Assume that a sample is used to estimate a population proportion
p. Find the 99% confidence interval for a sample of size
308 with 123 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.
Confidence interval for Population Proportion is given as below:
Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)
Where, P is the sample proportion, Z is critical value, and n is sample size.
We are given
x = 123
n = 308
P = x/n = 123/308 = 0.399350649
Confidence level = 99%
Critical Z value = 2.5758
(by using z-table)
Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)
Confidence Interval = 0.399350649 ± 2.5758* sqrt(0.399350649*(1 – 0.399350649)/308)
Confidence Interval = 0.399350649 ± 2.5758* 0.0279
Confidence Interval = 0.399350649 ± 0.0719
Lower limit = 0.399350649 - 0.0719 = 0.3275
Upper limit = 0.399350649 + 0.0719 = 0.4712
Confidence interval = (0.3275, 0.4712)
0.3275 < p < 0.4712
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