(3) Assume that a sample is drawn and z(a/2) = 0.025 and s = 20. Answer the foll
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Question
(3) Assume that a sample is drawn and z(a/2) = 0.025 and s = 20. Answer the following questions:(A) If the Maximum Error of Estimate is 0.04 for this sample, what would be the sample size?
(B) Given that the sample Size is 400 with this same z(a/2) and s, what would be the Maximum Error of Estimate?
(C) What happens to the Maximum Error of Estimate as the sample size gets smaller?
(C) What effect does the answer to C above have to the size of the confidence interval
(4) 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 standard deviation is 3.96 seconds.
Answer each of the following (show all work):
(A) How many measurements should be made in order to be 99% certain that the maximum error of estimation will not exceed 0.5 seconds?
(B) What sample size is required for a maximum error of 1.0 seconds?
Explanation / Answer
(3) z(a/2) = 0.025 and s = 20.
(A) Maximum Error of Estimate, E= 0.04
we know, E = z *e
where e = standard error = s/n
n = sample size
hence, 0.04=0.025 * (20/n)
hence, n = 156.25 156
hence the sample size would be 156
(B) sample Size , n = 400 with this same z(a/2) and s,
the Maximum Error of Estimate, E = z * e = 0.025 *(20/400) = 0.025
(C) the Maximum Error of Estimate gets bigger as the sample size gets smaller.
(C) effect the answer to C above have to the size of the confidence interval that means as Maximum Error of Estimate gets bigger, the confidence interval gets smaller.
(4) 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 standard deviation , s = 3.96 seconds.
(A) How many measurements should be made in order to be 99% certain that the maximum error of estimation will not exceed 0.5 seconds?
at 99% confidence for a two tail distribution, z = 2.58 (from z table)
maximum error of estimation maximum error of estimation , E = z *e = z *(s/n)
hence, 0.5 = 2.58 *(3.96/n)
hence n = 417.5 418
hence no. of measurements that should be made in order to be 99% certain that the maximum error of estimation will not exceed 0.5 seconds is 418
(B) What sample size is required for a maximum error of 1.0 seconds?
we know E = z *e = z *(s/n)
hence, 1 = 2.58*(3.96/n)
hence, n = 104.38 104
hence sample size required for a maximum error of 1.0 seconds is 104
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