Question

1. Assume that a sample is used to estimate a population proportion p. Find the 80% confidence interval for a sample of size 399 with 303 successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.
80% C.I. =

2. Assume that a sample is used to estimate a population proportion p. Find the 80% confidence interval for a sample of size 315 with 18.1% successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.
80% C.I. =

3. Assume that a sample is used to estimate a population proportion p. Find the 99.9% confidence interval for a sample of size 233 with 76.8% successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.
99.9% C.I. =

Answer 1

Solution:-

1. Given that n = 399,x = 303 , 80% confidence interval for the Z = 1.28

p = x/n = 303/399 = 0.759

q = 1-p = 1-0.759 = 0.241

80% confidence interval for the population proportion = p +/- Z*sqrt(pq/n)
= 0.759 +/- 1.28*sqrt(0.759*0.241/399)
= (0.732 , 0.786)

2. Given that n = 315, p = 0.181

q = 0.819

80% confidence interval for the population proportion = p +/- Z*sqrt(pq/n)
= 0.181 +/- 1.28*sqrt(0.181*0.819/315)
= 0.153,0.209

3. given that n = 233, p = 0.768, q = 0.232

99.9% confidence interval for the population proportion =
= 0.768 +/- 3.291*sqrt(0.768*0.232/233)
= 0.677 , 0.859