Question 552142: I'm needing some help with these last problems on this chapter. Thanks.
1.) Assume that the mean score on a certain aptitude test across the nation is 100, and that the standard deviation is 20 points. Find the probability that the mean aptitude test score for a randomly selected group of 150 8th graders is between 94 and 106.
2.) A 98% confidence interval estimate for a population mean was computed to be (33.7, 54.5). Determine the mean of the sample, which was used to determine the interval estimate (show all work).
3.) A confidence interval estimate for the population mean is given to be (39, 90, 48.11). If the standard deviation is 11.473 and the sample size is 52, answer the following (show all work). (a) determine the maximum error of the estimate, E. (b) determine the confidence level used for the given confidence interval.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 1.) Assume that the mean score on a certain aptitude test across the nation is 100, and that the standard deviation is 20 points. Find the probability that the mean aptitude test score for a randomly selected group of 150 8th graders is between 94 and 106.
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z(94) = (94-100)/20 = -6/20 = -0.3
z(106) = (106-100)/20 = 0.3
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P(94 < x < 106) = P(-0.3 < z < 0.3) = 0.2358
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2.) A 98% confidence interval estimate for a population mean was computed to be (33.7, 54.5). Determine the mean of the sample, which was used to determine the interval estimate (show all work).
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Solve:
xbar-ME = 33.7
xbar+ME = 54.5
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Add and solve for xbar:
2xbar = 88.2
x-bar = 44.1
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3.) A confidence interval estimate for the population mean is given
to be (39.90 , 48.11).
If the standard deviation is 11.473 and the sample size is 52, answer the following (show all work).
(a) determine the maximum error of the estimate, E.
Solve for "E"
xbar-E = 39.9
xbar+E = 48.11
Subtract and solve for "E":
2E = 8.21
E = 4.105
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(b) determine the confidence level used for the given confidence interval.
E = z*s/sqrt(n)
4.105 = z*11.473/sqrt(52)
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z = 4.105*sqrt(52)/11.473
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z = 2.5801
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P(z > 2.5801) = 0.0049
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Rounding you get Prob = 0.05
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Confidence level = 1-2*0.05 = 99%
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Cheers,
Stan H.
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