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Chapter 11: Problem 47

Construct the \(95 \%\) confidence intervals for the population variance and standard deviation for the following data, assuming that the respective populations are (approximately) normally distributed. a. \(n=10, s^{2}=7.2\) b. \(n=18, s^{2}=14.8\)

Short Answer

Expert verified
The \(95%\) confidence intervals for variance and standard deviation are for a: \([0.43, 2.67]\) and [0.66, 1.63], and for b: \([0.49, 1.75]\) and [0.70, 1.32] respectively.

Step by step solution

01

- Find the upper and lower bounds for first set of data

Let's start with the first set of data. Using a chi-square distribution table, find the chi-square values at \(5%\) and \(95%\) levels of significance for \(9\) degrees of freedom (\(n-1\)). The upper and lower bounds for the variance confidence interval are calculated by dividing the sample variance by the respective chi-square values. Use the square root of those values to get the standard deviation bounds.
02

- Calculate the confidence intervals for first data

For data set a with \(n=10, s^{2}=7.2\), find the chi-square values at \(5%\) and \(95%\) for \(9\) degrees of freedom. They are \(16.919\) and \(2.700\), respectively. The \(95%\) confidence interval for the variance is \(\left[\frac{7.2}{16.919}, \frac{7.2}{2.700}\right]\) and \(\left[\sqrt{\frac{7.2}{16.919}}, \sqrt{\frac{7.2}{2.700}}\right]\) for the standard deviation.
03

- Find the upper and lower bounds for second set of data

Next is the second data set. Similarly, find the chi-square values at \(5%\) and \(95%\) levels of significance for \(17\) degrees of freedom (\(n-1\)). The upper and lower bounds for the variance confidence interval are calculated by dividing the sample variance by the respective chi-square values. Use the square root of those bounds to get the standard deviation bounds.
04

- Calculate the confidence intervals for second data

For data set b with \(n=18, s^{2}=14.8\), find the chi-square values at \(5%\) and \(95%\) for \(17\) degrees of freedom. They are \(30.191\) and \(8.438\), respectively. The \(95%\) confidence interval for the variance is \(\left[\frac{14.8}{30.191}, \frac{14.8}{8.438}\right]\) and \(\left[\sqrt{\frac{14.8}{30.191}}, \sqrt{\frac{14.8}{8.438}}\right]\) for the standard deviation.

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Most popular questions from this chapter

Determine the value of \(\chi^{2}\) for 23 degrees of freedom and an area of \(.990\) in the left tail of the chi-square distribution curve.

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