st.statistics - Boxplot IQR and confidence interval

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 $overline{X} = left(X_1 + cdots + X_nright)/n$ 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
$$
Prleft( -A_n < frac{overline{X} - mu}{S/sqrt{n}} < A_n right) = frac{1}{2},
$$
or in other words
$$
Prleft(overline{X} - A_n frac{S}{sqrt{n}} < mu < overline{X} + A_n frac{S}{sqrt{n}}right) = frac{1}{2}.
$$
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.