Question

According to a Field Poll conducted, 79% of adults (actual results are 400 out of 506 surveyed) feel that "education and our schools" is one of the top issues facing the state. We wish to construct a 90% confidence interval for the true proportion of adults who feel that education and the schools is one of the top issues facing the state. Find a 90% confidence interval for the population proportion. Round your answers to three decimal places.

Answer 1

Solution :

Given that,

n = 506

x = 400

= x / n = 400 / 506 = 0.791

1 - = 1 - 0.791 = 0.209

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

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

= 1.645 * (((0.791 * 0.209) / 506)

= 0.030

A 90 % confidence interval for population proportion p is ,

- E < P < + E

0.791 - 0.030 < p < 0.791 + 0.030

0.761 < p < 0.821

The 90% confidence interval for the population proportion p is : ( 0.761 , 0.821)