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Construct a 99% confidence interval of the population proportion at the given level of confidence. x=240​, n=300 the lower bound is? the upper bound is?

1 Answer

1 vote

Answer: lower bound = 0.7404; upper bound = 0.8596

Explanation:

The proportion p for this population:

p =
(240)/(300)

p = 0.8

Confidence interval for proportion is calculated as:

p ± z-score.
\sqrt{(p(1-p))/(n) }

Z-score for a 99% confidence interval is: z = 2.58

Calculating:

0.8 ± 2.58.
\sqrt{(0.8(0.2))/(300) }

0.8 ± 2.58.
√(0.00053)

0.8 ± 2.58(0.0231)

0.8 ± 0.0596

This means that the lower limit of this interval is 0.7404 and upper bound is 0.8596

User Mdhale
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