Assume that a sample is used to estimate a population proportion
p. Find the 99.5% confidence interval for a sample of size
289 with 254 successes. Enter your answer as a tri-linear
inequality using decimals (not percents) accurate to three decimal
places.
< p <
Answer should be obtained without any preliminary rounding.
However, the critical value may be rounded to 3 decimal places.
Level of Significance, α = 0.005
Number of Items of Interest, x =
254
Sample Size, n = 289
Sample Proportion , p̂ = x/n =
0.879
z -value = Zα/2 = 2.807 [excel
formula =NORMSINV(α/2)]
Standard Error , SE = √[p̂(1-p̂)/n] =
0.0192
margin of error , E = Z*SE = 2.807
* 0.0192 = 0.0539
100% Confidence Interval is
Interval Lower Limit = p̂ - E = 0.879
- 0.0539 = 0.8250
Interval Upper Limit = p̂ + E = 0.879
+ 0.0539 = 0.9328
so, confidence interval is ( 0.825 < p <
0.933 )
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