Find the margin of error for each sample. then find an interval likely to contain the true population proportion. a. 15% of 457 teachers b. 56% of 87 musicians full answers

We will use a 95% confidence interval, for which the critical z-score is 1.96.

a) For n = 457, p = 0.15, mean x = 68.55, standard error = sqrt[p(1-p)/n] = 0.0167. Then the confidence interval is x +/- z*SE, which is68.55 +/- 1.96*0.0167 gives the interval (68.52, 68.58).
b) For n = 87, p = 0.56, man x = np = 48.72, SE = sqrt(0.56*0.44/87) = 0.1043The confidence interval will be 48.72 +/- 1.96*0.1043, which gives (48.52, 48.92).

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