Question
We wish to estimate what percent of adult residents in a certain county are parents. Out of 100 adult residents sam...
We wish to estimate what percent of adult residents in a certain county are parents. Out of 100 adult residents sampled, 88 had kids. Based on this, construct a 99% confidence interval for the proportion p of adult residents who are parents in this county. Express your answer in tri-inequality form. Give your answers as decimals, to three places. 0.796 <p < 0.9631 Express the same answer using the point estimate and margin of error. Give your answers as decimals, to three places. p=O * O *
Answers
Solution: Given that n = 100, x = 88
p = x/n = 88/100 = 0.88
q = 1-p = 1 - 0.88 = 0.1299% Confidence interval for Z = 2.576
99% Confidence interval for the population proportion = p +/- Z*Sqrt(pq/n)
= 0.88 +/- 2.576*sqrt(0.88*0.12/100)
= (0.796 , 0.964)=> ME = Z*Sqrt(pq/n) = 0.0837 = 0.084 (rounded)
=> p = 0.88 +/- 0.084
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