Question

In: Statistics and Probability

A random sample of 400 passengers of an airline is polled after their flights. Of those...

A random sample of 400 passengers of an airline is polled after their flights. Of those passengers, 300 say they will fly again with the same airline. Which of the following is the 99% confidence interval for the proportion of passengers that will fly again with the same airline?

Expert Solution

Solution :

Given that,

n = 400

x = 300

Point estimate = sample proportion = = x / n =300/400=0.75

1 - =0.25

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576 ( Using z table )

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

= 2.576* ( $\sqrt$ ((0.75*0.25) /400 )

E = 0.056

A 99% confidence interval is ,

- E < p < + E

0.75-0.056< p < 0.75+0.056

(0.694 , 0.806)

orchestra answered 2 years ago

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