/*! 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 27 Let \(X\) have the beta distribu... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(X\) have the beta distribution on \([0,1]\) with parameters \(\alpha=v_{1} / 2\) and \(\beta=v_{2} / 2\), where \(v_{1} / 2\) and \(v_{2} / 2\) are positive integers. Define \(Y=\) \((X / \alpha) /[(1-X) / \beta]\). Show that \(Y\) has the \(F\) distribution with degrees of freedom \(v_{1}, v_{2}\).

Short Answer

Expert verified
Y has an F-distribution with degrees of freedom \(v_1\) and \(v_2\).

Step by step solution

01

Understanding the Definitions

The beta distribution is a family of continuous probability distributions defined on the interval \([0,1]\). The parameters \(\alpha\) and \(\beta\) help to shape the distribution. When \(\alpha = v_1/2\) and \(\beta = v_2/2\), we use these parameters' properties in our transformation to solve the problem.
02

Define the Transformation

Given the transformation \(Y = \frac{(X/\alpha)}{((1-X)/\beta)}\), we need to determine the distribution of \(Y\). By rewriting, we get \(Y = \frac{X \beta}{(1-X) \alpha}\). This indicates that \(Y\) will be related to a distribution of the form \(\frac{U_1/v_1}{U_2/v_2}\) where \(U_1\) and \(U_2\) are chi-squared distributed random variables.
03

Relationship with the F-distribution

The F-distribution is expressed as the ratio of two scaled chi-squared distributions: \(F_{v_1,v_2} \sim \frac{U_1/v_1}{U_2/v_2}\). To show \(Y\) has an F-distribution, establish that \(X\) follows a beta distribution which can be rewritten in terms of similar chi-squared distributions due to its relation with such components.
04

Establish the Chi-squared Distribution

A Beta random variable \(X\) with parameters \(\alpha\) and \(\beta\) is equivalent to \(\frac{U_1}{U_1 + U_2}\) where \(U_1\) and \(U_2\) are independent chi-squared random variables with degrees of freedom \(2\alpha\) and \(2\beta\) respectively. Substituting \(\alpha = v_1/2\) and \(\beta = v_2/2\), we get \(U_1 \sim \chi^2_{v_1}\) and \(U_2 \sim \chi^2_{v_2}\).
05

Concluding the Transformation

Substitute back to show \(Y = \frac{U_1/v_1}{U_2/v_2}\). Thus, \(Y\) follows \(F\)-distribution with degrees of freedom \(v_1\) and \(v_2\). Therefore, \(Y\), given as \(\frac{X \beta}{(1-X) \alpha}\), confirms the transformation to the desired F-distribution.

