A survey of 1000 air travelers1 found that 60 % prefer a window seat.

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 StartItalic UpperWord SE EndItalic equals 0.015. Use a normal distribution to find a 95 % confidence interval for the proportion of air travelers who prefer a window seat. Round your answers to three decimal places. The 95 % confidence interval is