Variation of the coupon problem

[Smitz]

I have a problem which is basically a variation of the coupon problem... I can't get my head around it as I haven't done much stats .

I have a sample size of 900, say a pot of 900 balls, each with a unique id. Each week I am going to select a set number of balls randomly from the pot to test them. After testing, the balls go back into the pot. How many balls do I need to test every week so that after 3 years I can be, say, 90% confident that 90% of the sample have been tested in that period.

Any help would be greatly appreciated (especially a solution!)

Andrew

 

[EnumaElish]

This is related to the critical sample size problem:
http://en.wikipedia.org/wiki/Confidence_interval

E.g. to apply a "two-tailed" test to a given value x of a normally distributed random variable with mean m and variance σ², one can define z = (x - m) / (σ/√n) then solve Φ(z) = 0.975 for n, where Φ is the cumulative normal distribution function.

 

[tacman]

The variation of the coupon problem that you have described is known as the "balls in a jar" problem. In this case, the balls represent the coupons and the jar represents the population from which the coupons are being drawn. The goal is to determine how many balls need to be drawn (or coupons need to be tested) in order to have a certain level of confidence that a certain percentage of the population has been covered.

To solve this problem, we can use the concept of confidence intervals. A confidence interval is a range of values within which we can be confident that the true population parameter lies. In this case, our population parameter is the proportion of balls (or coupons) that have been tested.

To determine the sample size needed, we need to consider the following factors:
1. Confidence level: This is the level of certainty that we want to have in our results. In this case, you have specified a confidence level of 90%.
2. Margin of error: This is the maximum amount by which the true population proportion can differ from our sample proportion. In other words, it is the degree of precision we want in our estimate. For this problem, we can set the margin of error to be 5% (since you want to be 90% confident that 90% of the sample has been tested, we can allow for a 5% margin of error in either direction).
3. Population size: This is the total number of balls in the pot, which in your case is 900.

Using these factors, we can use a formula to calculate the sample size needed:
n = (Z^2 * p * q) / (e^2)
where:
n = sample size
Z = the z-score corresponding to the desired confidence level (for a 90% confidence level, Z = 1.645)
p = estimated proportion of the population that we want to capture (in this case, 90%)
q = 1-p (or the proportion of the population that we do not want to capture, which is 10% in this case)
e = margin of error (in decimal form, so 5% = 0.05)

Plugging in the values, we get:
n = (1.645^2 * 0.9 * 0.1) / (0.05^2) = 119.16

So, you would need to test 119 balls every week in