math

In a random sample of 75 individuals, it was found that 52 of them prefer coffee to tea. What is the margin of error for the true proportion of all individuals who prefer coffee?

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asked by LK
  1. Since we're speaking of proportion in this problem, the margin of error for the true proportion of all individuals who prefer coffee is E = (z critical value)*Sqrt(p-hat)*(1-p-hat)/n), where p-hat is the sample proportion x/n (=52/75 in this case) and n is the sample size (=75 in this case). The z-critical value depends on the level of confidence desired. For 95% confidence, that z-critical value would be 1.96. If the level of confidence is not specified, assuming that it's 95% is reasonable. A greater level of confidence would link to a larger z critical value and thus a greater margin of error, and vice versa.

    Assuming that the level of confidence here is 95%, E = margin of error = 1.96*Sqrt[(52/75)(1-(52/75)/75).

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    posted by chinah

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