Due in 1 hours, 9 minutes. Due Sun 03/11/2018 11:59 Show Intro/Instructi We wish

Question

Due in 1 hours, 9 minutes. Due Sun 03/11/2018 11:59 Show Intro/Instructi We wish to estimate what percent of adult residents in a certain county are parents. Out of 600 adult residents sampled 360 had kids. Based on this, construct a 99% confidence interval for the proportion, p ofadalt residents who are parents in this county Give your answers as decimals, to three places. Get help: Video Box 1. Enter your answer as an integer or decimal numbet Examples: 3, 4,5 5172 Enter DNE for Does Not Exist, oo for Infinity er your answer as an integer or decimal number Examples: 3,-4,5.5172 Enter DNE for Does Not Exist, oo for Infinity Points possible: 2 This is attempt 1 of 3. Post this question to forum Submit 8 5 6

Explanation / Answer

p = 360/600 = 0.6

Z for 99% confidence interval = Z0.005 = 2.58

Confidence interval = (p + Z0.005 * sqrt(p*(1-p)/n))

= (0.6 + 2.58 * sqrt (0.6 * 0.4 / 600))

= (0.6 + 0.0516)

= (0.5484, 0.6516)

0.5484 < p < 0.6516

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