Question

Data from 2011 shows that 25% of all flights from IAH departed late. An analyst took a simple random sample of 400 flights from IAH in 2019 and found that 92 flights departed late. Based on the sample data, construct a 95% confidence interval of true proportion of flights from IAH that departed late.

Answer 1

Solution :

Given that,

n = 400

x = 92

Point estimate = sample proportion = = x / n = 92/400 = 0.230

1 - = 1-0.230 = 0.77

At 95% confidence level

= 1-0.95% =1-0.95 =0.05

/2 =0.05/ 2= 0.025

Z/2 = Z0.025 = 1.960

Z/2 = 1.960

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

= 1.960 * ((0.230*(0.770) /400 )

= 0.041

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.230 -0.041 < p < 0.230+0.041

0.189 < p < 0.271  

( 0.189 ,0.271 )