test statistics and power function Let $X_1$ and $X_2$ be independent and uniformly distributed on $(\Theta , \Theta  + 1)$. Consider
the two tests with critical regions $C_1$ and $C_2$ given by $C_1 = \left \{ (x_1, x_2)| x_1 ≥ .95 \right \}$ and $C_2 =\left \{ (x_1,x_2)|x_1 +x_2 ≥c \right \}$ to test $H_0 :\Theta=0$ versus $H_1 :\Theta=\frac{1}{2}$.
In this question, if I want to find the value $c$ so that $C_2$ has same size as $C_1$, 
do I have to put $C_2 = 0.05$? 
I am confused how to get value $c$ to make these two have same size. 

Size of a test is the same as the probability of type-I error, i.e., probability of rejecting the null hypothesis when, in fact, it is true. So, you first calculate the size of the test using C.R. $C_1$. Then find the expression for the size of the test using C.R. $C_2$ in terms of $c$ and then equate it to the previously calculated size of the test. Solve for $c$. Note, $H_0$ states both $X_1$ and $X_2$ are independently Uniform over $(0,1)$.
$P_{H_0}(C_1) = P_{H_0}(X_1\ge.95) = 0.05$ which is the size of the test.
Now you find $P_{H_0}(C_2) = P_{H_0}(X_1+X_2\ge c)$ [Deduce the survival function of the sum of two independent Uniform$(0,1)$ variates. Note that the form will be different for $0\lt c\lt1$ and for $1\lt c\lt2$.]
Next you try solving for $c$ using the forms noting the ranges for $c$ (You'll find two solutions for $c$; but you have to invalidate one). There you go!
