Question

the diameters (in inches) of 17 randomly selected bolts produced by a machine are listed. use a 90% level of confidence to construct a confidence interval for (a) the population variance $\sigma^{2}$ and (b) the population standard deviation $\sigma$. interpret the results.
4.472 4.424 4.027 4.311 4.007 3.789
3.816 3.771 4.239 3.957 4.148 4.528
3.935 3.763 3.883 3.809 4.449
(a) the confidence interval for the population variance is (). (round to three decimal places as needed.)

Explanation:

Step1: Calculate sample variance \(s^{2}\)

First, find the sample mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\).
\(n = 17\)
\(\sum_{i=1}^{17}x_{i}=4.472 + 4.424+4.027+4.311+4.007+3.789+3.816+3.771+4.239+3.957+4.148+4.528+3.935+3.763+3.883+3.809+4.449 = 70.994\)
\(\bar{x}=\frac{70.994}{17}\approx4.176\)

Then, calculate \(s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}\)
\(\sum_{i = 1}^{17}(x_{i}-\bar{x})^{2}=(4.472 - 4.176)^{2}+(4.424 - 4.176)^{2}+\cdots+(4.449 - 4.176)^{2}\approx1.173\)
\(s^{2}=\frac{1.173}{16}\approx0.073\)

Step2: Determine critical values

For a \(90\%\) confidence level and \(n-1=16\) degrees of freedom, \(\alpha=1 - 0.90 = 0.10\), \(\frac{\alpha}{2}=0.05\)
From the \(\chi^{2}\) - distribution table, \(\chi_{L}^{2}=\chi_{0.05,16}^{2}=26.296\), \(\chi_{R}^{2}=\chi_{0.95,16}^{2}=7.962\)

Step3: Construct confidence interval for \(\sigma^{2}\)

The formula for the confidence interval of \(\sigma^{2}\) is \(\frac{(n - 1)s^{2}}{\chi_{L}^{2}}\leq\sigma^{2}\leq\frac{(n - 1)s^{2}}{\chi_{R}^{2}}\)
Substitute \(n = 17\), \(s^{2}=0.073\), \(\chi_{L}^{2}=26.296\), \(\chi_{R}^{2}=7.962\)
\(\frac{16\times0.073}{26.296}\leq\sigma^{2}\leq\frac{16\times0.073}{7.962}\)
\(\frac{1.168}{26.296}\leq\sigma^{2}\leq\frac{1.168}{7.962}\)
\(0.044\leq\sigma^{2}\leq0.147\)

Answer:

The confidence interval for the population variance is \((0.044,0.147)\)