Let $X_1, X_2, ..., X_n$ be a random sample from a $Gamma(\alpha, \beta)$ population. Assume $\alpha$ is known and $\beta$ is unknown. Consider testing $H_0: \beta = \beta_0$.
I'm trying to derive a Wald statistic for testing the null hypothesis, using the MLE in both the numerator and denominator but I am a bit stuck. I've computed $\hat{\beta}_{MLE} = \dfrac{\alpha}{\bar{x}}$. From here I know that Wald statistics take the form $Z_n = \dfrac{W_n-\theta_0}{S_n}$ where $W_n$ is the estimator of $\theta$, $\theta_0$ is a hypothesized value of $\theta$ and $S_n$ is the standard error of the estimator. I struggling with calculating the standard error of the estimator though. I know that the standard error is the square root of the variance of the MLE. From here, how should I go about calculating the variance? Should I be utilizing fisher information?
Based on your MLE, it appears that you are considering $\beta$ a rate parameter, namely $$f_{X_i}(x) = \frac{\beta^\alpha x^{\alpha-1} e^{-\beta x}}{\Gamma(\alpha)}.$$ Then the statistic $\bar X$ follows a gamma distribution with shape $n \alpha$ and rate $n \beta$. It follows that $Y = 1/\bar X$ is inverse gamma distributed with density $$f_Y(y) = \frac{f_{\bar X}(1/y)}{y^2} = \frac{(n \beta)^{n \alpha} y^{-(n \alpha - 1)} e^{-n \beta/y}}{y^2\Gamma(n\alpha)} = \frac{(n \beta/y)^{n \alpha} e^{-n \beta/y}}{y \Gamma(n \alpha)}.$$ I leave it as an exercise* to show that $$\operatorname{E}[Y] = \frac{n \beta}{n\alpha - 1}, \quad n\alpha > 1 \\ \operatorname{Var}[Y] = \frac{n^2 \beta^2}{(n \alpha - 1)^2 (n \alpha - 2)}, \quad n \alpha > 2.$$
From this, it is straightforward to compute the mean and variance of your MLE, and in turn the corresponding Wald statistic.
*Hint. There are a number of ways to do this, but the algebraic approach is to assume $\int_{y > 0} f_Y(y) \, dy = 1$ is a proper density, and then use this fact to compute $\operatorname{E}[Y^k] \int_{y > 0} y^k f_Y(y) \, dy$. Then substitute $k = 1$ and $k = 2$ to get the first two moments, from which the variance may be computed.