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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?

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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.

Non-symmetric confidence intervals are obtained when the MLE is not on a "usual" scale. For instance, in logistic regression, the model parameter is the log of the odds ratio, (significant if it contains 0). All confidence intervals for the log odds ratio are symmetric, but if you transform this to the odds ratio scale, the interval is not symmetric (but it is in some sense a "tighter" bound than any correct symmetric interval could be!).

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