# Prove minimal sample size

A point estimator for a Bernoulli probability $\theta$ is $\hat{\theta}$ is $\hat{\theta}=\frac{X}{n}$ where $X$ is the number of successes in $n$ Bernoulli trials. Using the normal approximation to the binomial for large $n$ (without continuity correction), show that the minimum sample size $n$ required to ensure that there is a probability of at least $1 − \alpha$ that $\hat{\theta}$ differs from $\theta$ by no more than $d$, is given by the smallest integer $n$ for which $$n\ge\theta(1-\theta)\left(\dfrac{z_{\alpha/2}}{d} \right)^2$$.

I tried to do it using Chebyshev's Inequality.

Pr($|\hat{\theta}-\theta| \geq d)\le 1-\alpha$

Pr($|\hat{\theta}-\theta|\ge d)\le \left(\dfrac{n\theta(1-\theta)}{d^2} \right)$

So, $1-\alpha=\left(\dfrac{n\theta(1-\theta)}{d^2} \right)$ and $n\ge \left(\dfrac{d^2(1-\alpha)}{\theta(1-\theta)} \right)$.

But I got stuck here and don't know how to proceed further. Or maybe Chebychev's Inequality isn't useful here?

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Use information from wikipedia, where this is solved for $\alpha=0.05$: http://en.wikipedia.org/wiki/Sample_size_determination#Estimating_proportions_and_means
$$n \geq \frac{Z_{\alpha/2}^2 \hat{\theta} (1 - \hat{\theta})}{ d^2}$$
which is maximized when $\hat{\theta} (1 - \hat{\theta})$ is maximized (use differential calculus), if you do not have a prior idea of $\hat{\theta}$.