Assume that a sample is used to estimate a population proportion p. Find the 99.9% confidence interval for a sample of size 391 with 304 successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.
Solution :
Given that,
Point estimate = sample proportion =
= x / n = 304 / 391 = 0.777
1 -
= 1 - 0.777 = 0.223
Z/2
= 3.291
Margin of error = E = Z
/ 2 *
((
* (1 -
)) / n)
= 3.291 * (((0.777
* 0.223) / 391)
Margin of error = E = 0.069
A 99.9% confidence interval for population proportion p is ,
- E < p <
+ E
0.777 - 0.069 < p < 0.777 + 0.069
0.708 < p < 0.846
The 99.9% confidence interval for the population proportion p is : 0.708 , 0.846
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