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Chapter 8: Problem 41

Suppose that \(Y\) is normally distributed with mean 0 and unknown variance \(\sigma^{2} .\) Then \(Y^{2} / \sigma^{2}\) has a \(x^{2}\) distribution with 1 df. Use the pivotal quantity \(Y^{2} / \sigma^{2}\) to find a a. \(95 \%\) confidence interval for \(\sigma^{2}\) b. \(95 \%\) upper confidence limit for \(\sigma^{2}\). c. \(95 \%\) lower confidence limit for \(\sigma^{2}\).

Short Answer

Expert verified
CI is \( \frac{Y^2}{5.024} < \sigma^2 < \frac{Y^2}{0.002} \); upper limit is \( \frac{Y^2}{3.841} \); lower limit is \( \frac{Y^2}{0.004} \).

Step by step solution

01

Identify Given Information and Distribution

The problem states that the random variable \( Y \) follows a normal distribution with mean \( 0 \) and variance \( \sigma^2 \). Consequently, \( Y^2/\sigma^2 \) follows a \( \chi^2 \) distribution with 1 degree of freedom (df). This is our pivotal quantity, which we'll use to construct the confidence interval and limits for \( \sigma^2 \).
02

Set Up Confidence Interval Using Chi-Squared Distribution

To find a \(95\%\) confidence interval for \( \sigma^2 \), begin by recognizing that since \( Y^2/\sigma^2 \) follows a \( \chi^2 \) distribution with 1 df, we use the critical values from the \( \chi^2 \) distribution table. Specifically, for a \(95\%\) confidence interval, find the critical values \( \chi^2_{\alpha/2,1} \) and \( \chi^2_{1-\alpha/2,1} \). Here, \( \alpha = 0.05 \), so \( \chi^2_{0.025,1} \) and \( \chi^2_{0.975,1} \) are the values we need.
03

Calculate Confidence Interval Bounds

Using the symmetry of the \( \chi^2 \) distribution with 1 df, we find \( \chi^2_{0.025,1} = 5.024 \) and \( \chi^2_{0.975,1} = 0.002 \). The \(95\%\) confidence interval for \( \sigma^2 \) becomes \[ \frac{Y^2}{\chi^2_{0.025,1}} < \sigma^2 < \frac{Y^2}{\chi^2_{0.975,1}} \]. This translates into \[ \frac{Y^2}{5.024} < \sigma^2 < \frac{Y^2}{0.002} \].
04

Determine Upper Confidence Limit

For a \(95\%\) upper confidence limit for \( \sigma^2 \), calculate based solely on the upper critical value of the \( \chi^2 \) distribution. Specifically, the \(95\%\) upper confidence limit is \[ \frac{Y^2}{\chi^2_{0.05,1}} \]. Look up \( \chi^2_{0.05,1} \) and use that value to find \[ \frac{Y^2}{3.841} \].
05

Determine Lower Confidence Limit

For the \(95\%\) lower confidence limit for \( \sigma^2 \), we use the lower critical value. Thus the lower confidence limit is \[ \frac{Y^2}{\chi^2_{0.95,1}} \]. The \( \chi^2_{0.95,1} \) is approximately \( 0.004 \), leading to \[ \frac{Y^2}{0.004} \].

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Key Concepts

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

Normal Distribution
Understanding the normal distribution is key in many statistical calculations, including the one in this exercise.
Normal distribution, also known as the Gaussian distribution, is a universal concept describing how values are distributed. It has a characteristic bell-shaped curve when plotted.

The key features of a normal distribution include:
  • Symmetry around the mean: The left half mirrors the right.
  • The mean, median, and mode are all equal in value.
  • The distribution is completely described by its mean and variance.
In the exercise, we have a normally distributed variable, \( Y \), with mean \( 0 \) and unknown variance \( \sigma^2 \).
These properties of the normal distribution allow us to apply further statistical techniques, such as finding the confidence interval for variance using other distributions like chi-squared.
Chi-Squared Distribution
When dealing with variance estimation, the chi-squared distribution is particularly useful, especially in the context of normal distributions.The chi-squared distribution is a special distribution used primarily in hypothesis testing and constructing confidence intervals for a population variance.
It is derived from the sum of the squares of independent standard normal variables.

The characteristics of the chi-squared distribution include:
  • It is skewed to the right, especially with fewer degrees of freedom (df).
  • It only takes positive values as it deals with squared terms.
  • With its degrees of freedom increasing, it approaches a normal distribution shape.
In our exercise, \( Y^2/\sigma^2 \) follows a chi-squared distribution with 1 degree of freedom, which becomes a pivotal tool for estimating variance via confidence intervals for \( \sigma^2 \).
Pivotal Quantity
Pivotal quantities are instrumental in constructing confidence intervals without relying on unknown parameters.
They are functions of data and parameters, but crucially, their distribution does not depend on these parameters. Thus, they provide a stable basis for inference.
  • They simplify the problem by reducing reliance on unknown parameters.
  • They can be used to set limits and intervals around estimates.
In the problem, \( Y^2/\sigma^2 \) is identified as a pivotal quantity, distributed as chi-squared.
This property allows us to directly derive confidence intervals for \( \sigma^2 \) by using discrete values from the chi-squared distribution, thus finding accurate estimates of variance.
Variance Estimation
Variance estimation is a critical task in statistics, portraying data spread. It helps in understanding how much the values differ from the mean. Smaller variances indicate that the data is more tightly clustered around the mean, while larger variances show more spread out data.

