Question

16 con in

We wish to estimate what percent of adult residents in a certain county are parents. Out of 600 adult residents sampled, 540 had kids. Based on this, construct a 95% 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.

p =  ±±

Answer 1

Solution :

Given that,

n = 600

x = 540

Point estimate = sample proportion = = x / n = 540 / 600 = 0.900

1 - = 1 - 0.900 = 0.100

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_{0.025 = 1.960}

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

= 1.96 (((0.900 * 0.100) / 600)

= 0.024

A 95% confidence interval for population proportion p is ,

± E

= 0.900 ± 0.024

= (0.876, 0.924)

b) - E < p < + E

= 0.900 - 0.024 < p < 0.900 + 0.024

c)   ± E

= 0.900 ± 0.024