Question

The New York Times conducted a poll on women’s issues in June of 1989. One question asked was, “Many women have better jobs and more opportunities than they did 20 years ago. Do you think women have had to give up too much in the process, or not? Of the 1,033 women who were asked, 800 said “Yes.” Take these 1,033 women to be a simple random sample of all adult women. Calculate a 95% confidence interval for the proportion of all adult women who would say “Yes” to this statement using two decimal places.

What's the center of the interval?

What's the margin of error?

What's the lower bound of the interval?

What's the upper bound of the interval?

Answer 1

Solution :

Given that,

n = 1033

x = 800

Point estimate = sample proportion = = x / n = 800 / 1033 =0.77

1 - = 1 - 0.77 = 0.23

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z0.025 = 1.960

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

= 1.96 (((0.77 * 0.23) / 1033)

= 0.03

A 95% confidence interval for population proportion p is ,

± E   

= 0.77  ± 0.03

= ( 0.74, 0.80 )

lower bound = 0.74

upper bound = 0.80