1
$\begingroup$

Hi

Just a short question. How are the IQR of the boxplot related to the confidence interval of a sample? Is the IQR actually the 50% confidence interval?

$\endgroup$

3 Answers 3

1
$\begingroup$

My answer didn't seem to score any points with anyone, and rdchat's answer is lousy, so let's look more closely.

Suppose $X_1,\dots,X_n$ are an i.i.d. sample from a normally distributed population with unknown mean $\mu$ and unknown variance $\sigma^2$, and we seek a confidence interval for the population mean. As usual, let $\overline{X} = \left(X_1 + \cdots + X_n\right)/n$ be the sample mean and $S^2 = 1/(n-1)\sum_{i=1}^n (X_i - \overline{X})^2$ be the (unbiased) sample variance. (BTW, unbiasedness is overrated. Everybody knows that, except some non-statisticians. Apparently there are lots of those.)

Now $$ \frac{\overline{X} - \mu}{\sigma/\sqrt{n}} $$ is normally distributed with mean 0 and variance 1, and $$ \frac{\overline{X} - \mu}{S/\sqrt{n}} $$ has a Student's t-distribution with $n-1$ degrees of freedom (called "Student's" because it's named after William Sealey Gosset, of course---and some of the content of this forum makes one suspect that a lot of people here don't know that standard bit of folklore). Go to the table and look up the value of $A_n$ for which $$ \Pr\left( -A_n < \frac{\overline{X} - \mu}{S/\sqrt{n}} < A_n \right) = \frac{1}{2}, $$ or in other words $$ \Pr\left(\overline{X} - A_n \frac{S}{\sqrt{n}} < \mu < \overline{X} + A_n \frac{S}{\sqrt{n}}\right) = \frac{1}{2}. $$ Then $$ \overline{X} \pm A_n \frac{S}{\sqrt{n}} $$ are the endpoints of a 50% confidence interval for $\mu$.

Important point: The length of the confidence interval goes to 0 as the sample size increases, since $\sqrt{n}$ is in the denominator (and $A_n$ approaches the value one would get for the normal distribution rather than for Student's distribution). But the sample quartiles do not get closer together in the limit as $n$ grows, since ("almost surely") they approach the population quartiles.

So the answer is NO.

$\endgroup$
1
$\begingroup$

First, the interquartile range (IQR) is the difference between the third and first quartiles, which is a single number, not an interval.

The interval $(Q_{1},Q_{3})$ might be considered a 50% confidence interval, but it's not a 50% confidence interval for the mean that you'd get from any of the usual formulas.

$\endgroup$
1
  • $\begingroup$ Saying this "might be considered a 50% confidence interval" makes no sense at all. I've expanded on this in another answer below. $\endgroup$ Jun 14, 2010 at 14:39
0
$\begingroup$

You'd hope that as the sample size grows the 50% confidence interval for a location parameter would shrink whereas the first and third quartiles of the sample would approach those of the population. So no, that's not a 50% confidence interval except in special cases, e.g. when the sample size is 2.

$\endgroup$

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.