Question

We wish to estimate what percent of adult residents in a certain county are parents. Out of 100 adult residents sampled, 74 had kids. Based on this, construct a 99% confidence interval for the proportion p of adult residents who are parents in this county.

Express your answer in tri-inequality form. Give your answers as decimals, to three places.

< p <  Express the same answer using the point estimate and margin of error. Give your answers as decimals, to three places.

Answer 1

Solution :

Given that,

n = 100

x = 74

Point estimate = sample proportion = = x / n = 74/100 = 0.740

1 - = 1 -0.740 = 0.260

At 99% confidence level

= 1-0.99% =1-0.99 =0.01

/2 =0.01/ 2= 0.005

Z/2 = Z0.005 = 2.576

Z/2 = 2.576

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

= 2.576 * ((0.74*(0.26) /100 )

= 0.113

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.740 -0.113 < p < 0.740+0.11

0.627< p < 0.853

(0.627 ,0.853 )