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 mean and sample variance

First, calculate the sample mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\).
Let \(x_{1}=4.472,x_{2}=4.424,\cdots,x_{17} = 4.449\), \(n = 17\).
\(\sum_{i=1}^{17}x_{i}=4.472 + 4.424+4.027+4.311+4.007+3.789+3.816+3.771+4.239+3.957+4.148+4.528+3.935+3.763+3.883+3.809+4.449=70.998\)
\(\bar{x}=\frac{70.998}{17}\approx4.176\)
Sample variance \(s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}\)
\(\sum_{i=1}^{n}(x_{i}-\bar{x})^{2}=(4.472 - 4.176)^{2}+(4.424 - 4.176)^{2}+\cdots+(4.449 - 4.176)^{2}\)
\(\sum_{i=1}^{n}(x_{i}-\bar{x})^{2}=6.791\)
\(s^{2}=\frac{6.791}{16}\approx0.424\)

Step2: Determine the critical values

For a \(90\%\) confidence level and \(n-1=16\) degrees of freedom.
The lower - tail critical value \(\chi_{1-\alpha/2}^{2}=\chi_{0.95}^{2}\), from the chi - square distribution table, \(\chi_{0.95}^{2}=7.962\)
The upper - tail critical value \(\chi_{\alpha/2}^{2}=\chi_{0.05}^{2}\), from the chi - square distribution table, \(\chi_{0.05}^{2}=26.296\)

Step3: Construct the confidence interval for the population variance

The formula for the confidence interval of the population variance \(\sigma^{2}\) is \(\frac{(n - 1)s^{2}}{\chi_{\alpha/2}^{2}}\leq\sigma^{2}\leq\frac{(n - 1)s^{2}}{\chi_{1-\alpha/2}^{2}}\)
Substitute \(n = 17\), \(s^{2}=0.424\), \(\chi_{0.05}^{2}=26.296\), \(\chi_{0.95}^{2}=7.962\)
\(\frac{16\times0.424}{26.296}\leq\sigma^{2}\leq\frac{16\times0.424}{7.962}\)
\(\frac{6.784}{26.296}\leq\sigma^{2}\leq\frac{6.784}{7.962}\)
\(0.258\leq\sigma^{2}\leq0.852\)

Answer:

\((0.258,0.852)\)