Assume that a sample is used to estimate a population proportion p. Find the 98% confidence interval for a sample of size 199 with 52 successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.
C.I. =
Solution :
Given that,
Point estimate = sample proportion =
= x / n = 52 / 199 = 0.261
1 -
= 1 - 0.261 = 0.739
Z/2
= 2.326
Margin of error = E = Z
/ 2 *
((
* (1 -
)) / n)
= 2.326 * (((0.261
* 0.739) /199 )
= 0.072
A 98% confidence interval for population proportion p is ,
- E < p <
+ E
0.261 - 0.072 < p < 0.261 + 0.072
0.189 < p < 0.333
C.I = 0.189 , 0.333
Get Answers For Free
Most questions answered within 1 hours.