Assume that a sample is used to estimate a population proportion p.

Question

Assume that a sample is used to estimate a population proportion p. Find the 90% confidence interval for a sample of size 234 with 11% successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.

C.I. = 

Answer 1

90% confidence interval for a sample proportion expressed as an open interval. Open interval means using parentheses to show that the endpoints are not included. (A closed interval would be expressed with brackets to include the endpoints)



x=234(0.11) = 26 (need to roundup)

n=234

phat = 0.11

Margin of error = z(α/2)*sqrt(phat*(1-phat)/n) = z(0.05)*sqrt(0.11*(1-0.11)/234) = 1.645(0.0205) = 0.0337

Lower bound = 0.11 - 0.0337 = 0.0763

Upper bound = 0.11 + 0.0337 = 0.1437

( 0.076, 0.144)

Let me know if you have any questions.

Dian