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Apply the Law of Large Numbers to show that \(\chi_{v}^{2} / v\) approaches 1 as \(v\) becomes large.

Short Answer

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\( \chi_{v}^{2} / v \) approaches 1 as \( v \) becomes large due to the Law of Large Numbers.

Step by step solution

01

Understand the Chi-Squared Distribution

A chi-squared random variable with "v" degrees of freedom, denoted by \( \chi_{v}^{2} \), is the sum of the squares of "v" independent standard normal random variables (i.e., Z-distributed variables whose mean is 0 and variance is 1). This means \( \chi_{v}^{2} = Z_1^2 + Z_2^2 + \, ... \, + Z_v^2 \).
02

Express Chi-Squared Distribution as Sample Mean

We need to show that \( \frac{\chi_{v}^{2}}{v} \rightarrow 1 \) as \( v \rightarrow \infty \). Notice that \( \frac{\chi_{v}^{2}}{v} \) can be expressed as the average of \( v \) independent \( Z^2 \) variables divided by \( v \), i.e., \( \frac{1}{v}(Z_1^2 + Z_2^2 + \, ... \, + Z_v^2) \). This is the sample mean of these squared normal variables.
03

Utilize the Law of Large Numbers

According to the Law of Large Numbers, the sample mean will converge to the expected value of the random variables from which the sample is drawn as the number of variables (or samples) becomes large. Here, the expected value of each \( Z_i^2 \) is 1, since \( Z_i \) is a standard normal variable with variance 1. Therefore, as \( v \to \infty \), the average \( \frac{\chi_{v}^{2}}{v} = \frac{1}{v}(Z_1^2 + Z_2^2 + \, ... \, + Z_v^2) \rightarrow 1 \).
04

Conclude Limit Behavior

Given that as \( v \to \infty \), the average sample mean of \( Z_i^2 \) terms converges to their expected mean, we conclude that \( \frac{\chi_{v}^{2}}{v} \rightarrow 1 \). Thus, the Law of Large Numbers perfectly illustrates and confirms that \( \chi_{v}^{2} / v \) approaches 1 as \( v \) becomes large.

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Key Concepts

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

Chi-Squared Distribution
The Chi-Squared distribution is a fundamental concept in statistics, especially when dealing with variances. It is denoted as \( \chi^2 \) and is the distribution of a sum of the squares of independent standard normal random variables. This distribution is frequently used in hypothesis testing and confidence interval estimation for variance. It's important to understand that a chi-squared random variable with \( v \) degrees of freedom is expressed as \( \chi^2_v = Z_1^2 + Z_2^2 + \ldots + Z_v^2 \), where each \( Z_i \) is a standard normal variable (mean \( 0 \), variance \( 1 \)).
Many properties make the chi-squared distribution unique and invaluable:
  • It is always non-negative because it involves squares of variables.
  • The shape of the distribution changes based on the degrees of freedom, \( v \).
  • For small \( v \), it's heavily skewed to the right, becoming more symmetric as \( v \) increases.
The Chi-Squared distribution is essential for understanding variances in data, particularly when identifying if a sample variance differs significantly from a hypothesized variance.
Degrees of Freedom
Degrees of freedom can sound a bit abstract, but they're a crucial component of many statistical analyses. They refer to the number of independent values or quantities that can vary in an analysis without violating any constraint. When working with the Chi-Squared distribution, the degrees of freedom \( v \) are particularly important, as they determine the form of the distribution.
Consider degrees of freedom as the number of pieces of information you have minus the number of parameters you need to estimate:
  • In the example of sum of squares of \( v \) standard normal variables forming a Chi-Squared distribution, \( v \) also corresponds to the degrees of freedom.
  • As degrees of freedom increase, the chi-squared distribution approximates a normal distribution, which is symmetrical.
Understanding degrees of freedom will help you make sense of how sample sizes impact statistical estimations and hypothesis tests.
Standard Normal Distribution
The Standard Normal Distribution is a special case of the normal distribution with a mean of 0 and a variance of 1. It's denoted as \( Z \). Every standard normal random variable \( Z \) plays a key role in creating more complex distributions like the Chi-Squared distribution.
Some key characteristics of the standard normal distribution include:
  • Bell-shaped and symmetric around the mean, making it a powerful tool for statistical analysis.
  • The central limit theorem suggests that, under certain conditions, the sum of a large number of random variables is approximately normally distributed, even if the original variables themselves are not normally distributed.
  • In the context of chi-squared distribution, the \( Z_i^2 \) terms used to create \( \chi^2 \) are squared values of standard normals.
The standard normal distribution is foundational for statistical procedures, allowing statisticians to make inferences about sample data through various transformations and standardizations.

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Most popular questions from this chapter

For males the expected pulse rate is \(70 / \mathrm{m}\) and the standard deviation is \(10 / \mathrm{m}\). For women the expected pulse rate is \(77 / \mathrm{m}\) and the standard deviation is \(12 / \mathrm{m}\). Let \(X=\) the sample average pulse rate for a random sample of 40 men and let \(\bar{Y}=\) the sample average pulse rate for a random sample of 36 women a. What is the approximate distribution of \(X\) ? Of \(Y\) ? b. What is the approximate distribution of \(\bar{X}-\bar{Y}\) ? Justify your answer. c. Calculate (approximately) the probability \(P(-2 \leq \bar{X}-\bar{Y} \leq 1)\). d. Calculate (approximately) \(P(X-Y \leq-15)\). If you actually observed \(X-Y \leq-15\), would you doubt that \(\mu_{1}-\mu_{2}=-7\) ? Explain.

Suppose the distribution of the time \(X\) (in hours) spent by students at a certain university on a particular project is gamma with parameters \(\alpha=50\) and \(\beta=2\). Because \(\alpha\) is large, it can be shown that \(X\) has approximately a normal distribution. Use this fact to compute the probability that a randomly selected student spends at most \(125 \mathrm{~h}\) on the project.

A large university has 500 single employees who are covered by its dental plan. Suppose the number of claims filed during the next year by such an employee is a Poisson rv with mean value \(2.3\). Assuming that the number of claims filed by any such employee is independent of the number filed by any other employee, what is the approximate probability that the total number of claims filed is at least 1200 ?

The tip percentage at a restaurant has a mean value of \(18 \%\) and a standard deviation of \(6 \%\). a. What is the approximate probability that the sample mean tip percentage for a random sample of 40 bills is between \(16 \%\) and \(19 \%\) ? b. If the sample size had been 15 rather than 40 , could the probability requested in part (a) be calculated from the given information?

A company maintains three offices in a region, each staffed by two employees. Information concerning yearly salaries (1000's of dollars) is as follows: $$ \begin{array}{lcccccc} \text { Office } & 1 & 1 & 2 & 2 & 3 & 3 \\ \text { Employee } & 1 & 2 & 3 & 4 & 5 & 6 \\ \text { Salary } & 29.7 & 33.6 & 30.2 & 33.6 & 25.8 & 29.7 \end{array} $$ a. Suppose two of these employees are randomly selected from among the six (without replacement). Determine the sampling distribution of the sample mean salary \(\bar{X}\). b. Suppose one of the three offices is randomly selected. Let \(X_{1}\) and \(X_{2}\) denote the salaries of the two employees. Determine the sampling distribution of \(X\). c. How does \(E(X)\) from parts (a) and (b) compare to the population mean salary \(\mu\) ?

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