A small pilot study estimated that 34% of Americans do not like the taste of cilantro. How large would the random sample need to be to obtain a margin of error of at most 0.03 with 95%?

For a population proportion, use n = z²p(1 − p)/E². With 95% confidence, z = 1.96; using p = 0.34 and margin of error E = 0.03 gives n = (1.96²)(0.34)(0.66)/(0.03²) ≈ 957.84. Since the sample size must be a whole number and must meet the margin of error, round up to 958, choice B.

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