Question

1.if 52% of 1100 randomly pulled voters of one state said that they support certain policy find 95% confidence interval?

2.Use confidence interval from problem 1 to tell if its likely that more than 50% of voters support that policy?

Answer 1

Solution :

Given that,

n = 1100

Point estimate = sample proportion = = 0.52

1 - = 1-0.52 = 0.48

At 95% confidence level

= 1-0.95% =1-0.95 =0.05

/2 =0.05/ 2= 0.025

Z/2 = Z_{0.025 = 1.960}

Z_/2 = 1.960

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

= 1.960 * ((0.52*(0.48) /1100 )

= 0.030

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.52 -0.030 < p < 0.52+0.030

( 0.49 ,0.55 )