Question

A study was conducted to determine the percent of children that want to grow up work in the same career as a parent. In a sample of 200 children, it was calculated that 43% wanted to eventually work in the same career as a parent. Construct the 95% confidence interval for the population proportion.

Solution:

phat = 0.43/200 = 0.00215
phat – qnorm(1.95/2)*sqrt(phat*(1-phat)/200) = -0.00426926

phat + qnorm(1.95/2)*sqrt(phat*(1-phat)/200) = 0.00856926

[-0.0043, 0.0086]

What is wrong with this solution?

Answer 1

Solution :

Given that,

n = 200

Point estimate = sample proportion = = 0.43

1 - = 1 - 0.43 = 0.57

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 (((0.43 * 0.57) / 200)

= 0.069

A 95% confidence interval for population proportion p is ,

± E

= 0.43  ± 0.069

= ( 0.361, 0.499 )