Suppose $X$ follows a chi-squared distribution with $r$ degrees of freedom, where $r \in \mathbb{Z}^+$ is unknown. Let $X_1, X_2, ..., X_n$ be a random sample of $X$.
(i) Suppose we have the following hypotheses: $H_0: r = 1$ against $H_1: r > 1$. We take $n = 10$ samples and reject $H_0$ if $\sum_{i = 1}^{10} x_i \geq 18.31$. Find the significance level of this test.
(ii) Compute $K(2)$, where $K(r)$, $r \in \{1, 2, 3, ...\}$ is the power function.
(iii) Find the best critical region for testing the hypotheses: $H_0: r = 1$ and $H_1: r = 2$ of the form $\{(x_1, ..., x_n) : \sum_{i = 1}^{n} ln(x_i) \geq c\}$ for some constant $c$. Hint: Use the Neyman-Pearson lemma.
This is what I did:
(i) Let $C = \{\sum_{i = 1}^{10} x_i: \sum_{i = 1}^{10} x_i \geq 18.31\}$ be the critical region. The sum $\sum_{i = 1}^{10} X_i$ is chi-squared with $10r$ degrees of freedom.
Taking $r = 1$, the significance level, $\alpha$, is
$$\alpha = P(X \in C) = P(\sum_{i = 1}^{10} X_i \geq 18.31) = \int_{18.31}^{\infty} \frac{1}{\Gamma (5) 2^{5}} x^{4} e^{-x/2} \,dx = 0.0499541$$
(ii) The power function is the probability that $X$ is in the critical region when the parameter, $r$ is some number. So
$$K(2) = P(X \in C | r = 2) = P(\sum_{i = 1}^{10} X_i \geq 18.31) = \int_{18.31}^{\infty} \frac{1}{\Gamma (20/2) 2^{20/2}} x^{20/2 - 1} e^{-x/2} \,dx = \int_{18.31}^{\infty} \frac{1}{\Gamma (10) 2^{10}} x^{9} e^{-x/2} \,dx = 0.567$$
(iii) By the Neyman-Pearson lemma
$$\frac{L(1)}{L(2)} = \frac{(2\pi)^{-n/2} \prod x_i^{-1/2} e^{\sum \frac{-x_i}{2}}}{\frac{1}{2^n} e^{\sum \frac{-x_i}{2}}} = \frac{2^{n/2} \prod x_i^{-1/2}}{\pi ^{n/2}} \leq k$$
Taking the natural log
$$\frac{n}{2}\ln(2) - \frac{n}{2}\ln(\pi) - \frac{1}{2} \sum_{i = 1}^{n} \ln(x_i) \leq \ln(k)$$
This is equivalent to:
$$\sum_{i = 1}^{n} \ln(x_i) \geq n\ln(2) - n\ln(\pi) - 2\ln(k)$$
This is of the required form where the RHS of the inequality is $c$.
Is what I have done correct? I am not quite sure about parts (i) and (ii).
they are both correct! usually in Statistics these questions are not answered solving the integral but reading the result on the tables.
(i) the significance level $\alpha=5\%$ is written in the red circle. for (ii) the table is not detailed enough...we have 19.337 and not 18.31 thus the result is a little bit greater than 50%. With a calculator or, as you correctly did, solving the integral you get the exact result of 56.70%