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 please

Respuesta :

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).