Assume that a sample is used to estimate a population proportion
p. Find the 99.5% confidence interval for a sample of size
248 with 196 successes. Enter your answer as an
open-interval (i.e., parentheses) using
decimals (not percents) accurate to three decimal places.
99.5% C.I. =
Answer should be obtained without any preliminary rounding.
However, the critical value may be rounded to 3 decimal places.
Solution :
Given that,
Point estimate = sample proportion = = x / n = 196 / 248 = 0.790
1 - = 1 - 0.790 = 0.21
Z/2 = 2.807
Margin of error = E = Z / 2 * (( * (1 - )) / n)
= 2.807 * (((0.790 * 0.21) / 248)
Margin of error = E = 0.073
A 99.5% confidence interval for population proportion p is ,
- E < p < + E
0.790 - 0.073 < p < 0.790 + 0.073
0.717 < p < 0.863
99.5% C.I. = (0.717 , 0.863)
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