Assume that a sample is used to estimate a population proportion p. Find the 95% confidence interval for a sample of size 238 with 159 successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places. 95% C.I. =
Sample proportion
= 0.668
95% confidence interval for p is
- Z
/2
* sqrt [
( 1 -
) / n] < p <
+ Z
/2
* sqrt [
( 1 -
) / n]
0.668 - 1.96 * sqrt ( 0.668 * ( 1 - 0.668) / 238 ) < p < 0.668 + 1.96 * sqrt ( 0.668 * ( 1 - 0.668) / 238 )
0.608 < p < 0.728
95% CI is ( 0.608 , 0.728 )
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