Question

A survey of 1000 air travelers 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 and interpret a 99% confidence interval for the proportion of air travelers who prefer a window seat.A survey of 1000 air travelers 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 and interpret a 99% confidence interval for the proportion of air travelers who prefer a window seat.

Answer 1

Solution :

n = 1000

= 0.60

1 - = 1 - 0.60 = 0.40

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

Standard error = (( * (1 - )) / n)

=(((0.60 * 0.40) / 1000)

=0.015

The standard error is SE = 0.015

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 2.576 * (((0.60 * 0.40) / 1000)

= 0.034

A 99 % confidence interval for population proportion p is ,

- E < P < + E

0.60 - 0.034 < p < 0.60 + 0.034

0.560 < p < 0.274  

( 0.560, 0.634 )