Question

A simple random sample of 400 persons is taken to estimate the percentage of Republicans in a large population. It turns out that 210 of the people in the sample are Republicans. True or False and explain.

a. The sample percentage is 52.5%; the SE for the sample percentage is 2.5%.

b. 52.5% ± 2.5% is a 75%-confidence interval for the population percentage.

c. 52.5% ± 5% is a 95%-confidence interval for the sample percentage.

d. 52.5% ± 5% is a 95%-confidence interval for the population percentage.

e. There is about a 95% probability for the percentage of Republicans in the population to be in the range 52.5% ± 5%

Answer 1

75%

Standard error of the mean = SEM = √x(N-x)/N³ = 0.025

α = (1-CL)/2 = 0.125

Standard normal deviate for α = Z_α = 1.150

Proportion of positive results = P = x/N = 0.525

Lower bound = P - (Z_α*SEM) = 0.496

Upper bound = P + (Z_α*SEM) = 0.554

95%

Standard error of the mean = SEM = √x(N-x)/N³ = 0.025

α = (1-CL)/2 = 0.025

Standard normal deviate for α = Z_α = 1.960

Proportion of positive results = P = x/N = 0.525

Lower bound = P - (Z_α*SEM) = 0.476

Upper bound = P + (Z_α*SEM) = 0.574

a. The sample percentage is 52.5%; the SE for the sample percentage is 2.5%. = true

b. 52.5% ± 2.5% is a 75%-confidence interval for the population percentage. = true

c. 52.5% ± 5% is a 95%-confidence interval for the sample percentage. = False

d. 52.5% ± 5% is a 95%-confidence interval for the population percentage. = true

e. There is about a 95% probability for the percentage of Republicans in the population to be in the range 52.5% ± 5% = true