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Finding a sample size to meet margin of error requirements

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Consider the case where we have n observations, sampled from a binomial pdf (our parameter estimator can therefore be given by X/n). Assume that the margin of error is 6%. How many observations do we need for the margin of error for the proportion to be cut in half (to 3%)?

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A step-by-step solution is provided. Given a margin of error requirement to be met, we find the minimum sample size needed. Word Attachment included with solution.

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A: A well-known formula for the margin of error from a proportion is given by

d=1.96/[2*sqrt(n)].

where d is the 'margin of error' and n is the sample size. We should note that 1.96 is chosen so that we may have a 95% probability of being within .06 (the margin of error). If you want to examine this more generally, it can be thought of as being z(alpha/2).

For our problem, we have that ...

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