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A box contains 100 balls. Each ball has a number from 1 to 10. How many balls should I draw (ball is put back in box after drawing) to predict the number of balls for each number with 95% certainty.

A prediction is correct if for each of the 10 numbers, the number of balls with that number is correctly predicted.

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What work have you done? What ideas do you have? What is this for? (Should it be tagged homework?) – David Manheim Nov 27 '12 at 19:56
It is not homework. I came to this problem after thinking about a question on a dutch forum from someone who wants to know how much time is spent on several activities by checking what his people are doing at regular intervals.… His question was how many times he should monitor them to know with 95% certainty the time the different activities take. I translated it into this problem with a box containing balls that are drawn. – wnvl Nov 27 '12 at 20:13
I don't need a complete solution, but would be very happy with an indication how the problem could be solved. – wnvl Nov 27 '12 at 20:14
Is Central Limit Theorem a good enough approximation? – Tunococ Nov 27 '12 at 20:36
@Tunococ I do not say a priori that CLT can not be used, but can you explain a little bit more how you would solve it using CLT before I answer if it is good enough. – wnvl Nov 27 '12 at 20:43
up vote 2 down vote accepted

You're basically asking for confidence intervals on the probabilities of a multinomial distribution.

Let $Q$ be the number of balls in the box (you said 100), of which there are distinct labels 1 to $m$ (you said 10). Let's say there are $N_i$ balls of type $i$, with $\sum_i N_i = Q$; we'll say $N = (N_1, \dots, N_m)$.

Say you draw $n$ balls and get a vector of counts for each type $x = (x_1, \dots, x_m)$. If $m = 3$ and you drew 10 with label 1, 2 with label 2, and 3 with label 3, you'd have $x = (10, 2, 3)$. $x$ is distributed according to $\text{Multinomial}(n, p_1, \dots, p_m)$, where $p_i = N_i / Q$.

After drawing $n$ balls, your maximum-likelihood prediction of the ball counts $\hat{N}$ is clearly $(Q x_1, \dots, Q x_m)$.

Now, you want to pick $n$ large enough such that your 95% confidence region on $N$ will contain only one integer for each component.

Confidence intervals on multinomials are actually somewhat complicated, because of the interactions between cells; here's some papers. R has a function gofCI to do this asymptotically, whose help page has some more references.

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Still reading the papers, but looks very promising. This is what I was looking for. Thanks. – wnvl Nov 28 '12 at 20:10

If I understand correctly, your problem is estimating the parameter of a Bernoulli distribution. I will attempt to find an approximate answer to the problem using CLT. (This is not an exact solution.)

Suppose we're interested in estimating the number of balls with number 1 written on them. Consider a draw a success if you draw a ball with 1 written on it. Then the actual rate of success is the number of balls with number 1 divided by 100. Let $p$ be this rate.

Let $X_n$ be the number of times 1 is drawn after $n$ drawings. Then CLT says that $\sqrt{n}\left(\frac{X_n}n - p\right)$ converges in distribution to a normal distribution with mean $0$ and variance $p(1 - p)$. A less formal statement is that the distribution of $\frac{X_n}n$ is roughly normal with mean $p$ and variance $\frac 1np(1 - p)$ for large $n$. So my approximation is that you estimate $p$ by $\hat p = \frac{X_n}n$, and also variance by $\hat p(1 - \hat p)$. Then you pick $n$ large enough such that the distribution $\mathcal{N}\left(0, \frac 1n\hat p(1 - \hat p)\right)$ has probability $0.95$ within the range $[-0.005, 0.005]$.

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Thanks for the effort. First remark is that I need probability 0.95 within the range [-0.5,0.5]. Second remark is that I want that the estimation for number 1 till 10 are correct. Third remark, you use p in your calculations, but what is the value of p? Fourth remark, the number of occurences of 1 till 10 are not independent, as the sum of the number of 1s, 2s, ..., 10s is equal to 100, this is not taken into account in your answer. – wnvl Nov 28 '12 at 16:38
I just realized that you replied to my answer, and I just discovered that you did not actually read my answer... – Tunococ Jan 12 '13 at 0:44

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