you know the drill by now. Let's see your attempt
The times to failure of certain electronic components in accelerate environment tests are 15, 28, 3, 12, 42, 19, 20, 2, 25, 30, 62, 12, 18, 16, 44, 65, 33, 51, 4, and 28 minutes. Looking upon these data as random sample from an exponential population, use the results of:
L = (1/θ^n)*e^[(-1/θ)*Σxi]
λ = [( x̅/θ)^n] * e^-[((n* x̅ )/θ) + n ]
x̅ * e^-( x̅ / θ ) ≤ K, where K = e*θ*k^(1/n)
and use the results of:
For large n, the distribution of -2*ln(λ) approaches, under very general conditions, the chi-square distribution with 1 degree of freedom.
use both of these results to test the null hypothesis θ = 15 minutes against the alternative hypothesis θ ≠ 15 minutes at the 0.05 level of significance. (Use ln(1.763) = 0.570)
I attempted this again:
H_0 : θ = 15
H_1 : θ ≠ 15
alpha = 0.05
Σxi = 529
n = 20
x-bar = 26.45
Then I just said to Reject H_0 if x is greater than or equal to 17.
So, P[ x ≥ 17 | H_0: θ = 15] = the integral from 17 to infinity of (1/15)*e^(-(1/15)*x) dx = 0.3220
I'm not sure if this is correct or if I am going in the right direction for solving this problem