Question

In a survey of

24562456

adults in a recent​ year,

13921392

say they have made a New​ Year's resolution.

Construct​ 90% and​ 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals.

Answer 1

a)

sample proportion, = 0.5668
sample size, n = 2456
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.5668 * (1 - 0.5668)/2456) = 0.01

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, Zc = Z(α/2) = 1.64

Margin of Error, ME = zc * SE
ME = 1.64 * 0.01
ME = 0.0164

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.5668 - 1.64 * 0.01 , 0.5668 + 1.64 * 0.01)
CI = (0.5504 , 0.5832)
Therefore, based on the data provided, the 90% confidence interval for the population proportion is 0.5504 < p < 0.5832 , which indicates that we are 90% confident that the true population proportion p is contained by the interval (0.5504 , 0.5832)

b)

sample proportion, = 0.5668
sample size, n = 2456
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.5668 * (1 - 0.5668)/2456) = 0.01

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

Margin of Error, ME = zc * SE
ME = 1.96 * 0.01
ME = 0.0196

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.5668 - 1.96 * 0.01 , 0.5668 + 1.96 * 0.01)
CI = (0.5472 , 0.5864)
Therefore, based on the data provided, the 95% confidence interval for the population proportion is 0.5472 < p < 0.5864 , which indicates that we are 95% confident that the true population proportion p is contained by the interval (0.5472 , 0.5864)

width of 95% confidence interval is more
for 90% = 0.5832 - 0.5504 = 0.0328
for 95% = 0.5864 - 0.5472 = 0.0392