Question

In a survey of 1007 adults, a poll? asked, "Are you worried or not worried about having enough money for? retirement?" Of the 1007 surveyed, 552 stated that they were worried about having enough money for retirement. Construct a 99?% confidence interval for the proportion of adults who are worried about having enough money for retirement.

A 99?% confidence interval for the proportion of adults who are worried about having enough money for retirement is ?(___,___?).

Answer 1

Solution :

Given that,

n = 1007

x = 552

= x / n = 552 / 1007 = 0.548

1 - = 1 - 0.548 = 0.452

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

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

= 2.576 * (((0.548 * 0.452) / 1007)

= 0.04

A 99% confidence interval for population proportion p is ,

- E < P < + E

0.548 - 0.04 < p < 0.548 + 0.04

0.508 < p < 0.588

(0.508 , 0.588)