In a Midwestern state the proportion of the population 21 years of age or older
ID: 2506834 • Letter: I
Question
In a Midwestern state the proportion of the population 21 years of age or older who have a high school diploma is ? = 0.9. 6 For samples of size n = 800 obtained from this population, the standard error of the sample proportion p? is, se(p?) = ______. a 0.0245 b 0.0205 c 0.0122 d 0.0106 7 The probability that in a sample of n = 800 the sample proportion of individuals with a higher education degree is between 0.88 and 0.92 is, P(0.88 < p? < 0.92) = _____. a 0.9817 b 0.9625 c 0.9412 d 0.9109 8 The proportion of p? values obtained from samples of size n = 800 that fall within In a Midwestern state the proportion of the population 21 years of age or older who have a high school diploma is ? = 0.9. 6 For samples of size n = 800 obtained from this population, the standard error of the sample proportion p? is, se(p?) = ______. a 0.0245 b 0.0205 c 0.0122 d 0.0106 7 The probability that in a sample of n = 800 the sample proportion of individuals with a higher education degree is between 0.88 and 0.92 is, P(0.88 < p? < 0.92) = _____. a 0.9817 b 0.9625 c 0.9412 d 0.9109 8 The proportion of p? values obtained from samples of size n = 800 that fall withinExplanation / Answer
(6) The standard error = sqrt(p*(1-p)/n)
=sqrt(0.9*0.1/800)
= 0.0106066
Answer: D. 0.0106
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(7)
P(0.88<phat<0.92) =P((0.88-0.9)/ 0.0106066 <(phat-p)/sqrt(p*(1-p)/n) <(0.92-0.9)/ 0.0106066)
=P(-1.89 <Z<1.89)
=0.9412 (from standard normal table)
Answer: c. 0.9412
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(8) P(-0.025 <phat-p<0.025)
=P(-0.025/0.0106066 <(phat-p)/sqrt(p*(1-p)/n) < 0.025/0.0106066)
=P(-2.36 <Z< 2.36)
=0.9818 (from standard normal table)
Answer: a. 0.9818
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(9) Given a=1-0.95=0.05, Z(0.025)=1.96 (from standard normal table)
So margin of error = Z*sqrt(p*(1-p)/n)
= 1.96*sqrt(0.9*0.1/800)
= 0.02078894
Answer: c. 0.021
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n=(Z/E)^2*p*(1-p)
=(1.96/0.03)^2*0.9*0.1
=384.16
Take n=385
Answer: d. 385
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