Question

Out of 200 people sampled, 122 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 = 200

x = 122

Point estimate = sample proportion = = x / n = 122/200=0.61

1 -   = 0.39

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

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

= 1.645 *((0.61*0.39) /200 )

E = 0.057

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.61-0.057 < p < 0.61+0.057

(0.553 , 0.667)