Math CentralQuandaries & Queries


Question from mini, a student:

Find the sample size needed to be 98% confident thata marketing survey on the proportion of shoppers who use the internet for holiday shopping is accurate within a margin of error of 0.02. Assume that the conditions for a binomial distribution are met, and that a current estimate for a sample proportion does not exist.

Hi Mini,

The confidence limits for the proportion p a binomial distribution is


where n is the sample size.

Since you want 98% confidence interval α is 0.02. Since the margin of error is 0.02 the bounds on the confidence interval are ±0.02. Since you have no previous estimate for p take p = 1/2.

Set the bounds on the confidence limits above to ±0.02, substitute the values you know and solve for n.

I hope this helps,

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