Question

construct the indicated confidence intervals for the population variance $\sigma^{2}$ and the population standard deviation $\sigma$. assume the sample is from a normally distributed population. $c = 0.95, s^{2}=15.06, n = 25$ a. confidence interval for $\sigma^{2}$: $(9.60, 24.74)$; confidence interval for $\sigma$: $(3.10, 4.97)$ b. confidence interval for $\sigma^{2}$: $(8.89, 27.55)$; confidence interval for $\sigma$: $(2.98, 5.25)$ c. confidence interval for $\sigma^{2}$: $(9.18, 29.15)$; confidence interval for $\sigma$: $(3.03, 5.40)$ d. confidence interval for $\sigma^{2}$: $(13.68, 26.10)$; confidence interval for $\sigma$: $(3.70, 5.11)$

Explanation:

Step1: Find the degrees of freedom and critical values

The degrees of freedom \(df=n - 1=25-1 = 24\).
For a confidence level \(c = 0.95\), the significance level \(\alpha=1 - c=1 - 0.95 = 0.05\).
\(\alpha/2=0.025\) and \(1-\alpha/2 = 0.975\).
From the \(\chi^{2}\) - distribution table, \(\chi_{L}^{2}=\chi_{1-\alpha/2,df}^{2}=\chi_{0.975,24}^{2}=12.401\) and \(\chi_{R}^{2}=\chi_{\alpha/2,df}^{2}=\chi_{0.025,24}^{2}=39.364\)

Step2: Calculate the confidence interval for \(\sigma^{2}\)

The formula for the confidence interval for \(\sigma^{2}\) is \(\frac{(n - 1)s^{2}}{\chi_{R}^{2}}<\sigma^{2}<\frac{(n - 1)s^{2}}{\chi_{L}^{2}}\)
Substitute \(n = 25\), \(s^{2}=15.06\), \(\chi_{L}^{2}=12.401\), \(\chi_{R}^{2}=39.364\)
\(\frac{(25 - 1)\times15.06}{39.364}<\sigma^{2}<\frac{(25 - 1)\times15.06}{12.401}\)
\(\frac{24\times15.06}{39.364}<\sigma^{2}<\frac{24\times15.06}{12.401}\)
\(\frac{361.44}{39.364}<\sigma^{2}<\frac{361.44}{12.401}\)
\(9.18<\sigma^{2}<29.15\)

Step3: Calculate the confidence interval for \(\sigma\)

Take the square - root of the endpoints of the confidence interval for \(\sigma^{2}\)
\(\sqrt{9.18}<\sigma<\sqrt{29.15}\)
\(3.03<\sigma<5.40\)

Answer:

C. Confidence interval for \(\sigma^{2}\): \((9.18,29.15)\); Confidence interval for \(\sigma\): \((3.03,5.40)\)