Question 232720: Given a level of confidence of 95% and a population standard deviation of 7, what other information is necessary:
(A) To find the Maximum Error of Estimate (E)?
(B) To find the sample size (n)?
(C) Given the above confidence level and population standard deviation, find the Maximum Error of Estimate (E) if n = 30. Show all your calculations. Show all work.
(D) For this same sample of n = 30, what is the width of the confidence interval around the population mean? Show all work.
(E) Given this same confidence level and standard deviation, find n if E = 2.8. (Always round to the nearest whole person.) Show all work.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Given a level of confidence of 95% and a population standard deviation of 7, what other information is necessary:
(A) To find the Maximum Error of Estimate (E)?
E depends on:
The z or t value associated with the degree of confidence
The standard deviation of the population or the sample
The sample size
E = z*s/sqrt(n)
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(B) To find the sample size (n)?
E, s, and z
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(C) Given the above confidence level and population standard deviation, find the Maximum Error of Estimate (E) if n = 30. Show all your calculations. Show all work.
Use the formula E = z*s/sqrt(30)
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(D) For this same sample of n = 30, what is the width of the confidence interval around the population mean? Show all work.
Use: x-bar-E < u < x-bar+E
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(E) Given this same confidence level and standard deviation, find n if E = 2.8. (Always round to the nearest whole person.) Show all work.
Use n = [z*s/E]^2
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Cheers,
Stan H.
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