STI survey funded by the U of Washington School of Medicine

Question

In a survey funded by the University of Washington School of Medicine, 600 of 1,500 adult Seattle residents said they did not believe they could come down with a sexually transmitted infection (STI). Construct a 90% confidence interval estimating the proportion of adult Seattle residents that do not believe they can contract an STI. Round the answers to three decimal places. Verify np^(1-p^) ≥ 10. Round your answer to one decimal place.  What is the 90% confidence interval to estimate the population proportion? Round your answer to three decimal places.

Answer 1

Explanation

 

N= sample size = 600

X = adult Seattle residents said they did not believe they could come down with a sexually transmitted infection (STI) = 1500

 

point estimate of population proportion = 

p​=nx

point estimate of population proportion = 0.4

n*p​(1−p​) = 1600*0.4*(1-0.4) = 384 ≥ 10

Standard error of the mean = SEM = √x(N-x)/N³ = 0.0126

α = (1-CL)/2 = 0.0500

Standard normal deviate for α = Z_α = 1.6449

Proportion of positive results = P = x/N = 0.4000

Lower bound = P - (Z_α*SEM) = 0.3792

Upper bound = P + (Z_α*SEM) = 0.4208

 

The 90% confidence interval to estimate the population proportion is 0.379 < p < 0.421.