Question

In a survey of 2241 U.S Adults in a recent year, 1322 say they have made a new years resolution.

This is a proportion confidence interval problem. Construct 90% and 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence interval.

Answer 1

90%

Standard error of the mean = SEM = √x(N-x)/N³ = 0.010

α = (1-CL)/2 = 0.050

Standard normal deviate for α = Z_α = 1.645

Proportion of positive results = P = x/N = 0.590

Lower bound = P - (Z_α*SEM) = 0.573

Upper bound = P + (Z_α*SEM) = 0.607

95%

Standard error of the mean = SEM = √x(N-x)/N³ = 0.010

α = (1-CL)/2 = 0.025

Standard normal deviate for α = Z_α = 1.960

Proportion of positive results = P = x/N = 0.590

Lower bound = P - (Z_α*SEM) = 0.570

Upper bound = P + (Z_α*SEM) = 0.610