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Chapter 8: Problem 97

Suppose that \(S^{2}\) is the sample variance based on a sample of size \(n\) from a normal population with unknown mean and variance. Derive a \(100(1-\alpha) \%\) a. upper confidence bound for \(\sigma^{2}\). b. lower confidence bound for \(\sigma^{2}\).

Short Answer

Expert verified
Upper bound is \(\frac{(n-1)S^2}{\chi^2_{n-1}(1-\alpha)}\), lower bound is \(\frac{(n-1)S^2}{\chi^2_{n-1}(\alpha)}\)."

Step by step solution

01

Understanding the Problem

We want to find confidence bounds for the population variance \(\sigma^2\), given that we have a sample variance \(S^2\) from a normal population. The sample size is \(n\). Confidence bounds are intervals where we expect the true variance \(\sigma^2\) to lie with certain confidence. We'll first derive the upper bound and then the lower bound for \(\sigma^2\).
02

Theoretical Background

In a normal distribution, the sample variance \((n-1)S^2/\sigma^2\) follows a chi-square distribution with \((n-1)\) degrees of freedom, i.e., \(\frac{(n-1)S^2}{\sigma^2} \sim \chi^2_{n-1}\). This property is useful for constructing confidence intervals and bounds for \(\sigma^2\).
03

Upper Confidence Bound Derivation

To derive the upper bound, we use the \(\chi^2\) distribution to state that \(\frac{(n-1)S^2}{\sigma^2} \leq \chi^2_{n-1}(1-\alpha)\) with probability \(1-\alpha\). Rearranging, we get \(\sigma^2 \leq \frac{(n-1)S^2}{\chi^2_{n-1}(1-\alpha)}\).
04

Lower Confidence Bound Derivation

For the lower confidence bound, consider that \(\frac{(n-1)S^2}{\sigma^2} \geq \chi^2_{n-1}(\alpha)\) with probability \(\alpha\). Rearranging, this becomes \(\sigma^2 \geq \frac{(n-1)S^2}{\chi^2_{n-1}(\alpha)}\).
05

Constructing the Bounds

The derived bounds provide the interval boundaries. The \(100(1-\alpha)\%\) upper confidence bound for \(\sigma^2\) is \(\frac{(n-1)S^2}{\chi^2_{n-1}(1-\alpha)}\), while the lower confidence bound is \(\frac{(n-1)S^2}{\chi^2_{n-1}(\alpha)}\). These results depend on the chi-square distribution quantiles.

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Key Concepts

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

Chi-square Distribution
The chi-square distribution is a fundamental concept in statistics, often used when dealing with sample variances and constructing confidence intervals for the population variance. This distribution arises when each of the numbers is squared, following a normal distribution. It plays a significant role in our original exercise, particularly in determining the confidence bounds for the variance.
  • The chi-square distribution is asymmetric and skews to the right.
  • It is defined by \(n-1\) degrees of freedom, which ties directly to the sample size in focus.
  • As the degrees of freedom increase, the chi-square distribution increasingly resembles a normal distribution.
By leveraging the properties of this distribution, we can derive both upper and lower confidence bounds for the variance. Understanding it allows us to confidently assert if our sample variance reflects the population variance accurately. This makes it an essential tool in statistical inference.
Sample Variance
Sample variance, denoted as \(S^2\), is a key statistic that represents the dispersion of sample values around the mean in a set of observations. It is a crucial estimator of the population variance when samples are drawn from a normally distributed population.A key point about sample variance:
  • It is calculated by summing the squared differences between each observation and the sample mean, then dividing by \(n-1\) (where \(n\) is the sample size).
  • This adjustment, \(n-1\), instead of \(n\), is made to correct the bias in the estimation of the population variance from a small sample, a concept known as Bessel's correction.
The sample variance serves as an unbiased, efficient, and consistent estimator for the population variance, making it fundamental in statistical practice and theoretical approaches.
Population Variance
Population variance provides a measure of how data points in the entire population are spread out around the mean. Unlike sample variance, it is a parameter that describes the dispersion in a whole dataset rather than just a part of it.
  • It is usually denoted by \(\sigma^2\).
  • The calculation involves summing the squared differences of each data point from the population mean, then dividing by the number of data points \(N\). This means complete data without adjustments like \(n-1\).
Population variance is crucial when modeling data as it provides insights into the variability inherent in the entire dataset, rather than sampling fluctuations. Accurately estimating population variance from sample variance is a frequent goal in data analysis and is central to our confidence interval development.
Degrees of Freedom
Degrees of freedom (d.o.f.) is an intriguing concept in statistics that essentially tells us how many values in a calculation are free to vary. Interpreted in the context of variance, it influences both the calculation of sample variance and the chi-square distribution.
  • In our context, it often represents \(n-1\), where \(n\) is the sample size. The \(-1\) accounts for the number of parameters estimated (e.g., the mean).
  • Degrees of freedom are critical in shaping the chi-square distribution, which we use for confidence interval calculations.
Understanding degrees of freedom is imperative when working with statistical tests and models, ensuring correct interpretations and robust statistical practices. It directly impacts our ability to construct meaningful confidence intervals, as seen in the textbook problem's solutions.

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Most popular questions from this chapter

Suppose that \(Y_{1}, Y_{2}, Y_{3}\) denote a random sample from an exponential distribution with density function $$f(y)=\left\\{\begin{array}{ll} \left(\frac{1}{\theta}\right) e^{-y / \theta}, & y>0 \\ 0, & \text { elsewhere } \end{array}\right.$$ Consider the following five estimators of \(\theta\) : $$\hat{\theta}=Y_{1}, \quad \hat{\theta}_{2}=\frac{Y_{1}+Y_{2}}{2}, \quad \hat{\theta}_{3}=\frac{Y_{1}+2 Y_{2}}{3}, \quad \hat{\theta}_{4}=\min \left(Y_{1}, Y_{2}, Y_{3}\right), \quad \hat{\theta}_{5}=\bar{Y}$$ a. Which of these estimators are unbiased? b. Among the unbiased estimators, which has the smallest variance?

Exercise 8.129 suggests that \(S^{2}\) is superior to \(S^{\prime 2}\) in regard to bias and that \(S^{\prime 2}\) is superior to \(S^{2}\) because it possesses smaller variance. Which is the better estimator? [Hint: Compare the mean square errors.]

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The ages of a random sample of five university professors are \(39,54,61,72,\) and \(59 .\) Using this information, find a \(99 \%\) confidence interval for the population standard deviation of the ages of all professors at the university, assuming that the ages of university professors are normally distributed.
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