\). Within what limits may \(\underline{Y}\) vary without our having to reject \(\mathrm{H}_{0}\) ?

All boxes of a particular type of detergent indicate that they contain 21 ounces of detergent. A government agency receives many consumer complaints that the boxes contain less than 21 ounces. To check the consumers' complaints at the \(5 \%\) level of significance, the government agency buys 100 boxes of this detergent and finds that \(\underline{\mathrm{X}}=20.5\) ounces and \(\mathrm{S}=2\) ounces. Should the government agency order the seller to put more detergent into its boxes?

A toothpaste manufacturer wants to determine whether the addition of a certain chemical to his toothpaste will increase its ability to fight tooth decay. Two groups of 100 people each are chosen and an experiment is conducted. Of the 100 people using the toothpaste with the chemical addition for a specified period of time, the manufacturer finds that they have an average of 8 cavities and a standard deviation of 3 cavities. The average number of cavities of the 100 people using the toothpaste without the chemical addition is 9 with a standard deviation of \(4 .\) Can we conclude at the \(1 \%\) level of significance that the addition of the chemical increased the toothpaste's ability to fight tooth decay?

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