Question

Assume that a sample is used to estimate a population proportion p. Find the 80% confidence interval for a sample of size 324 with 67% successes.

A. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.

Confidence interval =

B. Express the same answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

< p <

C. Express the same answer using the point estimate and margin of error. Give your answers as decimals, to three places.

p =    ±

Answer 1

A)

80% confidence interval for p is

- Z/2 * sqrt [ ( 1 - ) / n ] < p < + Z/2 * sqrt [ ( 1 - ) / n ]

0.67 - 1.2816 * sqrt [ 0.67 ( 1 - 0.67) /324) < p < 0.67 + 1.2816 * sqrt [ 0.67 ( 1 - 0.67) /324)

0.637 < p < 0.703

95% CI is ( 0.637 , 0.703 )

b)

80% confidence interval for p is

- Z/2 * sqrt [ ( 1 - ) / n ] < p < + Z/2 * sqrt [ ( 1 - ) / n ]

0.67 - 1.2816 * sqrt [ 0.67 ( 1 - 0.67) /324) < p < 0.67 + 1.2816 * sqrt [ 0.67 ( 1 - 0.67) /324)

0.637 < p < 0.703

c)

E = Z/2 * sqrt [ ( 1 - ) / n ]

= 0.67 1.2816 * sqrt [ 0.67 ( 1 - 0.67) /324)

= 0.67 0.033