In order to solve the following inequality for $n$, where $0 < \theta < 1$, they use a trick that I do not understand

$n \geq (\frac{1.96}{0.03})^2 \theta(1 - \theta)$

Since $\theta$ is unknown we take $\theta=0.5$ so the inequality is true for all $0 < \theta < 1$. Thus

$n \geq (\frac{1.96}{0.03})^2 (0.5)^2 = 1067.1$

Why are they allowed to set $\theta = 0.5$? Why does that value of $\theta$ make the inequality hold for all $0 < \theta <1$?

  • $\begingroup$ In general, drawing a little picture is a good idea. In this case, plotting $x \mapsto x(1-x)$ will give a hint. $\endgroup$ – copper.hat Jun 9 at 20:56

The value of $\theta(1-\theta)$ is maximal when we set $\theta=\frac12$. That is the reasoning justifying that step.

(Think of a sketch of the parabola $x(1-x)$).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.