Question

4. According to one source, 50% of plane crashes are due at least in part to pilot error (http://www.planecrashinfo.com). Suppose that in a random sample of 100 separate airplane accidents, 62 of them were due to pilot error (at least in part). a. Obtain a 95% CI for the percentage of plane crashes due at least in part to pilot error. b. If a 95% CI is desired with a width of 10%, what sample size will accomplish this regardless of value of p^ ..

Answer 1

a)

Sample proportion $\hat{p}$ = 62 / 100 = 0.62

95% confidence interval for p is

$\hat{p}$ - Z$\alpha$/2 * sqrt [ $\hat{p}$ ( 1 - $\hat{p}$ ) / n] < p < $\hat{p}$ + Z$\alpha$/2 * sqrt [ $\hat{p}$ ( 1 - $\hat{p}$ ) / n]

0.62 - 1.96 * sqrt ( 0.62 (1 - 0.62) / 100) < p < 0.62 + 1.96 * sqrt ( 0.62 (1 - 0.62) / 100)

0.525 < p < 0.715

95% CI is ( 0.525 , 0.715 )

b)

Margin of error = Width of CI / 2 = 0.10 / 2 = 0.05

Sample size = Z²$\alpha$/2 * p ( 1 - p) / E²

= 1.96² * 0.62 * 0.38 / 0.05²

= 362.03

= 363 (Rounded up to nearest integer)