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Problem 4 (20 pts) Let $X_1, \dots, X_n$ be a random sample from a normal distribution $N(\mu, \sigma^2)$. We want to test $H_0: \sigma^2 = \sigma_0^2$ vs $H_1: \sigma^2 \neq \sigma_0^2$. a. Show that the likelihood ratio test of level $\alpha$ rejects the null hypothesis if, and only if, $\sum_{i=1}^n (X_i - \bar{X})^2/\sigma_0^2 \leq c_1$ or $\sum_{i=1}^n (X_i - \bar{X})^2/\sigma_0^2 \geq c_2$, where $c_1$ and $c_2$ satisfy, (i) $F(c_2) - F(c_1) = 1 - \alpha$ where F is the distribution function of the $\chi^2$ distribution with d.f $n - 1$. (ii) $c_1 - c_2 = n \log c_1/c_2$. b. Show that $c_{1n} = n - \sqrt{2n z_{\alpha/2}}$ and $c_{2n} = n + \sqrt{2n z_{\alpha/2}}$ approximately satisfy (i) and (ii) in a sense that the ratio $\frac{c_{1n} - c_{2n}}{n \log c_{1n}/c_{2n}} \to 1$ as $n \to \infty$ where $z_\alpha$ denotes the upper $\alpha$th quantile of the standard normal distribution. [Hint. For (i), consider the normal approximation of $\chi^2$ distribution when d.f is getting large.] c. Justify to use the equal tail test which rejects the null hypothesis if, and only if, $\sum_{i=1}^n (x_i - \bar{x})^2/\sigma_0^2 \leq \chi_{n-1}^2(1 - \alpha/2)$ or $\sum_{i=1}^n (x_i - \bar{x})^2/\sigma_0^2 \geq \chi_{n-1}^2(\alpha/2)$ where $\chi_{n-1}^2(\alpha)$ denotes the upper $\alpha$th quantile of $\chi^2$ distribution with d.f $n - 1.$

          Problem 4 (20 pts) Let $X_1, \dots, X_n$ be a random sample from a normal distribution $N(\mu, \sigma^2)$. We want to test
$H_0: \sigma^2 = \sigma_0^2$ vs $H_1: \sigma^2 \neq \sigma_0^2$.
a. Show that the likelihood ratio test of level $\alpha$ rejects the null hypothesis if, and only if,
$\sum_{i=1}^n (X_i - \bar{X})^2/\sigma_0^2 \leq c_1$ or $\sum_{i=1}^n (X_i - \bar{X})^2/\sigma_0^2 \geq c_2$, where $c_1$ and $c_2$ satisfy,
(i) $F(c_2) - F(c_1) = 1 - \alpha$ where F is the distribution function of the $\chi^2$ distribution with d.f $n - 1$.
(ii) $c_1 - c_2 = n \log c_1/c_2$.
b. Show that $c_{1n} = n - \sqrt{2n z_{\alpha/2}}$ and $c_{2n} = n + \sqrt{2n z_{\alpha/2}}$ approximately satisfy (i) and (ii) in a sense that the ratio
$\frac{c_{1n} - c_{2n}}{n \log c_{1n}/c_{2n}} \to 1$ as $n \to \infty$
where $z_\alpha$ denotes the upper $\alpha$th quantile of the standard normal distribution. [Hint. For (i), consider the normal approximation of $\chi^2$ distribution when d.f is getting large.]
c. Justify to use the equal tail test which rejects the null hypothesis if, and only if,
$\sum_{i=1}^n (x_i - \bar{x})^2/\sigma_0^2 \leq \chi_{n-1}^2(1 - \alpha/2)$ or $\sum_{i=1}^n (x_i - \bar{x})^2/\sigma_0^2 \geq \chi_{n-1}^2(\alpha/2)$
where $\chi_{n-1}^2(\alpha)$ denotes the upper $\alpha$th quantile of $\chi^2$ distribution with d.f $n - 1.$

Problem 4 (20 pts) Let X1, …, Xn be a random sample from a normal distribution N(μ, σ^2). We want to test
H0: σ^2 = σ0^2 vs H1: σ^2 ≠σ0^2.
a. Show that the likelihood ratio test of level α rejects the null hypothesis if, and only if,
∑i=1^n (Xi - X̅)^2/σ0^2 ≤ c1 or ∑i=1^n (Xi - X̅)^2/σ0^2 ≥ c2, where c1 and c2 satisfy,
(i) F(c2) - F(c1) = 1 - α where F is the distribution function of the χ^2 distribution with d.f n - 1.
(ii) c1 - c2 = n log c1/c2.
b. Show that c1n = n - √(2n zα/2) and c2n = n + √(2n zα/2) approximately satisfy (i) and (ii) in a sense that the ratio
(c1n - c2n)/(n log c1n/c2n)→ 1 as n →∞
where zα denotes the upper αth quantile of the standard normal distribution. [Hint. For (i), consider the normal approximation of χ^2 distribution when d.f is getting large.]
c. Justify to use the equal tail test which rejects the null hypothesis if, and only if,
∑i=1^n (xi - x̅)^2/σ0^2 ≤χn-1^2(1 - α/2) or ∑i=1^n (xi - x̅)^2/σ0^2 ≥χn-1^2(α/2)
where χn-1^2(α) denotes the upper αth quantile of χ^2 distribution with d.f n - 1.

Added by Luis F.

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