# Obtaining confidence interval from Wald's test?

I have the following Wald test statistic: $$(\beta - p)^2 \frac{2n}{\beta^2 (1- \beta)}$$ where $\beta$ is my MLE and $p$ is the actual parameter. I wish to contruct a 95 % confidence interval for p, but I am unsure how. I know the statistic is approximately $\chi$^2 distributed (in this case, df = 1), but I cannot understand what "type" of confidence interval I should construct (symmetric, right? left?) and nor do I understand the answer in my lecture notes: $$\beta \pm 1.96 \sqrt{ \beta^2 ( 1- \beta) /2n}$$ This looks something like a normal distribution though? Where did it come from?

Your notes are wrong, or at least they do not match any known form of a test statistic with which I am familiar. Your theory is right. We use the hat, written \hat{\beta} = $\hat{\beta}$ to denote estimators whereas $\beta$ represents an assumed parameter, unmeasured as does $p$.
Coincidentally, if you were interested in calculating a 95% interval and test for the sample proportion from an experiment of IID Bernoulli($p$) random variables, the test statistic is: $$W = \frac{(\hat{p} - p)^2}{\hat{p}(1-\hat{p})/n}$$
which has an asymptotic $\chi^2_1$ distribution under the null hypothesis.
The Wald test is intrinsically related to the confidence interval. In general, the Wald test takes the form: $W = \left(\hat{\beta} / \mbox{SE} ( \hat{\beta} )\right)^2$ where $\mbox{SE}$ denotes some estimate of the standard error of the estimator $\hat{\beta}$ which may be a maximum likelihood estimator. Similarly, the 95% CI is constructed: $\hat{\beta} \pm 1.96 \cdot \mbox{SE} \left( \hat{\beta} \right)$. Basically, the Wald test is significant if any only if the 95% CI does not contain 0.