I have that the UMP is:
$1$ if $\sum_{i=1}^{n} x_i^2 >c$ and $0$ if $\sum_{i=1}^{n} x_i^2 <c$,
where $X$ consists of $n$ iid random variables with distribution $Normal(0,θ)$ where $θ$ is the variance which is $>0$ and the hypothesis are $H_0:θ=θ_0$ vs $H_1:θ=θ_1$ with $θ_1 > θ_0$ . I am then told $θ_0 = 3, θ_1 = 4, α = 0.1 $ and $ n = 10$, and that I need to find what $c$ is.
To do this, I have said that $θ\sum_{i=1}^{10} x_i^2 / θ$ ∼ $θχ^2(10)$ but I am unsure on how to advance from here. Any help much appreciated!
UPDATED:
Part b) Calculate the probability of Type 1 and Type 2 error using the values given in the previous part.
I calculated c to be $48$ when completing the last part. Knowing the definition of T1&2 errors, I say
$P$($\sum_{i=1}^{10} x_i^2 >48 | Variance = 3$) is the probability of Type 1 errors, and
$P(\sum_{i=1}^{10} x_i^2 <48 | Variance = 4$) is the probability of Type 2 errors. However, although I understand the intuition, I am unsure on how to calculate these probabilities at the moment, or if this is even the right approach.
As you have mentioned, $$\frac1{\theta}\sum\limits_{i=1}^{10} X_i^2\sim \chi^2_{10}$$
Size of the UMP test is $0.1$, so using the above fact you have
$$P_{H_0}\left[\sum_{i=1}^{10} X_i^2>c\right]=P_{H_0}\left[\frac13\sum_{i=1}^{10} X_i^2>\frac c3\right]=P\left[\chi^2_{10}>\frac c3\right]=0.1$$
This indeed gives $$\frac c3=\chi^2_{0.1,10}\approx 16\implies c\approx 48$$
Your formulae for the probabilities of type I error and type II error are correct.
The former is just the size of the test, i.e. $0.1$.
And the latter is given by
\begin{align} P_{H_1}\left[\sum_{i=1}^{10} X_i^2<48\right]&=P_{H_1}\left[\frac14\sum_{i=1}^{10} X_i^2<12\right] \\&=P\left[\chi^2_{10}<12\right] \\&\approx 0.71494 \end{align}