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Key Concepts

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

F Distribution
The **F Distribution** is a continuous probability distribution that arises frequently in statistics, particularly in the context of variance analysis and hypothesis testing. It characterizes the ratio of two independent chi-squared distributed random variables, each divided by their respective degrees of freedom. In simpler terms, it's used to compare two variances and determine if they significantly differ.
When we say a random variable follows an F distribution, denoted as \( F_{v_1,v_2} \), it means this variable is the ratio \( \frac{U_1/v_1}{U_2/v_2} \), where \(U_1\) and \(U_2\) are independent chi-squared random variables with degrees of freedom \(v_1\) and \(v_2\), respectively. This specific ratio is crucial in the design of experiments and analysis of variance (ANOVA).
  • The F distribution is asymmetric and positively skewed, making it suitable for testing non-negative variances.
  • It's used in comparing statistical models, often in the context of finding which model better exaplins or fits the data.
Overall, understanding the F distribution allows researchers to conduct robust tests of variance, which is particularly useful in fields such as biology, engineering, and economics.
Chi-Squared Distribution
The **Chi-Squared Distribution** is another vital continuous probability distribution in statistics. It represents the sum of the squares of independent standard normal random variables, often symbolized as \( \chi^2 \). This distribution is used extensively in hypothesis testing and construction of confidence intervals, especially when dealing with variance of a normal distribution.
  • The chi-squared distribution is always non-negative and skewed to the right, especially for lower degrees of freedom.
  • The shape of the chi-squared distribution curve depends on the degrees of freedom, \( k \). As \( k \) increases, the distribution becomes more symmetric and approaches a normal distribution.
This distribution is fundamental in the development of other distributions as well, such as the F-distribution. In the problem at hand, the chi-squared distribution helps in framing the beta distribution in terms of more primitive distribution components, paving the way to identifying \( Y \) as an F-distribution. Understanding its properties is key to various analyses, including goodness-of-fit tests and tests for independence.
Continuous Probability Distributions
**Continuous Probability Distributions** are fundamental concepts in probability and statistics, describing phenomena that can take any value within a given range. Unlike discrete probability distributions, which are concerned with particular outcomes, continuous distributions describe the probabilities of outcomes within intervals, making them essential for modeling real-world continuous data.
Examples of continuous probability distributions include:
  • **Normal Distribution:** Often used in natural and social sciences to represent real-valued random variables with a bell-shaped probability density function.
  • **Exponential Distribution:** Describes time between events in a Poisson process, crucial in fields like queuing theory and reliability testing.
  • **Beta Distribution:** Flexible in modeling probabilities and proportions, emphasizing finite-range scenarios as highlighted in the exercise above.
Understanding continuous probability distributions allows for precise modeling and analysis in numerous fields ranging from physics to finance. They form the bedrock for advanced statistical methods and complex decision-making models, making their study indispensable for aspiring statisticians and data scientists.

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Most popular questions from this chapter

A random sample of 15 automobile mechanics certified to work on a certain type of car was selected, and the time (in minutes) necessary for each one to diagnose a particular problem was determined, resulting in the following data: \(\begin{array}{llllllll}30.6 & 30.1 & 15.6 & 26.7 & 27.1 & 25.4 & 35.0 & 30.8 \\\ 31.9 & 53.2 & 12.5 & 23.2 & 8.8 & 24.9 & 30.2 & \end{array}\) Use the Wilcoxon test at significance level 10 to decide whether the data suggests that true average diagnostic time is less than 30 minutes.

The accompanying data resulted from an experiment to compare the effects of vitamin \(\mathrm{C}\) in orange juice and in synthetic ascorbic acid on the length of odontoblasts in guinea pigs over a 6-week period ("The Growth of the Odontoblasts of the Incisor Tooth as a Criterion of the Vitamin C Intake of the Guinea Pig," J. Nutrit., 1947: 491-504). Use the Wilcoxon rank-sum test at level \(.01\) to decide whether true average length differs for the two types of vitamin \(C\) intake. Compute also an approximate \(P\)-value. [Hint: See Exercise 14.] \(\begin{array}{lrrrrrr}\text { Orange Juice } & 8.2 & 9.4 & 9.6 & 9.7 & 10.0 & 14.5 \\ & 15.2 & 16.1 & 17.6 & 21.5 & & \\ \text { Ascorbic Acid } & 4.2 & 5.2 & 5.8 & 6.4 & 7.0 & 7.3 \\ & 10.1 & 11.2 & 11.3 & 11.5 & & \end{array}\)

Assume a random sample \(X_{1}, X_{2}, \ldots, X_{n}\) from the Poisson distribution with mean \(\lambda\). If the prior distribution for \(\lambda\) has a gamma distribution with parameters \(\alpha\) and \(\beta\), show that the posterior distribution is also gamma distributed. What are its parameters?

