Question

A sample of n=600 homeowners living in a particular community showed that 60% of them had lived in the community at least five years. A confidence interval for the population proportion of homeowners who have lived in the community at least five years is given by:

.6 ± 1.645 ( .6 ) ( .4 ) 600 ⟹.6 ± 1.645 ( .02 ) ⟹.6 ± .0329 ⟹.5671 to .6329 or 56.71% to 63.29%.

The estimate used in making the confidence interval is __________ .

Answer 1

Solution :

Given that,

n = 600

Point estimate = sample proportion = = 0.60

1 - = 1-0.60 - 0.40

Z/2 = 1.645

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

= 1.645 * ((0.60*(0.40) /300 )

= 0.0329

A 90% confidence interval for population proportion p is ,

E

0.60-0.0329 < p < 0.60+0.0329

( 0.5671 ,0.6329 ) or 56.71% to 63.29%

The estimate used in making the confidence interval is ( 0.5671 to 0.6329 )