5
$\begingroup$

When calculating a confidence interval for proportion, given a sample of the population, why is it "ok" that the standard deviation used is an estimate in itself? I mean, the actual standard deviation of $\hat{p} = \frac{x}{n}$ is $\sqrt{\frac{p(1-p)}{n}}$, how come it can be replaced by $\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$?

Thanks!

$\endgroup$
1
  • 1
    $\begingroup$ What would you use in place of this formula? Remember, you don't know $p$. Whatever you answer, bear in mind there are clear criteria for confidence intervals: the first is that they should have at least the intended coverage of the parameter; the second is that subject to this, they should actually attain the nominal coverage for some possible parameter values; the third is that subject to the first two constraints, the interval should tend to be narrow. Any satisfactory explanation would have to refer to at least one (if not all three) of these criteria. $\endgroup$
    – whuber
    Commented Jul 3, 2017 at 16:35

3 Answers 3

1
$\begingroup$

When we don't know a quantity, we estimate it using sample statistics. For example, when making confidence interval around the mean of a normal distribution, the standard deviation for the mean $\mu$ is $\sigma/\sqrt{n}$. Of course, we generally don't know this, so we replace it with $s/\sqrt{n}$, where $$s = \sqrt{\dfrac{1}{n-1}\sum_{i=1}^{n}(x_i - \bar{x})^2 }\,. $$

The reason we can do this "replacement" is because as $n$ increases, $s$ converges to $\sigma$. That is, it becomes closer and closer to $\sigma$.

With the same logic, for the confidence interval of a proportion, the standard deviation of the sample proportion is $\sqrt{p(1-p)/n}$. The quantity that we don't know, we replace with its estimate, because as $n$ increases $\hat{p}$ converges to $p$. In addition, for a function $g$, $g(\hat{p})$ converges to $g(p)$. Thus,

$$\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}} \text{ converges to } \sqrt{\dfrac{p(1-p)}{n}}\,.$$

So $\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$ is a good replacement for $ \sqrt{\dfrac{p(1-p)}{n}} \,.$

$\endgroup$
1
$\begingroup$

It could be viewed as an application of the plug-in principle: basically, you present the value you are interested in estimating as a functional of the underlying distribution $F(x)$, and then replace $F(x)$ with the empirical distribution function $F_n(x)$.

Under some conditions on the functional, the resulting estimator converges to the true parameter value, and is asymptotically normal.

$\endgroup$
1
$\begingroup$

The maximum likelihood estimate of $p$ is $\hat {p}$. The variance of the average is $p(1-p)/n$. Its MLE is $\hat {p} (1-\hat{p})/n$. Therefore the MLE of its standard error is $\sqrt{\hat {p} (1-\hat{p})/n}$. That is, the estimate with $n$ is maximum likelihood because functions of MLEs are (usually) themselves also MLEs.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.