Question

Out of 100 people sampled, 28 had kids. Based on this, construct a 90% confidence interval for the true population proportion of people with kids.

Give your answers as decimals, to three places

____ < p < ____

Answer 1

Solution :

Given that,

n = 100

x = 28

Point estimate = sample proportion = = x / n = 28 / 100 = 0.28

1 - = 1 - 0.28 = 0.72

At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645}

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

= 1.645 (((0.28 * 0.72) / 100 )

= 0.074

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.28 - 0.074 < p < 0.28 + 0.074

0.206 < p < 0.354