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.

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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 \(c\).
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/
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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