Question

A Retirement Confidence Survey of 1153 workers and retirees in the United States 25 years of age and older conducted by the Employee Benefit Research Institute in January 2012 found that 537 had less than $10,000 in savings. Use this sample to construct a 99% confidence interval for the proportion of all workers and retirees in the U.S. with less than $10,000 in savings. The sample proportion is p^ = .

(Round to 3 decimal places, if applicable. Use this rounded value in all subsequent calculations.)

Conditions: np^(1-p^) = (Round to 3 decimal places, if applicable.) so conditions for constructing the confidence interval (are / are not) satisfied for the given data. The appropriate critical value for the desired interval is zα/2 = . Based on the given data, we are 99% confident that the proportion of all workers and retirees in the U.S. with less than $10,000 in savings is between (lower bound) and (upper bound). (Round your answers to 3 decimal places, if applicable.

Answer 1

sample proportion, 537/1153 = 0.466

n*pcap*(1-pcap) = 1153 * 0.466 * (1-0.466) = 286.917
o conditions for constructing the confidence interval are satisfied for the given data.

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, Zc = Z(α/2) = 2.58

sample proportion, = 0.466
sample size, n = 1153
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.466 * (1 - 0.466)/1153) = 0.0147

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.466 - 2.58 * 0.0147 , 0.466 + 2.58 * 0.0147)
CI = (0.428 , 0.504)