Solution Found!
Let S2 be the variance of a random sample of size n from N(, 2). Using the fact that (n
Chapter 7, Problem 7.1-16(choose chapter or problem)
Let \(S^2\) be the variance of a random sample of size n from \(N(\mu, \sigma^2)\). Using the fact that \((n − 1)S^2/\sigma^2\) is \(\chi^2(n-1)\), note that the probability
\(P\left[a \leq \frac{(n-1) S^{2}}{\sigma^{2}} \leq b\right]=1-\alpha\),
where \(a=\chi_{1-\alpha / 2}^{2}(n-1)\) and \(b=\chi_{\alpha / 2}^{2}(n-1)\). Rewrite the inequalities to obtain
\(P\left[\frac{(n-1) S^{2}}{b} \leq \sigma^{2} \leq \frac{(n-1) S^{2}}{a}\right]=1-\alpha\),
If n = 13 and \(12 S^{2}=\sum_{i=1}^{13}\left(x_{i}-\bar{x}\right)^{2}=128.41\), show that [6.11, 24.57] is a 90% confidence interval for the variance \(\sigma^2\). Accordingly, [2.47, 4.96] is a 90% confidence interval for \(\sigma\).
Questions & Answers
QUESTION:
Let \(S^2\) be the variance of a random sample of size n from \(N(\mu, \sigma^2)\). Using the fact that \((n − 1)S^2/\sigma^2\) is \(\chi^2(n-1)\), note that the probability
\(P\left[a \leq \frac{(n-1) S^{2}}{\sigma^{2}} \leq b\right]=1-\alpha\),
where \(a=\chi_{1-\alpha / 2}^{2}(n-1)\) and \(b=\chi_{\alpha / 2}^{2}(n-1)\). Rewrite the inequalities to obtain
\(P\left[\frac{(n-1) S^{2}}{b} \leq \sigma^{2} \leq \frac{(n-1) S^{2}}{a}\right]=1-\alpha\),
If n = 13 and \(12 S^{2}=\sum_{i=1}^{13}\left(x_{i}-\bar{x}\right)^{2}=128.41\), show that [6.11, 24.57] is a 90% confidence interval for the variance \(\sigma^2\). Accordingly, [2.47, 4.96] is a 90% confidence interval for \(\sigma\).
ANSWER:Step 1 of 2
Confidence interval =
Significance Level =
