Could you please take a look if my answer makes sense- I don't have a solution to this practice question. Also, I am not quite sure how to read the data in part (c).

Question

Let $\displaystyle Y_1,...Y_n$ be a random sample from an exponential distribution with density function

$\displaystyle f_Y(y)={\theta}e^{-{\theta}y}, y>0$

where $\displaystyle \theta>0$ is an unknown parameter.

(a) Obtain the likelihood ratio test statistic for testing $\displaystyle H_0:\theta=\theta_0$ against $\displaystyle H_1:\theta=\ne{\theta_0}$.

(b) State the assymptotic distribution of twice the logarithm of the likelihood ratio test statistic under Ho.

(c) If $\displaystyle n=100,Y=31, \theta_0=3$, will you reject the null hypothesis Ho at a significance level of 5%?

Answer.

(a)

the MLE $\displaystyle \hat{\theta}}$for $\displaystyle \theta$ is $\displaystyle \frac{1}{\bar{y}}$ (I checked against a few sources).

Therefore my likelihood ratio

$\displaystyle r=\frac{L_y(\hat{\theta};y)}{L_Y(\theta_0;y)}=\fra c{\hat{\theta}e^{-\hat{\theta}\Sigma_{i=1}^nY_i}}{{\theta_0}e^{{-\theta_0}\Sigma_{i=1}^nY_i}}=\frac{{(\frac{1}{\bar {Y}})^n}e^{-\frac{1}{\bar{Y}}n{\bar{Y}}}}{{\theta_0}^ne^{{-\theta_0}n\bar{Y}}}=(\frac{1}{\bar{Y}\theta_0})^ne ^{n{\bar{Y}(-\frac{1}{\bar{Y}}+\theta_0)}$

(b) the assymptotic distribution

$\displaystyle 2ln(r(Y))->\chi^2_1$ (coverges in distribution)

(c) calculate r from the given test results.

Now I am not sure what is "Y" - if that is $\displaystyle \Sigma_{i=1}^nY_i$ then why the MLE realisation ($\displaystyle \frac{1}{31/100}$) is so far away from $\displaystyle \theta_0=3$?

Ideally I would use (b) to test the hypothesis.

Does the part (c) data make sense to you?