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Chapter 14: Problem 485

Consider a distribution \(\mathrm{N}\left(\mu, \sigma^{2}\right)\) where \(\mu\) is known but \(\sigma^{2}\) is not. Devise a method of producing a confidence interval for \(\sigma^{2}\)

Short Answer

Expert verified
A method for constructing a confidence interval for the unknown variance \(\sigma^2\) in a normal distribution with a known mean \(\mu\) involves using the Chi-squared distribution. Given a sample, we can calculate the sum of the squared differences between each sample point and the mean, and scale this by the unknown variance to create the scaled Chi-squared variable \(Q\). Then, we find the upper and lower limits corresponding to the desired confidence level, and finally invert the \(Q\) variable to express the confidence interval for \(\sigma^2\) as: $$\left[\frac{\sum_{i=1}^n (X_i - \mu)^2}{U}, \frac{\sum_{i=1}^n (X_i - \mu)^2}{L}\right]$$

Step by step solution

01

Identifying the Relevant Distribution

Since we are given a normal distribution and want to find a confidence interval for the variance, we will be using the Chi-squared distribution. The sample variance will follow a scaled chi-squared distribution. If we have a sample from a normal distribution with known mean \(\mu\) but unknown variance \(\sigma^2\), we can define the scaled chi-squared variable as: $$ Q = \frac{\sum_{i=1}^n (X_i - \mu)^2}{\sigma^2} $$ where \(X_i\) are the sample points, and \(n\) is the sample size. Q will have a Chi-squared distribution with \(n\) degrees of freedom.
02

Determine the Confidence Interval for Q

We need to find the range of values for Q such that we can say with some level of confidence (\(1-\alpha\)) that it contains the true value of Q. This involves finding the lower limit, \(L\), and the upper limit, \(U\), for the Chi-squared distribution corresponding to the confidence level. For a \((1-\alpha)\) confidence interval, we have: $$P\left( L \le Q \le U \right) = 1-\alpha$$ Here, \(\alpha\) is the significance level and \(L\) and \(U\) are the values of Q that correspond to the lower and upper tail probabilities \((\frac{\alpha}{2}, 1-\frac{\alpha}{2})\) respectively. We can now find \(L\) and \(U\) by using the Chi-squared distribution table or software for a given number of degrees of freedom \((n)\) and significance level \((\alpha)\).
03

Construct the Confidence Interval for \(\sigma^2\)

To create a confidence interval for \(\sigma^2\), we need to invert the definition of our scaled Chi-squared variable \(Q\). Since we know that: $$L \le Q = \frac{\sum_{i=1}^n (X_i - \mu)^2}{\sigma^2} \le U$$ We can now obtain the confidence interval for the variance by dividing the sum of squares by \(U\) and \(L\), respectively, and reversing the inequalities: $$\frac{\sum_{i=1}^n (X_i - \mu)^2}{U} \le \sigma^2 \le \frac{\sum_{i=1}^n (X_i - \mu)^2}{L}$$ Thus, the confidence interval for the unknown variance \(\sigma^2\) is: $$\left[\frac{\sum_{i=1}^n (X_i - \mu)^2}{U}, \frac{\sum_{i=1}^n (X_i - \mu)^2}{L}\right]$$

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Key Concepts

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

Chi-squared distribution
When studying various types of data, the Chi-squared distribution plays a crucial role, particularly in scenarios dealing with variance and goodness-of-fit tests. It is a statistical distribution that arises frequently when dealing with normally distributed data. In essence, if we take several independent variables that are normally distributed, square them, and sum them up, the resulting distribution will be a Chi-squared distribution.

It's imperative to understand that the Chi-squared distribution is characterized by its degrees of freedom, which typically equate to the number of independent variables in the sum. In the context of constructing a confidence interval for variance, we use the Chi-squared distribution because it allows us to relate the sum of squared deviations from the mean (which forms a part of the variance calculation) to a probability distribution. Since variance is effectively a scaled sum of squared deviations, a Chi-squared distribution becomes the natural choice.
Normal distribution
The normal distribution, also known as the Gaussian distribution, is a bell-shaped curve that is symmetrical around the mean. It describes a continuous probability distribution for a random variable. Within the field of statistics, it's one of the most commonly assumed distributions for modeling various phenomena due to its natural occurrence in many biological, social, and physical processes.

