In: Statistics and Probability
Question:
Give an example of a hypergeometric random variable for which a binomial random variable is NOT a good approximation. You must describe each of the following:
i. the experiment
ii. a random variable X from the experiment and what X represents
iii. the probability mass function (PMF) of X
iv. a binomial random variable that approximates X and its parameters
v. the PMF of the binomial random variable and why it's a good estimate of the PMF of X
Solution:-
The Binomial Approximation to the Hypergrometric
Suppose we still have the population of size N with M units labelled as "success" and N-M labelled as "failure", but now we take a sample of size n is drawn with replacement. Then, with each draw, the units remaining to be drawn look the same: still M "successes" and N-M "failures". Thus, the probability drawing a "success" on each single draw is
and this doesn't change. When we were drawing without replacement, the proportions of successes would change, depending on the result of previous draws. For example, if we were to a obtain a "success" on the first draw, then the proportion of "successes" for the second draw would be (M-1)/(N-1), whereas if we were to obtain a "failure" on the first draw the proportion of successes for the second draw would be M/(N-1).
The random variable Y is defined as the number of "successes" in the sample, when we are drawing with replacement. Then Y is a binomial random variable.
The probability mass function for Y is
and otherwise.
Proposition: if the population size in such a way that the proportion of successes , and n is held constant, then the hypergeometric probability mass function approaches the binomial probability mass function.
In practice, this means that we can approximate the hypergeometric probabilities with binomial probabilities, provided . As a rule of thumb, if the population size is more than 20 times the sample size (N > 20 n), then we amy use binomial probabilities in place of hypergeometric probabilities.
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