/*! 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 13 Consider a random sample of size... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider a random sample of size \(n\) from a continuous distribution having median 0 so that the probability of any one observation being positive is .5. Disregarding the signs of the observations, rank them from smallest to largest in absolute value, and let \(W=\) the sum of the ranks of the observations having positive signs. For example, if the observations are \(-.3,+.7,+2.1\), and \(-2.5\), then the ranks of positive observations are 2 and 3 , so \(W=5\). In Chapter \(15, W\) will be called Wilcoxon's signed-rank statistic. W can be represented as follows: a. Determine \(E(Y)\) and then \(E(W)\) using the equation for \(W\). [Hint: The first \(n\) positive integers sum to \(n(n+1) / 2 .]\) b. Determine \(V\left(Y_{i}\right)\) and then \(V(W)\). [Hint: The sum of the squares of the first \(n\) positive integers can be expressed as \(n(n+1)(2 n+1) / 6\).] $$ \begin{aligned} W &=1 \cdot Y_{1}+2 \cdot Y_{2}+3 \cdot Y_{3}+\cdots+n \cdot Y_{n} \\ &=\sum_{i=1}^{n} i \cdot Y_{i} \end{aligned} $$ where the \(Y_{i}\) 's are independent Bernoulli rv's, each with \(p=.5\left(Y_{i}=1\right.\) corresponds to the observation with rank \(i\) being positive).

Short Answer

Expert verified
\(E(W) = \frac{n(n+1)}{4}, V(W) = \frac{n(n+1)(2n+1)}{24}.\)

Step by step solution

01

Understand the Wilcoxon Signed-Rank Statistic Definition

The sum of the ranks of the positive observations, denoted as \( W \), is called the Wilcoxon signed-rank statistic. The focus is on ranks and is calculated based on the absolute values of observations. Positive observations are assigned ranks, and their sum is \( W \).
02

Formulate the Expected Value of \( Y_i \)

Given that each \( Y_i \) is a Bernoulli random variable with probability \( p = 0.5 \), the expected value \( E(Y_i) \) is calculated as follows: \[ E(Y_i) = 1 \cdot 0.5 + 0 \cdot 0.5 = 0.5. \]
03

Calculate the Expected Value of \( W \)

Using \( W = \sum_{i=1}^{n} i \cdot Y_{i} \), the expected value \( E(W) \) becomes: \[ E(W) = E\left( \sum_{i=1}^n i \cdot Y_i \right) = \sum_{i=1}^n i \cdot E(Y_i) = \sum_{i=1}^n i \cdot 0.5. \]By the summation formula, sum of the first \( n \) positive integers is \( \frac{n(n+1)}{2} \), hence \[ E(W) = 0.5 \cdot \frac{n(n+1)}{2} = \frac{n(n+1)}{4}. \]
04

Determine the Variance of \( Y_i \)

The variance of a Bernoulli random variable \( Y_i \) with parameter \( p = 0.5 \) is computed as:\[ V(Y_i) = p(1-p) = 0.5 \cdot 0.5 = 0.25. \]
05

Calculate the Variance of \( W \)

Since \( W = \sum_{i=1}^{n} i \cdot Y_i \), its variance is:\[ V(W) = V\left(\sum_{i=1}^n i \cdot Y_i \right) = \sum_{i=1}^n i^2 \cdot V(Y_i) = \sum_{i=1}^n i^2 \cdot 0.25. \]Using the formula for the sum of squares \( \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6} \), we get:\[ V(W) = 0.25 \cdot \frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)(2n+1)}{24}. \]

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Key Concepts

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

Probability Theory and Random Variables
Probability theory is the mathematical framework that helps us understand uncertainty, randomness, and the likelihood of different outcomes. It provides the backbone for statistical analysis, enabling us to describe the behavior of random variables and compute quantities like probabilities, expected values, and variances.

In the context of the Wilcoxon signed-rank test, a continuous random variable represents each observation in a sample. These observations can be positive or negative, and we are interested in their ranks and signs. The test uses this probabilistic understanding to make inferences about a population's median.

In the given exercise, observations are assumed to be from a distribution with a median of zero, implying that each observation has a 50% probability of being positive. This fundamental probability concept helps us calculate the expected value of the sum of ranks of positive observations.
Understanding Bernoulli Random Variables
To better grasp the Wilcoxon signed-rank test, it's essential to understand Bernoulli random variables. Named after the Swiss mathematician Jacob Bernoulli, these are the simplest kinds of random variables, which can take only two possible outcomes, typically 0 or 1. They are fundamental in probability theory.

In our exercise, each observation is represented by a Bernoulli random variable \(Y_i\), where \(Y_i = 1\) if the observation is positive and \(Y_i = 0\) if it is not. With the probability of each observation being positive set at \(p = 0.5\), we can calculate the expected value and variance for each \(Y_i\).
  • The expected value \(E(Y_i) = 0.5\), reflecting the 50% chance of a positive observation.
  • The variance \(V(Y_i) = 0.25\), indicating the variability in whether an observation is positive or not.
This simple Bernoulli process facilitates understanding how the sum of ranks \(W\) behaves and allows for the calculation of its mean and variance.
Basics of Statistical Analysis and Wilcoxon Signed-Rank Test
Statistical analysis involves the collection, exploration, and interpretation of data. The Wilcoxon signed-rank test falls under non-parametric statistical tests, used to assess whether the median of differences between pairs is significantly different from zero.

