In Exercise 4.2.18 we found a confidence interval for the variance o using the variance S? of a random sample of size n arising fromN (u, o2), where the mean u is unknown. In testing Ho : o? = of against Ho : o? > of, use the critical region defined by n-Ds > c. That is, reject Ho and accept H1 if S > co/(n – 1). If n = 13 and the significance level a = 0.025, determine c.
In Exercise 4.2.18 we found a confidence interval for the variance o using the variance S? of a random sample of size n arising fromN (u, o2), where the mean u is unknown. In testing Ho : o? = of against Ho : o? > of, use the critical region defined by n-Ds > c. That is, reject Ho and accept H1 if S > co/(n – 1). If n = 13 and the significance level a = 0.025, determine c.
Advanced Engineering Mathematics
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Chapter2: Second-order Linear Odes
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Please see attached mathematical statistics question and additional info below. If n = 13 and the significance level alpha = 0.025, how to determine c?
![In Exercise 4.2.18, we found a confidence interval for the variance \(\sigma^2\) using the variance \(S^2\) of a random sample of size \(n\) arising from \(N(\mu, \sigma^2)\), where the mean \(\mu\) is unknown. In testing \(H_0 : \sigma^2 = \sigma_0^2\) against \(H_1 : \sigma^2 > \sigma_0^2\), use the critical region defined by
\[
\frac{(n-1)S^2}{\sigma_0^2} \geq c.
\]
That is, reject \(H_0\) and accept \(H_1\) if \(S^2 \geq c\sigma_0^2/(n-1)\). If \(n = 13\) and the significance level \(\alpha = 0.025\), determine \(c\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5aa4cb66-a2f7-4199-af14-322b31fac275%2F1e992d37-e5ea-4b50-9013-1720c56af76a%2Flvqt3s4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:In Exercise 4.2.18, we found a confidence interval for the variance \(\sigma^2\) using the variance \(S^2\) of a random sample of size \(n\) arising from \(N(\mu, \sigma^2)\), where the mean \(\mu\) is unknown. In testing \(H_0 : \sigma^2 = \sigma_0^2\) against \(H_1 : \sigma^2 > \sigma_0^2\), use the critical region defined by
\[
\frac{(n-1)S^2}{\sigma_0^2} \geq c.
\]
That is, reject \(H_0\) and accept \(H_1\) if \(S^2 \geq c\sigma_0^2/(n-1)\). If \(n = 13\) and the significance level \(\alpha = 0.025\), determine \(c\).
![**4.2.18.** Let \( X_1, X_2, \cdots, X_n \) be a random sample from \( N(\mu, \sigma^2) \), where both parameters \(\mu\) and \(\sigma^2\) are unknown. A confidence interval for \(\sigma^2\) can be found as follows.
We know that \((n-1)S^2/\sigma^2\) is a random variable with a \(\chi^2(n-1)\) distribution. Thus we can find constants \(a\) and \(b\) so that \(P((n-1)S^2/\sigma^2 < b) = 0.975\) and \(P(a < (n-1)S^2/\sigma^2 < b) = 0.95\).
---
**(a)** Show that this second probability statement can be written as
\[ P((n-1)S^2/b < \sigma^2 < (n-1)S^2/a) = 0.95. \]
**(b)** If \(n = 9\) and \(s^2 = 7.93\), find a 95% confidence interval for \(\sigma^2\).
**(c)** If \(\mu\) is unknown, how would you modify the preceding procedure for finding a confidence interval for \(\sigma^2\)?
---
**Solution.**
**(a)**
\[ a < (n-1)S^2/\sigma^2 \iff \sigma^2 < (n-1)S^2/a; \]
and
\[ (n-1)S^2/\sigma^2 < b \iff \sigma^2 > (n-1)S^2/b. \]
Hence the second probability statement can be written as
\[ P((n-1)S^2/b < \sigma^2 < (n-1)S^2/a) = 0.95. \]
---
**(b)** Since \(\alpha = 0.05\), \(n = 9\), \(b = \chi^2_{\alpha/2,n-1} = \chi^2_{0.025,8} = 17.54\), and \(a = \chi^2_{1-\alpha/](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5aa4cb66-a2f7-4199-af14-322b31fac275%2F1e992d37-e5ea-4b50-9013-1720c56af76a%2F31x08zr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**4.2.18.** Let \( X_1, X_2, \cdots, X_n \) be a random sample from \( N(\mu, \sigma^2) \), where both parameters \(\mu\) and \(\sigma^2\) are unknown. A confidence interval for \(\sigma^2\) can be found as follows.
We know that \((n-1)S^2/\sigma^2\) is a random variable with a \(\chi^2(n-1)\) distribution. Thus we can find constants \(a\) and \(b\) so that \(P((n-1)S^2/\sigma^2 < b) = 0.975\) and \(P(a < (n-1)S^2/\sigma^2 < b) = 0.95\).
---
**(a)** Show that this second probability statement can be written as
\[ P((n-1)S^2/b < \sigma^2 < (n-1)S^2/a) = 0.95. \]
**(b)** If \(n = 9\) and \(s^2 = 7.93\), find a 95% confidence interval for \(\sigma^2\).
**(c)** If \(\mu\) is unknown, how would you modify the preceding procedure for finding a confidence interval for \(\sigma^2\)?
---
**Solution.**
**(a)**
\[ a < (n-1)S^2/\sigma^2 \iff \sigma^2 < (n-1)S^2/a; \]
and
\[ (n-1)S^2/\sigma^2 < b \iff \sigma^2 > (n-1)S^2/b. \]
Hence the second probability statement can be written as
\[ P((n-1)S^2/b < \sigma^2 < (n-1)S^2/a) = 0.95. \]
---
**(b)** Since \(\alpha = 0.05\), \(n = 9\), \(b = \chi^2_{\alpha/2,n-1} = \chi^2_{0.025,8} = 17.54\), and \(a = \chi^2_{1-\alpha/
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