Question

the number of hours of reserve capacity of 10 randomly selected automotive batteries is shown to the right.
1.78 1.85 1.53 1.67 1.77
1.91 1.38 1.53 1.49 2.08
assume the sample is taken from a normally distributed population. construct 98% confidence intervals for (a) the population variance \\( \sigma^{2} \\) and (b) the population standard deviation \\( \sigma \\).
(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 \(\bar{x}=\frac{1.78 + 1.85+1.53+1.67+1.77+1.91+1.38+1.53+1.49+2.08}{10}=1.699\)
Then, \(s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}\)
\(\sum_{i=1}^{10}(x_{i}-\bar{x})^{2}=(1.78 - 1.699)^{2}+(1.85 - 1.699)^{2}+(1.53 - 1.699)^{2}+(1.67 - 1.699)^{2}+(1.77 - 1.699)^{2}+(1.91 - 1.699)^{2}+(1.38 - 1.699)^{2}+(1.53 - 1.699)^{2}+(1.49 - 1.699)^{2}+(2.08 - 1.699)^{2}\)
\(\sum_{i = 1}^{10}(x_{i}-\bar{x})^{2}=0.006561+0.022801+0.028561+0.000841+0.005041+0.044521+0.101761+0.028561+0.043681+0.145161 = 0.42749\)
\(s^{2}=\frac{0.42749}{9}\approx0.0475\)

Step2: Determine critical values

For a \(98\%\) confidence interval and \(n=10\), the degrees of freedom \(df=n - 1=9\)
\(1-\alpha=0.98\), so \(\alpha = 0.02\), \(\frac{\alpha}{2}=0.01\)
From the \(\chi^{2}\) - distribution table, \(\chi_{L}^{2}=\chi_{1-\frac{\alpha}{2},df}^{2}=\chi_{0.99,9}^{2}=2.088\), \(\chi_{R}^{2}=\chi_{\frac{\alpha}{2},df}^{2}=\chi_{0.01,9}^{2}=21.666\)

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

The formula for the confidence interval of \(\sigma^{2}\) is \(\frac{(n - 1)s^{2}}{\chi_{R}^{2}}\leq\sigma^{2}\leq\frac{(n - 1)s^{2}}{\chi_{L}^{2}}\)
Substitute \(n = 10\), \(s^{2}=0.0475\), \(\chi_{L}^{2}=2.088\), \(\chi_{R}^{2}=21.666\)
\(\frac{(10 - 1)\times0.0475}{21.666}\leq\sigma^{2}\leq\frac{(10 - 1)\times0.0475}{2.088}\)
\(\frac{0.4275}{21.666}\leq\sigma^{2}\leq\frac{0.4275}{2.088}\)
\(0.020\leq\sigma^{2}\leq0.205\)

Answer:

\((0.020,0.205)\)