Question

A survey of 1000 air travelers1 found that 60% prefer a window seat. The sample size is large enough to use the normal distribution, and a bootstrap distribution shows that the standard error is .SE is = 0.015. Use a normal distribution to find a 99% confidence interval for the proportion of air travelers who prefer a window seat. Round your answers to three decimal places.

The 99% confidence interval, value

Answer 1

Solution:

n= 1000, = 60%=0.60, C= 99%

formula for standard error is

= 0.0154919334

= 0.015

thus standard error is = 0.015

now formula for confidence interval is

Where Zc is the Z critical value for C= 99%

Zc= 2.58

0.560 < P < 0.639

Thus 99% confidence interval is ( 0.560 , 0.639 )

Therefore we are 99% confident that the the proportion of air travelers who prefer window seat lies between 0.560 to  0.639

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