Given a level of confidence of 95% and a population standard deviation of 7, wha

Question

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 = 45. Show all your calculations. Show all work.

(D) For this same sample of n = 45, 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 = 3.0. (Always round to the nearest whole person.) Show all work. 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 = 45. Show all your calculations. Show all work.

(D) For this same sample of n = 45, 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 = 3.0. (Always round to the nearest whole person.) Show all work.

Explanation / Answer

We have calculated MoE above as ±2.04 So width of confidence interval is 2.04+2.04 = 4.08 around mean (E) Given this same confidence level and standard deviation, find n if E = 3.0. (Always round to the nearest whole person.) Show all work. Given Conf level = 95%. So z = 1.96 MoE = s*z = 3 So sample std dev s = MoE/z = 3/1.96 = 1.53
Now Sample Sd s = Population SD/vn SO n = s^2/s^2 = 7^2/1.53^2 = 49/2.34 = 20.92 ˜ 21 So n = 21 when E=3

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