A survey of 1000 air travelers 1 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 = 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 is to .

answer: .75 km

step-by-step explanation:

by definition, an arc subtended by a central angle of some given measure has the same measure. so .

multiply then divide

step-by-step explanation:

step by step solution :

step   1   :

1.1     4 = 22

(4)3 = (22)3 = 26

equation at the end of step   1   :

  ((23) • 26)2

step   2   :

equation at the end of step   2   :

  (23 • 26)2

step   3   :

final result :

  218

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