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Chapter 6: Problem 47

Why should \(\chi_{v}^{2}\) be approximately normal for large \(v\) ? What theorem applies here, and why?

Short Answer

Expert verified
The Central Limit Theorem ensures that \(\chi_v^2\) is approximately normal for large \(v\) due to the sum of many independent variables.

Step by step solution

01

Identify the Distribution

The symbol \(\chi_v^2\) represents a chi-squared distribution with \(v\) degrees of freedom. A chi-squared distribution is derived from the sum of the squares of \(v\) independent standard normal random variables.
02

Understand the Central Limit Theorem

The Central Limit Theorem (CLT) states that when independent random variables are added, their properly normalized sum tends toward a normal distribution, even if the original variables themselves are not normally distributed.
03

Apply the Central Limit Theorem

As the degrees of freedom \(v\) increases, the chi-squared distribution becomes the sum of many independent normal variables. According to the CLT, this sum, thus \(\chi_v^2\), should become approximately normal.
04

Relation to Chi-Squared Distribution

Specifically, for large \(v\), the chi-squared distribution \(\chi_v^2\) can be approximated by a normal distribution with mean \(v\) and variance \(2v\), given by \(N(v, 2v)\). This reflects in the trait of normal approximation accurately provided by the CLT.

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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. It arises from the sum of the squares of independent standard normal random variables. This means if you take several random variables that follow a normal distribution (mean = 0, variance = 1), square each one, and then add them up, the result will follow a chi-squared distribution.

This distribution is very useful in a wide range of applications, particularly in hypothesis testing and construction of confidence intervals.
  • It's denoted by \(\chi_v^2\), where \(v\) stands for the degrees of freedom.
  • The shape of the chi-squared distribution depends on the degrees of freedom. More degrees lead to a more symmetric shape.
  • A chi-squared distribution with low degrees of freedom is skewed right, but as degrees increase, the distribution becomes more symmetric.
Degrees of freedom
Degrees of freedom are a crucial part of understanding distributions in statistics. When we talk about degrees of freedom, we're referring to the number of independent values or quantities that can vary in an analysis.

Let's break it down:
  • In the case of a chi-squared distribution, the degrees of freedom \(v\) are determined by the number of squares you summed to get the chi-squared value.
  • The degrees of freedom impact the shape of the chi-squared distribution. The higher the degrees of freedom, the more the distribution resembles a normal distribution.
A higher number of degrees of freedom essentially means averaging more variables, which leads to a smoother, more bell-shaped curve.
Normal distribution approximation
Normal distribution approximation is an important statistical concept used to simplify the analysis of more complex distributions like the chi-squared distribution.

Thanks to the Central Limit Theorem, when the degrees of freedom increase, the chi-squared distribution starts to resemble a normal distribution with mean \(v\) and variance \(2v\). This is because:
  • With a large \(v\), we sum a large number of independent variables, each contributing slightly to the total.
  • The CLT states that the distribution of the sum approaches a normal distribution, simplifying theoretical and practical work.
This normal approximation is useful because the normal distribution is mathematically well-behaved and easier to work with for calculations involving probabilities and statistical testing.
Theorems in statistics
Theorems in statistics such as the Central Limit Theorem (CLT) provide foundational insights that enable us to make sense of complex data distributions. The CLT is a powerful statistical tool because:
  • It explains that, under certain conditions, the sum of a large enough number of random variables is approximately normally distributed, regardless of the original distribution.
  • This theorem applies to chi-squared distributions by stating that as the number of degrees of freedom (equivalently, the size of the dataset) becomes large, the distribution approximates a normal distribution.
This approximation might seem abstract, but it's invaluable in practical terms. Theorems like the CLT allow statisticians to predict, innovate, and create models based on real-world data, making statistical methods more robust and widely applicable.

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Most popular questions from this chapter

It is known that \(80 \%\) of all brand A DVD players work in a satisfactory manner throughout the warranty period (are "successes"). Suppose that \(n=10\) players are randomly selected. Let \(X=\) the number of successes in the sample. The statistic \(X / n\) is the sample proportion (fraction) of successes. Obtain the sampling distribution of this statistic. [Hint: One possible value of \(X / n\) is \(.3\), corresponding to \(X=3\). What is the probability of this value (what kind of random variable is \(X)\) ?]

Let \(\mu\) denote the true \(\mathrm{pH}\) of a chemical compound. A sequence of \(n\) independent sample pH determinations will be made. Suppose each sample \(\mathrm{pH}\) is a random variable with expected value \(\mu\) and standard deviation .1. How many determinations are required if we wish the probability that the sample average is within \(.02\) of the true \(\mathrm{pH}\) to be at least \(.95\) ? What theorem justifies your probability calculation?

A concert has three pieces of music to be played before intermission. The time taken to play each piece has a normal distribution. Assume that the three times are independent of each other. The mean times are 15,30 , and \(20 \mathrm{~min}\), respectively, and the standard deviations are 1,2, and \(1.5 \mathrm{~min}\), respectively. What is the probability that this part of the concert takes at most \(1 \mathrm{~h}\) ? Are there reasons to question the independence assumption? 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.

Suppose we take 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. We now disregard the signs of the observations, rank them from smallest to largest in absolute value, and then 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 \(14, W\) will be called Wilcoxon's signed-rank statistic. W can be represented as follows: $$ \begin{aligned} W &=1 \cdot Y_{1}+2 \cdot Y_{2}+3 \cdot Y_{3}+\cdots+n \cdot Y_{n} \\ &=\sum_{i=1}^{n} i \cdot Y_{i} \end{aligned} $$ where the \(Y_{i}^{\prime}\) s are independent Bernoulli rv's, each with \(p=.5\left(Y_{i}=1\right.\) corresponds to the observation with rank \(i\) being positive). Compute the following: a. \(E\left(Y_{i}\right)\) and then \(E(W)\) using the equation for \(W\) [Hint: The first \(n\) positive integers sum to \(n(n+1) / 2 .]\) b. \(V\left(Y_{i}\right)\) and then \(V(W)\) [Hint: The sum of the squares of the first \(n\) positive integers is \(n(n+1)(2 n+1) / 6 .]\)
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