The exercise involves calculating confidence intervals for the variance \( \sigma^2 \) by using the chi-squared distribution and the pivotal quantity \( Y^2/\sigma^2 \).
  • Confidence intervals provide a range within which the true variance likely falls.
  • Upper and lower limits provide bounds, helping to understand the possible variance spread.
These estimates guide decisions under uncertainty by offering a probabilistic insight into the variability of the data in question.
By constructing these intervals and limits, we can have a better understanding of the true population variance with a given level of confidence.

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Most popular questions from this chapter

If \(Y_{1}, Y_{2}, \ldots, Y_{n}\) denote a random sample from an exponential distribution with mean \(\theta\), then \(E\left(Y_{i}\right)=\theta\) and \(V\left(Y_{i}\right)=\theta^{2} .\) Thus, \(E(\bar{Y})=\theta\) and \(V(\bar{Y})=\theta^{2} / n,\) or \(\sigma_{\bar{Y}}=\theta / \sqrt{n} .\) Suggest an unbiased estimator for \(\theta\) and provide an estimate for the standard error of your estimator.

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The EPA has set a maximum noise level for heavy trucks at 83 decibels (dB). The manner in which this limit is applied will greatly affect the trucking industry and the public. One way to apply the limit is to require all trucks to conform to the noise limit. A second but less satisfactory method is to require the truck fleet's mean noise level to be less than the limit. If the latter rule is adopted, variation in the noise level from truck to truck becomes important because a large value of \(\sigma^{2}\) would imply that many trucks exceed the limit, even if the mean fleet level were 83 dB. A random sample of six heavy trucks produced the following noise levels (in decibels): \(\begin{array}{llllll}85.4 & 86.8 & 86.1 & 85.3 & 84.8 & 86.0 .\end{array}\) Use these data to construct a \(90 \%\) confidence interval for \(\sigma^{2}\), the variance of the truck noiseemission readings. Interpret your results.

Suppose that we take a sample of size \(n_{1}\) from a normally distributed population with mean and variance \(\mu_{1}\) and \(\sigma_{1}^{2}\) and an independent of sample size \(n_{2}\) from a normally distributed population with mean and variance \(\mu_{2}\) and \(\sigma_{2}^{2} .\) If it is reasonable to assume that \(\sigma_{1}^{2}=\sigma_{2}^{2},\) then the results given in Section 8.8 apply. What can be done if we cannot assume that the unknown variances are equal but are fortunate enough to know that \(\sigma_{2}^{2}=k \sigma_{1}^{2}\) for some known constant \(k \neq 1 ?\) Suppose, as previously, that the sample means are given by \(\bar{Y}_{1}\) and \(\bar{Y}_{2}\) and the sample variances by \(S_{1}^{2}\) and \(S_{2}^{2}\), respectively. a. Show that \(Z^{\star}\) given below has a standard normal distribution. $$Z^{*}=\frac{\left(\bar{Y}_{1}-\bar{Y}_{2}\right)-\left(\mu_{1}-\mu_{2}\right)}{\sigma_{1} \sqrt{\frac{1}{n_{1}}+\frac{k}{n_{2}}}}$$ b. Show that \(W^{\star}\) given below has a \(\chi^{2}\) distribution with \(n_{1}+n_{2}-2\) df. $$W^{*}=\frac{\left(n_{1}-1\right) S_{1}^{2}+\left(n_{2}-1\right) S_{2}^{2} / k}{\sigma_{1}^{2}}$$ c. Notice that \(Z^{\star}\) and \(W^{\star}\) from parts (a) and (b) are independent. Finally, show that $$T^{*}=\frac{\left(\bar{Y}_{1}-\bar{Y}_{2}\right)-\left(\mu_{1}-\mu_{2}\right)}{S_{p}^{*} \sqrt{\frac{1}{n_{1}}+\frac{k}{n_{2}}}}, \quad \text { where } S_{p}^{2 *}=\frac{\left(n_{1}-1\right) S_{1}^{2}+\left(n_{2}-1\right) S_{2}^{2} / k}{n_{1}+n_{2}-2}$$ has a \(t\) distribution with \(n_{1}+n_{2}-2\) df. d. Use the result in part (c) to give a \(100(1-\alpha) \%\) confidence interval for \(\mu_{1}-\mu_{2},\) assuming that \(\sigma_{2}^{2}=k \sigma_{1}^{2}\) e. What happens if \(k=1\) in parts \((\mathrm{a}),(\mathrm{b}),(\mathrm{c}),\) and \((\mathrm{d}) ?\)

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