Assume that a sample is used to estimate a population proportion p. Find the margin of error (E) that corresponds to a sample of size 123 with 25 successes at a confidence level of 98%.

Answer 1

The margin of error for a sample proportion p is ME = z√[p(1-p)/n]

p = 25/123, z = 2.326 (for 98% confidence level), and n = 123:

ME = z√[p(1-p)/n] = 2.326√[(25/123)(1-25/123)/123] ≈ 0.0844

Thus, the margin of error is about 0.0844

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