Question
In: Statistics and Probability
Data from 2011 shows that 25% of all flights from IAH departed late. An analyst took...
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.
Solutions
Expert Solution
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 * (
= 0.041
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 )
orchestra answered 2 years agoRelated Solutions
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