Question 1200394: Assume that a sample is used to estimate a population proportion p. Find the 90% confidence interval for a sample of size 372 with 96 successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.
C.I. = ___________
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! sample size is 372 with 96 successes.
p = 96 / 372
q = 1 - 96 / 372 = 276 / 372
standard error = sqrt((96/372 * 276/372) / 372) .0226869309.
critical z at 90% two tailed confidence intervall = plus or minus z = 1.644853626.
when z = - 1.644853626, z = (x - p) / s becomes:
-1.644853626 = (x - 96/372) / .0226869309.
solve for x to get:
x = -1.64485 * .0226869309 + 96/372 = .2207478357.
when z = 1.644853626:
x = 1.64485 * .0226869309 + 96/372 .2953811966.
your interval at 90% confidence interval is .221 to .295.
in interval notation, that would be (.221,.295), i believe.
|
|
|
