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We have a population where the mean is 50.5 and the variance is 833.25, we need to calculate the sample size given a 95% confidence and the sample mean lying 3 units of the population mean.

So if I do it with duplicates allowed i get n > 355 and if I don't I am supposed to get n > 79. What equation do we have to use? I used the same equation and method and god 355 for the second one instead of 79, but basically I don't even know what I am doing. Can someone explain to me so I can understand what is being done and why?

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Sample size $355$ is about right for sampling with replacement. For sampling without replacement, which I am guessing is meant by the unusual expression "no duplicates," one can give an answer only if the total population is of finite known size. The answer depends very much on the population size. If that is large (a few thousand) then it will not matter much whether one is sampling with or without replacement. – André Nicolas Aug 18 at 0:30
do you know what equation we must use? the one i've seen uses N and n and it's completely different from the one i've used before. – Arma Bingo Aug 18 at 0:36
Not off-hand. If the actual question were given, I would probably know what to do. As it is, there was only the vague no duplicates which is not enough. – André Nicolas Aug 18 at 0:40
well, I'll just give you the equation then... both uses the confidence interval equation, but in the second one we use: $n =$ $\frac{N*Var(x)}{(N-1)B^{2}/k^{2} + Var(x)}$ I have no idea where that comes from. In the first one we have $n = StandardDev(x)*Z/number_of_units$ it's a lot more straightforward. – Arma Bingo Aug 18 at 0:45
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I was just going to post something similar. For the binomial distribution, one makes a correction to the variance by multiplying by $\frac{N-n}{N-1}$, where $N$ is the population size and $n$ the sample size. – André Nicolas Aug 18 at 0:47
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