Consider a single value, \(X\), from a \(\BinomDistn(n, \pi)\) distribution. The MLE of \(\pi\) is
\[ \hat{\pi} \;\;=\;\; \frac X n \]The binomial variable \(X\) has mean and variance
\[ E[X] = n\pi \spaced{and} \Var(X) = n\pi(1 - \pi) \]Standardising \(X\) (subtracting its mean and dividing by its standard deviation) gives a distribution that is approximately normal as the sample size, \(n\), increases. Therefore
\[ \frac{X - n\pi}{\sqrt{n \pi(1 - \pi)}} \;\;\underset {\text{approx}}{\sim} \;\; \NormalDistn(0, 1) \]This can be used as a pivot. An approximate 95% confidence interval is therefore the solution to
\[ -1.96 \;\;\lt\;\; \frac{x - n\pi}{\sqrt{n \pi(1 - \pi)}} \;\;\lt\;\; 1.96 \]Question
A retail clothing outlet has collected the following data from random sampling of invoices for T-shirts over the past month.
Small | Medium | Large | XL | Total | |
---|---|---|---|---|---|
North Island | 2 | 15 | 24 | 9 | 50 |
South Island | 4 | 17 | 23 | 6 | 50 |
Find a 95% confidence interval for the probability that a T-shirt purchased from one of the store's North Island shops is Small.
(Solved in full version)
Using a pivot, the above 95% confidence interval is
\[ 0.011 \;\;\lt\;\; \pi \;\;\lt\;\; 0.135 \]whereas the conventional Wald-type 95% confidence interval would be
\[ -0.014 \;\;\lt\;\; \pi \;\;\lt\;\; 0.094 \]This includes impossible negative values for \(\pi\), so the confidence interval found from a pivot is better.