Question

According to a poll, 78% of California adults (397 out of 506 surveyed) feel that education is one of the top issues facing California. We wish to construct a 90% confidence interval for the true proportion of California adults who feel that education is one of the top issues facing California.

What is a point estimate for the true population proportion?

Answer 1

Solution :

Given that,

n = 506

x = 397

Point estimate = sample proportion = = x / n = 78%=0.78

1 -   = 1-0.78 =0.22

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

Z/2 = Z0.01 = 2.326 ( Using z table    )

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

= 2.326 (((0.78*0.22) /506 )

E = 0.043

A 98% confidence interval is

- E < p < + E

0.78-0.043 < p < 0.78+0.043

0.737<p<0.823