Question

We wish to estimate what percent of adult residents in a certain county are parents. Out of 400 adult residents sampled, 128 had kids. Based on this, construct a 90% confidence interval for the proportion pp of adult residents who are parents in this county.

Give your answers as decimals, to 4 places.

Answer 1

Solution:

Given:

n =Sample size = Number of adult residents = 400

x = Number of adult residents in a certain county are parents ( had kids) = 128

thus sample proportion of adult residents in a certain county are parents ( had kids) is:

We have to construct a 90% confidence interval for the proportion p of adult residents who are parents in this county.

Formula:

where

Z_c is z critical value for c = 90% confidence level.

Find Area = ( 1 + c ) / 2 = ( 1 + 0.90) / 2 = 1.90 / 2 = 0.9500

Look in z table for Area = 0.9500 or its closest area and find corresponding z value.

Area 0.9500 is in between 0.9495 and 0.9505 and both the area are at same distance from 0.9500

Thus we look for both area and find both z values

Thus Area 0.9495 corresponds to 1.64 and 0.9505 corresponds to 1.65

Thus average of both z values is : ( 1.64+1.65) / 2 = 1.645

Thus Z_c = 1.645

Thus

Thus

Thus we are 90% confident that the true population proportion of adult residents who are parents in this county is between 0.2816 and 0.3584.