# Large Sample Confidence Intervals for a Population Mean and Proportion

Posted by Beetle B. on Sun 16 July 2017

$$\newcommand{\Cov}{\mathrm{Cov}}$$ $$\newcommand{\Corr}{\mathrm{Corr}}$$ $$\newcommand{\Sample}{X_{1},\dots,X_{n}}$$

For any distribution, if $$n$$ is large, use $$Z=\frac{\bar{X}-\mu}{S/\sqrt{n}}$$ By the Central Limit Theorem, it is approximately normal. Then $$\bar{x}\pm z_{\alpha/2}s/\sqrt{n}$$ will give you a confidence interval for $$\mu$$. It is approximately $$100(1-\alpha)$$

$$n>40$$ is a good size to justify this approximation. Note that this actually has 2 random variables: $$\bar{X},S$$, which is why $$n$$ is a bit larger than our previous rule for the CLT.

There is no way to estimate $$n$$ in advance, though. Just be conservative on $$\sigma$$.

For large $$n$$, the upper confidence bound for $$\mu<\bar{x}+z_{\alpha}s/\sqrt{n}$$.

For large $$n$$ the lower confidence bound for $$\mu>\bar{x}-z_{\alpha}s/\sqrt{n}$$.

## A General Large Sample Confidence Interval

Need to estimate $$\theta$$. Let $$\hat{\theta}$$ satisfy:

1. It is approximately normal.
2. It is unbiased.
3. The expression for $$\sigma_{\hat{\theta}}$$ is known.

Then standardizing $$\hat{\theta}$$ gives:

\begin{equation*} P\left(-z_{\alpha/2}<\frac{\hat{\theta}-\theta}{\sigma_{\hat{\theta}}}<z_{\alpha/2}\right)\approx1-\alpha \end{equation*}

If $$\hat{\theta}$$ involves an unknown (e.g. $$\sigma$$), see if substituting $$s_{\hat{\theta}}$$ works. If $$\sigma_{\hat{\theta}}$$ involves $$\theta$$, try replacing $$\theta$$ with $$\hat{\theta}$$.

## A Large Sample Confidence Interval for a Population Proportion

Recall that if $$n<<N$$, and $$np\ge10$$, and $$nq\ge10$$, then $$X$$, the number of successes in the sample, can be treated as a normal distribution. $$p$$ is unknown. But if we use $$\hat{p}=\frac{X}{n}$$, then it is approximately normal. We know that $$E(\hat{p})=p$$ and $$\sigma_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}$$. Standardizing $$\hat{p}$$ and computing the CI:

\begin{equation*} p=\frac{p+\frac{z_{\alpha/2}^{2}}{2n}\pm z_{\alpha/2}\sqrt{\frac{\hat{p}\hat{q}}{n}+\frac{z_{\alpha/2}^{2}}{4n^{2}}}}{1+\frac{z_{\alpha/2}^{2}}{n}} \end{equation*}

If $$n$$ is large, this is roughly $$\hat{p}\pm z_{\alpha/2}\sqrt{\frac{\hat{p}\hat{q}}{n}}$$

But the full form should be used. It is often accurate even when $$n\hat{p}\ge 10$$ or $$n\hat{q}\ge 10$$ does not apply.

If $$w$$ is the desired width, then $$n\approx\frac{4z_{\alpha/2}^{2}\hat{p}\hat{q}}{w^{2}}$$. But we don’t know $$\hat{p}\hat{q}$$. However, we do know it is maximum when each is 0.5.