Question

A survey (simple random sample) of 440 population members classify 170 in a category.

(a) Find a 95% confidence interval for the population proportion in the category using the precise method.
(b) Find the confidence level that the population proportion is less than 0.40 using the Agresti-Coull estimator.
(c) Find the sample size necessary to calculate a 95% confidence interval for population proportion of width 0.05

Answer 1

A)we want to find the confidence interval for the population proportion and the z

value for the 95% is 1.96

here random sample of pop=440

formula for CI for 95%=1.96*stnd dev/sqrtof sample size

so[ 1.96*(440/170)]/sqrt 440

=5.0729/20.97

=.2419

b)the population proportion is less than .40 and we are to find confidence level for it using the Agresti-Coull estimator

using binomial prpoption p=.40 n=440 the confidence level will be 90%

c)if confidence interval=95%

population proprtion of width=alpha =.05

then the sample size =n=(1.96*1.96)*.24*.24/0.05^2

=88 thanks