Question 663663: These problems I have been working on all weekend, and I do not understand. My grade and passing the course depends on this. HOWEVER, I have not received a single reply. WHY? Very unusual, from the tutors here! I need your help, if I could pay I would! It is that important.
Problem 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 100. (Points:3)
Solution what I am doing right or wrong: z=(100)=(100-100)/20=-0/20=
Problem2:.Assume that a sample is drawn and z(a/2) =1.96 and σ=35. Answer the following questions:
If the Maximum Error of Estimate is 0.02 for this sample, what would be the sample size? ( I got this far: N=zs/E)^2N= (1.96 *35/)(.02^2=) don’t know if I am doing it right)
Given that the sample SIZE is 400 with this same z(a/2) and σ, what would be the Maximum Error of Estimate? (My workings are E= zs/sqrt(n) E= 1.96*35/sqrt400=.606)
What happens to the Maximum Error of Estimate as the sample size gets smaller? (I have no idea)
What effect does the answer to C above have to the size of the confidence interval? (My work; As sample gets larger the ME gets smaller so the CI gets smaller.) (Points:8)
Problem:3 By measuring the amount of time it takes a component of a product to move from one workstation to the next, an engineer has estimated that the SD is 4.34 seconds. Answer all of the question by showing your work:
How many measurements should be made in order to be 90% CERTAIN that the ME of estimation will not exceed 2.0 seconds?
What sample size is required for a ME of 0.5 seconds? (Points:6)
(My work: a) 1-a=0.90 =2.0 ) TOTALLY LOST
Problem 3: A 99% confidence interval estimate for a population mean was computed to be (47.1, 65.9). Determine the mean of the sample, which was used to determine the interval
Estimate. (Must show all work) Points: 4) ( z=0.4761 0.6591) from table 3- I really don’t know) just guessing …
Problem:4: A study was conducted to estimate the mean amount spent on birthday gifts for a typical family having 2 children. A sample of 185 was taken, and the mean amount spent was $244.16. Assuming a SD equal to $38.98, find the 98% confidence interval for μ, the mean for all such families. (Show ALL work) (Points 4). (Don’t know where to start as I am very confused)
Please answer me this time around....
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Problem 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 100. (Points:3)
Solution what I am doing right or wrong: z=(100)=(100-100)/20=-0/20=0
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Note: In a continuous distribution the probability of gettin ANY exact
score is zero.
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Problem2:.Assume that a sample is drawn and z(a/2) =1.96 and σ=35. Answer the following questions:
If the Maximum Error of Estimate is 0.02 for this sample, what would be the sample size? ( I got this far:
N = (zs/E)^2
N = (1.96 *35/0.02)^2
= 11,764,900
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Given that the sample SIZE is 400 with this same z(a/2) and σ, what would be the Maximum Error of Estimate?
E= zs/sqrt(n)
E= 1.96*35/sqrt(400) = 3.43
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What happens to the Maximum Error of Estimate as the sample size gets smaller?
E = z*s/sqrt(n)
Notice that E and sqrt(n) are indirectly related.
So, as n gets smaller, E must get larger.
Try it with numbers and you will see what I mean.
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What effect does the answer to C above have to the size of the confidence interval? (My work; As sample gets larger the ME gets smaller so the CI gets smaller.) (Points:8)
Correct
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Problem:3 By measuring the amount of time it takes a component of a product to move from one workstation to the next, an engineer has estimated that the SD is 4.34 seconds. Answer all of the question by showing your work:
How many measurements should be made in order to be 90% CERTAIN that the ME of estimation will not exceed 2.0 seconds?
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n = [z*s/E]^2 = [1.645*4.34/2]^2 = 13 when rounded up
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What sample size is required for a ME of 0.5 seconds? (Points:6)
Same formula with 0.5 replacing 2.
Ans: 204 when rounded up
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Problem 3: A 99% confidence interval estimate for a population mean was computed to be (47.1, 65.9). Determine the mean of the sample, which was used to determine the interval Estimate.
Note: The width of the CI is ALWAYS 2*ME
2*ME = 65.9-47.1 = 18.8
ME = 9.4
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Note The sum of the CI limits is ALWAYS 2*(sample mean)
2(sample mean) = 65.9+47.1 = 113
sample mean = 56.5
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Problem:4: A study was conducted to estimate the mean amount spent on birthday gifts for a typical family having 2 children. A sample of 185 was taken, and the mean amount spent was $244.16. Assuming a SD equal to $38.98, find the 98% confidence interval for μ, the mean for all such families. (Show ALL work) (Points 4). (Don’t know where to start as I am very confused)
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x-bar = 244.16
ME = 2.3263*38.98/sqrt(185) = 6.67
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98% CI: 244.16-6.67 < u < 244.16+6.67
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Cheers,
Stan H.
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