Question

A flight is said to be on time if it lands within 15 min. of its published arrival.

a.) What is the minimum sample size needed to be 95% confident within 2 percentage points of the actual proportion of on-time flights?

b) A recent study of 150 flights found that 84% of Southwest Airlines flights arrive on time. Rounding to the nearest tenth of a percent, we can say with 95% confidence that the proportion of soutwest flights that arrive on time is between ____ and ____ with a margin of error ____.

Answer 1

Solution :

Given that,

a)

= 0.5

1 - =0.5

margin of error = E = 0.02

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.96 / 0.02)² * 0.5*0.5

= 2401

sample size = 2401

b)

n = 150

Point estimate = sample proportion = = 0.84

1 - = 0.16

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.96 * (((0.84*0.16) / 150)

= 0.059

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.84 - 0.059 < p < 0.84 + 0.059

0.781 < p < 0.899

we can say with 95% confidence that the proportion of southwest flights that arrive on time is between 0.781 and 0.899 with a margin of error 0.059.