How many data we need to be 95% confidence to tell A from B? Here is the problem:
We know that 


*

*50% probability a College student can answer question correctly from a question set.

*25% probability a High School student can answer question correctly from the same question set.


Now, how many questions we need ask a person (who is either College or HighSchool), so that we can decide whether he is College or HighSchool with 95% confidence ?
Thanks a lot  
PS: This is actually an interview question that my friend asked me, so I do not know how to define "confidence". Apparently the definition is part of the problem.
 A: Here's my attempt. Let's let $X$ be the number of correct answers for the college student, so $X\sim \mathrm{Bin}(n_1, \frac{1}{2})$. Then let's let $Y$ be the number of correct answers for a high school student, so $Y\sim \mathrm{Bin}(n_2, \frac{1}{4})$. Since we'll be doing the experiment on one individual, we can assume $n_1=n_2=n$.
We know $E(X)=\frac{n}{2}$ and $Var(X)=\frac{n}{4}$, so $\sigma_X=\frac{\sqrt{n}}{2}$. Similarly, $E(Y)=\frac{n}{4}$ and $\sigma_Y=\frac{\sqrt{3n}}{4}$.
If $n$ is sufficiently large, then both $X$ and $Y$ are approximately normal, so we'd expect roughly 95% of $X$ values to be in $\frac{n}{2}\pm2\frac{\sqrt{n}}{2}$, and 95% of $Y$ values to be in $\frac{n}{4}\pm2\frac{\sqrt{3n}}{4}$.
If we want to be 95% certain, we just make the lower limit of the college confidence interval above the upper limit of the high school interval. In other words, $$\frac{n}2+2\frac{\sqrt{3n}}4\le\frac{n}2-2\frac{\sqrt{n}}2$$
Solving for $n$, we get $n\ge(4+2\sqrt3)^2$, so $n\ge56$.
