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Chapter 9: Problem 595

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
To find a confidence interval for \(\sigma^2\), compute the sample variance, \(s^2\), and the Chi-squared test statistic, \(\chi^2\), with \(n-1\) degrees of freedom. Find the Chi-squared critical values for a specified confidence level, say \((1-\alpha)100\%\), and construct the confidence interval for \(\sigma^2\) using these critical values, the sample variance, and the sample size. The \((1-\alpha)100\%\) confidence interval for \(\sigma^2\) is given by: \[\frac{(n-1)s^2}{\chi^2_{1-\frac{\alpha}{2}}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{\frac{\alpha}{2}}}\]

Step by step solution

01

Sample Variance

Compute the sample variance, \(s^2\), from the given data. The sample variance is calculated as: \[s^2 = \frac{1}{n-1}\sum_{i=1}^{n}(x_i - \bar{x})^2\] where \(n\) is the sample size, \(\bar{x}\) is the sample mean, and \(x_i\) are the individual data points.
02

Compute the Chi-squared Test Statistic

Once we have the sample variance \(s^2\), we can compute the Chi-squared test statistic, \(\chi^2\), using the following formula: \[\chi^2 = \frac{(n-1)s^2}{\sigma^2}\] Since \(\sigma^2\) is unknown, we cannot compute the exact value of the Chi-squared test statistic. However, we know that \(\chi^2\) follows a Chi-squared distribution with \(n-1\) degrees of freedom.
03

Find Chi-squared Critical Values

We then need to find the critical values of the Chi-squared distribution for a specified confidence level, say \((1-\alpha)100\%\), where \(\alpha\) is the chosen significance level. Look up the Chi-squared distribution table or use statistical software to find the critical values \(\chi^2_{\frac{\alpha}{2}}\) and \(\chi^2_{1-\frac{\alpha}{2}}\) such that: \[P\left(\chi^2_{\frac{\alpha}{2}}\leq\chi^2\leq\chi^2_{1-\frac{\alpha}{2}}\right) = 1-\alpha\] These critical values correspond to the lower and upper bounds of the Chi-squared distribution for our \((1-\alpha)100\%\) confidence interval.
04

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

Finally, we can construct the confidence interval for the unknown variance \(\sigma^2\) using the critical values found in Step 3, as well as our sample variance \(s^2\) and sample size \(n\). The \((1-\alpha)100\%\) confidence interval for \(\sigma^2\) is given by the following inequalities: \[\frac{(n-1)s^2}{\chi^2_{1-\frac{\alpha}{2}}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{\frac{\alpha}{2}}}\] By solving for \(\sigma^2\) in these inequalities, we obtain the confidence interval for the unknown variance \(\sigma^2\).

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Most popular questions from this chapter

During the \(1976-77\) season Coach Jerry Tarkanian outfitted his University of Nevada at Las Vegas basketball team with new sneakers. The 16 member team had an average size of \(14.5\) and a standard deviation of 5 . Find a 90 percent confidence interval for the mean sneaker size of all collegiate basket- ball players. Assume the population is normal and the variance is not known.

The makers of a certain brand of car mufflers, claim that the life of the mufflers has a variance of \(.8\) year. A random sample of 16 of these mufflers showed a variance of 1 year. Using a \(5 \%\) level of significance, test whether the variance of all the mufflers of this manufacturer exceeds \(.8\) year.

Find the expected values of the random variables \(\mathrm{X}\) and \(\mathrm{Y}\) if \(\quad \operatorname{Pr}(\mathrm{X}=0)=1 / 2 \quad\) and \(\operatorname{Pr}(\mathrm{X}=1)=1 / 2\) and \(\operatorname{Pr}(\mathrm{Y}=1)=1 / 4 \quad\) and \(\operatorname{Pr}(\mathrm{Y}=2)=3 / 4\). Compare the sum of \(\mathrm{E}(\mathrm{X})+\mathrm{E}(\mathrm{Y})\) with \(\mathrm{E}(\mathrm{X}+\mathrm{Y})\) if \(\operatorname{Pr}(\mathrm{X}=\mathrm{x}, \mathrm{Y}=\mathrm{y})=\operatorname{Pr}(\mathrm{X}=\mathrm{x}) \operatorname{Pr}(\mathrm{Y}=\mathrm{y})\)

Two independent reports on the value of a tincture for treating a disease in camels were available. The first report made on a small pilot series showed the new tincture to be probably superior to the old treatment with a Yates' \(\mathrm{X}^{2}\) of \(3.84, \mathrm{df}=1, \alpha=.05 .\) The second report with a larger trial gave a "not significant" result with a Yates \(\mathrm{X}^{2}=2.71, \mathrm{df}=1\), \(\alpha=.10 .\) Can the results of the 2 reports be combined to form a new conclusion?

Harvey of Brooklyn surveyed a random sample of 625 students at SUNY-Stony Brook. Being a pre-medical student, he hoped that most students would major in the social sciences rather than the natural sciences, thus provide him with less competition. To Harvey's dismay, \(60 \%\) of the students he surveyed were majoring in the natural sciences. Construct a \(95 \%\) confidence interval for \(p\), the population proportion of students majoring in the natural sciences.
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