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In lecture today, my professor stated that the Wald confidence interval for the logit of a binomial parameter $\pi$ can be manipulated to obtain a Wald confidence interval for the parameter itself. I have been trying to figure out how this is possible but have had lackluster results.

The interval for the logit:

$\log(\frac{\hat \pi}{1-\hat \pi})-z_{\frac{\alpha}{2}}\cdot \frac{1}{\sqrt{n\hat \pi(1-\hat \pi)}}<\log(\frac{\pi}{1-\pi})<\log\left(\frac{\hat \pi}{1-\hat \pi}\right)+z_{\frac{\alpha}{2}}\cdot \frac{1}{\sqrt{n\hat \pi(1-\hat \pi)}}$

The interval for the binomial parameter:

$\hat \pi-z\sqrt{\frac{1}{n}\hat \pi(1-\hat \pi)}<\pi<\hat \pi+z\sqrt{\frac{1}{n}\hat \pi(1-\hat \pi)}$

I tried taking the exponential function on the logit interval:

$\frac{\hat \pi}{1-\hat \pi}\cdot \exp(-z_{\frac{\alpha}{2}}\cdot\frac{1}{\sqrt{n\hat \pi(1-\hat \pi)}})<\frac{\pi}{1-\pi}<\frac{\hat \pi}{1-\hat \pi}\cdot \exp(z_{\frac{\alpha}{2}}\cdot\frac{1}{\sqrt{n\hat \pi(1-\hat \pi)}})$

And then multiplying by $1-\pi$:

$\hat \pi \cdot \exp(-z_{\frac{\alpha}{2}}\cdot\frac{1}{\sqrt{n\hat \pi(1-\hat \pi)}})<\pi<\hat \pi \cdot \exp(z_{\frac{\alpha}{2}}\cdot\frac{1}{\sqrt{n\hat \pi(1-\hat \pi)}})$

However, I am a bit confused as this doesn't look like the interval for the binomial parameter. What am I doing wrong? I'd appreciate any guidance. Thank you in advance.

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Your CI does not have $\pm$ in it, so what? It is still has the (large sample) 95% coverage, because the underlying one for the logit parameter does. It is asymmetric, and a lot of people find that appealing for a parameter that has a restricted range. –  StasK Feb 13 at 4:36

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My guess is that your professor had in mind the delta-method: if you have a sequence $\sqrt{n}(X_n-\mu) \rightarrow N(0,v)$ in distribution, then for a continuously differentiable $f(x)\in C^1_{\{\mu\}}$, $\sqrt{n}(f(X_n)-f(\mu))\rightarrow N(0,(f'(\mu))^2 v)$. So if $\pi$ is the binomial parameter, and $\theta$ is its logit (odds ratio) transform $\theta = \ln \frac{\pi}{1-\pi}$, we have $$\pi(\theta) = \frac 1{1+\exp(-\theta)},$$ $$\pi'(\theta) = \frac{\exp(-\theta)}{(1+\exp(-\theta))^2}=\pi(\theta)(1-\pi(\theta)),$$ and if $$\sqrt{n}(\hat\theta-\theta_0) \sim N(0,v),$$ then $$\sqrt{n}(\hat\pi-\pi_0) \sim N(0,\pi_0^2(1-\pi_0)^2v).$$ Now, since $v=1/\pi(1-\pi)$, we have $$\sqrt{n}(\hat\pi-\pi_0) \sim N(0,\pi_0 (1-\pi_0) )$$ which of course makes perfect sense from the point of view of the central limit theorem. So if you have the logit estimate $\hat\theta_n$ with a standard error $s_n$, then the binomial parameter estimate is $$\hat\pi_n = \frac1{1+\exp(-\hat\theta_n)}$$ with a standard error $$s_n \frac{\exp(-\hat\theta_n)}{(1+\exp(-\hat\theta_n))^2}.$$ In practice, the logit transform has better properties and achieves approximate asymptotic normality faster than the binomial parameter itself, so inference on the logit scale is preferable. Also, the standard error may be derived by extraneous variance estimation methods, such as the bootstrap, especially when the data are not i.i.d., so $v$ is different from $1/\pi(1-\pi)$ (usually larger with positively correlated data).

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