Question

According to the U.S. Census Bureau, 43% of men who worked at home were college graduates. In a sample of 525 women who worked at home, 140 were college graduates. What is the upper bound for the 99% confidence interval for the proportion of women who work at home who are college graduates? Round to three decimal places (for example: 0.419). Write only a number as your answer.

Answer 1

Solution :

Given that,

n = 525

= 43% = 0.43

1 - = 1 - 0.43 = 0.57

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

= 0.01

Z = Z_{0.01 = 2.326}

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

= 2.326 * (((0.43 * 0.57) / 525) = 0.050

A 99 % confidence interval for population proportion p is ,

+ E

0.43 + 0.050

0.48

upper bound = 0.48