Question

Out of 300 people sampled, 246 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

Answer 1

Solution :

Given that,

n = 300

x = 246

Point estimate = sample proportion = = x / n = 246/300=0.82

1 -   = 1- 0.82 =0.18

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z  0.05 = 1.645 ( Using z table )

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

= 1.645 *((0.82*0.18) / 300)

E = 0.036

A 90% confidence interval proportion p is ,

- E < p < + E

0.82-0.036 < p < 0.82+0.036

0.784< p < 0.856