Both a gravimetric and a spectrophotometric method are under consideration for determining phosphate content of a particular material. Twelve samples of the material are obtained, each is split in half, and a determination is made on each half using one of the two methods, resulting in the following data: $$ \begin{aligned} &\begin{array}{l|cccc} \text { Sample } & 1 & 2 & 3 & 4 \\ \hline \text { Gravimetric } & 54.7 & 58.5 & 66.8 & 46.1 \\ \hline \text { Spectrophotometric } & 55.0 & 55.7 & 62.9 & 45.5 \end{array}\\\ &\begin{array}{l|cccc} \text { Sample } & 5 & 6 & 7 & 8 \\ \hline \text { Gravimetric } & 52.3 & 74.3 & 92.5 & 40.2 \\ \hline \text { Spectrophotometric } & 51.1 & 75.4 & 89.6 & 38.4 \end{array}\\\ &\begin{array}{l|cccc} \text { Sample } & 9 & 10 & 11 & 12 \\ \hline \text { Gravimetric } & 87.3 & 74.8 & 63.2 & 68.5 \\ \hline \text { Spectrophotometric } & 86.8 & 72.5 & 62.3 & 66.0 \end{array} \end{aligned} $$ Use the Wilcoxon test to decide whether one technique gives on average a different value than the other technique for this type of material.

The single-factor ANOVA model considered in Chapter 11 assumed the observations in the \(i\) th sample were selected from a normal distribution with mean \(\mu_{i}\) and variance \(\sigma^{2}\), that is, \(X_{i j}=\mu_{i}+\varepsilon_{i j}\) where the \(\varepsilon\) 's are normal with mean 0 and variance \(\sigma^{2}\). The normality assumption implies that the \(F\) test is not distribution-free. We now assume that the \(\varepsilon\) 's all come from the same continuous, but not necessarily normal, distribution, and develop a distribution-free test of the null hypothesis that all \(I \mu_{i}\) 's are identical. Let \(N=\sum J_{i}\), the total number of observations in the data set (there are \(J_{i}\) observations in the \(i\) th sample). Rank these \(N\) observations from 1 (the smallest) to \(N\), and let \(\bar{R}_{i}\) be the average of the ranks for the observations in the ith sample. When \(H_{0}\) is true, we expect the rank of any particular observation and therefore also \(\bar{R}_{i}\) to be \((N+1) / 2\). The data argues against \(H_{0}\) when some of the \(\bar{R}_{i}\) 's differ considerably from \((N+1) / 2\). The Kruskal-Wallis test statistic is $$ K=\frac{12}{N(N+1)} \sum J_{i}\left(\bar{R}_{i}-\frac{N+1}{2}\right)^{2} $$ When \(H_{0}\) is true and either (1) \(I=3\), all \(J_{i} \geq 6\) or (2) \(I>3\), all \(J_{i} \geq 5\), the test statistic has approximately a chi-squared distribution with \(I-1\) df. The accompanying observations on axial stiffness index resulted from a study of metal-plate connected trusses in which five different plate lengths-4 in., 6 in., 8 in., 10 in., and 12 in. were used ("Modeling Joints Made with LightGauge Metal Connector Plates," Forest Products \(J ., 1979: 39-44)\). \(\begin{array}{lllll}i=1(4 \text { in. }): & 309.2 & 309.7 & 311.0 & 316.8 \\\ & 326.5 & 349.8 & 409.5 & \\ i=2(6 \text { in. }): & 331.0 & 347.2 & 348.9 & 361.0 \\ & 381.7 & 402.1 & 404.5 & \\ i=3(8 \text { in. }): & 351.0 & 357.1 & 366.2 & 367.3 \\ & 382.0 & 392.4 & 409.9 & \\ i=4(10 \text { in. }): & 346.7 & 362.6 & 384.2 & 410.6 \\ & 433.1 & 452.9 & 461.4 & \\ i=5(12 \text { in. }): & 407.4 & 410.7 & 419.9 & 441.2 \\ & 441.8 & 465.8 & 473.4 & \end{array}\) Use the \(K-W\) test to decide at significance level \(.01\) whether the true average axial stiffness index depends somehow on plate length.

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