In: Statistics and Probability
1. We wish to estimate what percent of adult residents in a
certain county are parents. Out of 600 adult residents sampled, 330
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 = ±±
2. We wish to estimate what percent of adult residents in Ventura County like chocolate. Out of 600 adult residents sampled, 454 like chocolate. Based on this, construct a 90% confidence interval for the proportion (p) of adult residents who like chocolate in Ventura County.
Confidence interval =
< p <
p = ±±
Solution :
1) Given that,
n = 600
x = 330
Point estimate = sample proportion = = x / n = 330 / 600 = 0.550
1 - = 1 - 0.550 = 0.450
At 95% confidence level
= 1 - 95%
=1 - 0.95 =0.05
/2
= 0.025
Z/2
= Z0.025 = 1.960
Margin of error = E = Z / 2 * (( * (1 - )) / n)
= 1.96 (((0.550 * 0.450) / 600)
= 0.040
A 95% confidence interval for population proportion p is ,
- E < p < + E
0.550 - 0.040 < p < 0.550 + 0.040
( 0.510 < p < 0.590 )
± E
= 0.550 ± 0.040
= ( 0.510, 0.590 )
2) Given that,
n = 600
x = 454
Point estimate = sample proportion = = x / n = 454 / 600 = 0.757
1 - = 1 - 0.757 = 0.243
At 90% confidence level
= 1 - 90%
=1 - 0.90 =0.10
/2
= 0.05
Z/2
= Z0.05 = 1.645
Margin of error = E = Z / 2 * (( * (1 - )) / n)
= 1.645 (((0.757 * 0.243) / 600 )
= 0.029
A 90% confidence interval for population proportion p is ,
- E < p < + E
0.757 - 0.029 < p < 0.757 + 0.029
( 0.728 < p < 0.786 )
± E
= 0.757 ± 0.029
= ( 0.728, 0.786 )