Date:, Quiz # 10 Section Use the one-proportion z-interval procedure to find the
ID: 2907404 • Letter: D
Question
Date:, Quiz # 10 Section Use the one-proportion z-interval procedure to find the required confidence interval. 1) 1) A researcher wishes to the proportion of adults in the city of Darby who are vegetarian. confidence interval tor the proportion of all adults in the city of Darby that are vegetarians ) 0.0521 to 0.1039 random sample of 713pdults from this city, the propottion that are vegetarian is.0.078 Find a B) 0.0680 to 0.0880 D) 00619 to 0.0941 C) 0.0298 to 0.1262 estimated from the sample described. Approximate the 95% confidence interval. Round A population proportion is to be your answer to four decimal places, if necessary 2) A medical researcher wishes to estimate what proportion of babies born at a particular hospital2) are born by Caesarean section. In a random sample of 49 births at the hospital, 32% were Caesarean sections. Find the 95% confidence interval for the population proportion. A) 0.3010 PExplanation / Answer
Solution:-
1) Given that p = 0.078 , n = 713
=> 99% confidence interval for the proportion of all adults in the city
= p +/- Z*sqrt(pq/n)
= 0.078 +/- 2.576*sqrt(0.078*0.922/713)
= 0.0521 , 0.1039
=> option A. 0.0521 to 0.1039
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2) 95% confidence interval for the population propertion :
= 0.32 +/- 1.96*sqrt(0.32*0.68/49)
= 0.1894 , 0.4506
=> option D)
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3) p = X/n = 2250/4000 = 0.5625
99% confidence interval = 0.5625 +/- 2.576*sqrt(0.5625*0.4375/4000)
= 0.5625 +/- 0.0202
=> option D) 0.5625 +/- 0.0202
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4) Given that n = 10, df = n-1 = 9 , x = 11.08 , s = 2.3734
t = 2.262
95% confidence interval for the mean time for all players = X +/- t*s/sqrt(n)
= 11.08 +/-2.262*2.3734/sqrt(10)
= 9.3823 , 12.7777
=> option D. 9.40 to 12.80 minutes
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5) mean = upper+lower/2 = (25+7)/2 = 16
E = (u - L)/2 = (25-7)/2 = 9
=> option B)
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