In this exercise, we focus on the Wilcoxon signed-rank statistic \(W\), which is derived from ranking the observations based on their absolute values and summing the ranks of the positive ones.
  • It provides insights into the central tendency of a dataset without assuming a normal distribution. This feature makes it very useful for data that does not meet the normal distribution assumptions.
  • The calculation of the expected value \(E(W)\) and variance \(V(W)\) of \(W\) helps us understand its distribution under the null hypothesis.
By using the rank sums, this test reduces the impact of outliers and skewed data, offering a robust method for analyzing paired sample differences.

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Most popular questions from this chapter

Suppose the expected tensile strength of type-A steel is \(105 \mathrm{ksi}\) and the standard deviation of tensile strength is \(8 \mathrm{ksi}\). For type-B steel, suppose the expected tensile strength and standard deviation of tensile strength are \(100 \mathrm{ksi}\) and \(6 \mathrm{ksi}\), respectively. Let \(\bar{X}=\) the sample average tensile strength of a random sample of 40 type-A specimens, and let \(\bar{Y}=\) the sample average tensile strength of a random sample of 35 type-B specimens. a. What is the approximate distribution of \(\bar{X}\) ? Of \(\bar{Y}\) ? b. What is the approximate distribution of \(\bar{X}-\bar{Y}\) ? Justify your answer. c. Calculate (approximately) \(P(-1 \leq \bar{X}-\bar{Y} \leq 1)\). d. Calculate \(P(\bar{X}-\bar{Y} \geq 10)\). If you actually observed \(\bar{X}-\bar{Y} \geq 10\), would you doubt that \(\mu_{1}-\mu_{2}=5\) ?

A service station has both self-service and full-service islands. On each island, there is a single regular unleaded pump with two hoses. Let \(X\) denote the number of hoses being used on the self-service island at a particular time, and let \(Y\) denote the number of hoses on the full-service island in use at that time. The joint pmf of \(X\) and \(Y\) appears in the accompanying tabulation. \begin{tabular}{ll|ccc} \(p(x, y)\) & & 0 & 1 & 2 \\ \hline \multirow{x}{*}{\(x\)} & 0 & \(.10\) & \(.04\) & \(.02\) \\ & 1 & \(.08\) & \(.20\) & \(.06\) \\ & 2 & \(.06\) & \(.14\) & \(.30\) \end{tabular} a. What is \(P(X=1\) and \(Y=1)\) ? b. Compute \(P(X \leq 1\) and \(Y \leq 1)\). c. Give a word description of the event \(\\{X \neq 0\) and \(Y \neq 0\\}\), and compute the probability of this event. d. Compute the marginal pmf of \(X\) and of \(Y\). Using \(p_{X}(x)\), what is \(P(X \leq 1)\) ? e. Are \(X\) and \(Y\) independent rv's? Explain.

A shipping company handles containers in three different sizes: (1) \(27 \mathrm{ft}^{3}(3 \times 3 \times 3)\), (2) \(125 \mathrm{ft}^{3}\), and (3) \(512 \mathrm{ft}^{3}\). Let \(X_{i}(i=1,2,3)\) denote the number of type \(i\) containers shipped during a given week. With \(\mu_{i}=E\left(X_{i}\right)\) and \(\sigma_{i}^{2}=V\left(X_{i}\right)\), suppose that the mean values and standard deviations are as follows: $$ \begin{array}{lll} \mu_{1}=200 & \mu_{2}=250 & \mu_{3}=100 \\ \sigma_{1}=10 & \sigma_{2}=12 & \sigma_{3}=8 \end{array} $$ a. Assuming that \(X_{1}, X_{2}, X_{3}\) are independent, calculate the expected value and variance of the total volume shipped. [Hint: Volume \(=27 X_{1}+125 X_{2}+512 X_{3}\).] b. Would your calculations necessarily be correct if the \(X_{i}^{\prime}\) s were not independent? Explain.

Annie and Alvie have agreed to meet for lunch between noon (0:00 P.M.) and 1:00 P.M. Denote Annie's arrival time by \(X\), Alvie's by \(Y\), and suppose \(X\) and \(Y\) are independent with pdf's $$ \begin{aligned} &f_{X}(x)=\left\\{\begin{array}{cc} 3 x^{2} & 0 \leq x \leq 1 \\ 0 & \text { otherwise } \end{array}\right. \\ &f_{X}(y)=\left\\{\begin{array}{cl} 2 y & 0 \leq y \leq 1 \\ 0 & \text { otherwise } \end{array}\right. \end{aligned} $$ What is the expected amount of time that the one who arrives first must wait for the other person? [Hint: \(h(X, Y)=|X-Y| .]\)

Two different professors have just submitted final exams for duplication. Let \(X\) denote the number of typographical errors on the first professor's exam and \(Y\) denote the number of such errors on the second exam. Suppose \(X\) has a Poisson distribution with parameter \(\mu_{1}, Y\) has a Poisson distribution with parameter \(\mu_{2}\), and \(X\) and \(Y\) are independent. a. What is the joint pmf of \(X\) and \(Y\) ? b. What is the probability that at most one error is made on both exams combined? c. Obtain a general expression for the probability that the total number of errors in the two exams is \(m\) (where \(m\) is a nonnegative integer). [Hint: \(A=\) \(\\{(x, y): x+y=m\\}=\\{(m, 0),(m-1,1), \ldots\), \((1, m-1),(0, m)\\}\). Now sum the joint pmf over \((x, y) \in A\) and use the binomial theorem, which says that $$ \sum_{k=0}^{m}\left(\begin{array}{c} m \\ k \end{array}\right) a^{k} b^{m-k}=(a+b)^{m} $$ for any \(a, b .]\)

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