Question

In a survey of 3130 Adult 1402 say they have to start paying bills online in the last year construct a 99% confidence in server for the population portion. Interpretare the resul

Answer 1

Solution :

Given that,

n = 3130

x = 1402

Point estimate = sample proportion = = x / n = 1402 / 3130 = 0.452

1 - = 1 - 0.452 = 0.548

At 99% confidence level

= 1 - 99%  

= 1 - 0.99 =0.01

/2 = 0.005

Z/2 = Z0.005 = 2.576

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

= 2.576 (((0.452 * 0.548) / 3130)

= 0.023

A 95% confidence interval for population proportion p is ,

± E  

0.452 ± 0.023

(0.429 , 0.475)