So I am given that the distribution is exponential
$f(x|\lambda) = \lambda e^{-\lambda x}$ for $x > 0$
I am also given that there is an independent sample size $n$, $X_1, X_2, ..., X_n$
I am given that the null hypothesis $H_0 : \lambda = 1$ and the alternative hypothesis $H_a : \lambda > 1$. So I noticed that this is a composite hypothesis so I have the use the uniformly most powerful test rather than Neymann Pearson.
I am asked to show that the uniformly most powerful test of $H_0$ is of the form “reject $H_0$ if $T < c$”, for some value c, where T is the sample total.
Then the next part says; suppose that $n = 100$, and the chosen significance level is $0.05$. Find, approximately, the value of c for this test
So to start this off I have done the likelihood ratio
$LR = \frac {l(\lambda_a)}{l(\lambda_0)}$
and I have found $l(\lambda_0) = e^{-\sum_{i=0}^n x_i} = e^{-n\bar x}$
I then find $l(\lambda_a) = \lambda_a^n e^{-\lambda_a\sum_{i=0}^n x_i}$
This gives me a likelihood ratio
$LR = \lambda_a^ne^{-(\lambda_a-1)n\bar x} = \lambda_a^ne^{-(\lambda_a-1)T}$
Where T is the sum total.
This satisfies the statement 'reject $H_0$ if $T < c$'
now onto the next part with $n = 100$ with significance level 0.05I don't really know what to do here so any hints would be extremely useful.
I guess that $\lambda_a^ne^{-(\lambda_a-1)n\bar x} = \lambda_a^ne^{-(\lambda_a-1)T} < K$ for some K
and that $T = 100\bar x$
ANy tips would be great. Thank you