A desk wants to estimate the share of orders that are aggressive. The half-width is h = 2 × √(p(1 − p)/n). Using the worst case p = 0.5, what is the smallest integer n guaranteeing h ≤ 0.02?

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A desk wants to estimate the share of orders that are aggressive. The half-width is $h = 2 \times \frac{p ( 1 - p )}{n}$ . Using the worst case $p = 0.5$ , what is the smallest integer $n$ guaranteeing $h \leq 0.02$ ?

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Hint

The worst case maximizes $p (1 - p)$ .

Solution

With $p = 0.5$ we have $p (1 - p) = 0.25$ , so $p (1 - p) = 0.5$ .
The condition becomes $2 \times \frac{0.5}{n} \leq 0.02$ .
This means $n \geq \frac{1}{0.02} = 50$ .
Squaring gives $n \geq 2500$ .
So the smallest $n$ is $2500$ .

Answer: $2500$

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