11.20a: Here n = 60, k = 44
Conservative 95% C.I.: pi = p +/- (1.96)SE = 44/60 +/- (1.96)(1/2sqrt[60]) = .7333 +/- .1265
or .6068 <= pi <= .8599
11.20b: Bootstrapping 95% C.I.: pi = p +/- (1.96)SE = 44/60 +/- (1.96)(sqrt[(44/60)(16/60)/60]) = .7333 +/- .1119
or .6214 <= pi <= .8452
11.22a:
Conservative 95% C.I.: pi = p +/- (1.96)SE = p +/- (1.96)(1/2sqrt[n])
To make the width of this interval no more than .10, we solve
(1.96)(1/2sqrt[n]) < .05
(1.96)(1/2)(1/.05) < sqrt[n], or 19.6 < sqrt[n] or n > 384.16, which means that the minimum sample size is 385.
11.22b: If pi <= .25, we can use the bootstrapping CI to solve
(1.96)(sqrt[(.25)(.75)]/sqrt[n]) < .05
(1.96)(sqrt[(.25)(.75)]/(.05) < sqrt[n]
16.9741 < sqrt[n]
288.12 < n, which means that the minimum sample size is 289.
11.32: Here p = 40/144 = .278. Using Table 5, we get the 95% CI of .21 < pi < .37 (approximately. It depends on the thinness of the lines you draw).