When we are given a dataset assumed to be normally distributed, the calculation of variance becomes inherently meaningful as it reflects the spread of the data around the mean. The normal distribution's properties facilitate the derivation of a confidence interval for the variance, a process that involves the use of the Chi-squared distribution when the mean is known but the variance is to be estimated.
Significance level

Understanding Significance Level

The significance level, denoted by \(\alpha\), is a critical concept in statistics that indicates the threshold for rejecting a null hypothesis. It defines how confident we want to be in our results, and typically, it is chosen as 0.05 (or 5%), representing a 95% confidence level. When computing confidence intervals, the significance level determines how narrow or wide the interval will be鈥攖he lower the significance level, the wider the confidence interval, signifying greater uncertainty in our estimate.

Impact on Interval Estimation

In the context of our variance estimation scenario, the significance level tells us what percentage of the time the true variance will lie outside our confidence interval. By setting the significance level, we are effectively controlling the type I error rate in our interval estimation. We select values from the Chi-squared distribution that correspond to the tails defined by this significance level to construct our confidence interval.
Degrees of freedom
Degrees of freedom commonly abbreviated as 'df', are a measure of the amount of 'free' variability in a dataset or in the estimation of a statistical parameter. Simply said, they represent the number of independent values that can vary in the analysis without breaking any constraints.

In the situation of estimating a variance using a sample, the degrees of freedom are related to the number of observations we have. In the Chi-squared distribution, degrees of freedom are vitally important because they shape the distribution's form and spread. When we estimate the sample variance, we are imposing a constraint (the sample mean), and thus, we lose one degree of freedom. Therefore, if we use all 'n' data points to estimate the variance, the degrees of freedom for the Chi-squared distribution will be \(n - 1\). Understanding the concept of degrees of freedom helps us correctly apply and interpret the Chi-squared distribution tables or software outputs when constructing confidence intervals for variance.

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Most popular questions from this chapter

A factory has a machine that cuts spaghetti and noodles into appropriate lengths for packaging. It can be adjusted to cut any given length. The variability of the machine is determined to be a standard deviation of o. \(1 \mathrm{~cm}\). The machine is set to cut pasta \(30 \mathrm{~cm}\) in length and you take a number of 36 noodle samples. According to the normal distribution table, how often will \(\underline{X}\) be within \(0.0167 \mathrm{~cm}\) of \(\mu\) ?

A survey was conducted in 1970 to determine the average hourly earnings of a female sales clerk employed by a department store in metropolitan Los Angeles. A simple random sample of 225 female clerks was selected and the following information obtained: \(\mathrm{X}=\) hourly wage rate earned by female sales clerk, $$ \sum \mathrm{x}=\$ 450.00, \Sigma(\mathrm{x}-\overline{\mathrm{x}})^{2}=\$ 2016.00 $$ What is the 99 confidence interval estimate of the average hourly wage rate?

There was an old woman who lived in a shoe. She had so many children she didn't know what to do, so she measured their heights. First she sampled 13 girls and found a sample mean \(\underline{\mathrm{X}}_{\mathrm{G}}\), or \(31.4\) inches. Next she sampled 7 boys and calculated a sample mean \(\underline{\mathrm{X}}_{\mathrm{B}}\) of \(38.1\) inches. From the U.S. Statistical Abstract she found that the variance in girls' heights is \(100 \mathrm{in}^{2}\) and 80 in \(^{2}\) in boys. Help the old girl find a 95 per cent confidence interval for the difference in height between the heights of her boys and her girls, \(\mu_{\mathrm{B}}-\mu_{\mathrm{G}}\). Assume normality.

In a public opinion poll, 320 out of 400 persons interviewed supported the administration's policy on disarmament. (a) Establish a \(95 \%\) confidence interval estimate of the proportion of persons supporting the government's stand on disarmament. (b) What can be concluded with \(.99\) confidence about the maximum error in our estimate if the proportion of those supporting the administration policy is estimated as \(80 \%\) ? (c) With what degree of confidence can we assert that the proportion of persons supporting the administration policy is somewhere between \(77 \%\) and \(83 \% ?\)

In the Idaho State Home for Runaway Girls, 25 residents were polled as to what age they ran away from home. The sample mean was 16 years old with a standard deviation of \(1.8\) years. Establish a \(95 \%\) confidence interval for \(\mu\), the mean age at which runaway girls leave home in Idaho.
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