To estimate the average time it takes to assemble a certain computer component,

Question

To estimate the average time it takes to assemble a certain computer component, the. industrial engineer at an electronics firm timed 64 technicians in the performance of this task, getting a mean of 12.50 minutes and a variance of 4.00 minutes. (a) What can one assert with 98% confidence about the maximum error if 1250 is used as a point estimate for the true average time it takes to assemble the computer component? (b) Construct a 98% confidence interval for the true average time it takes to assemble the computer component. (c) How large a sample is needed so that the engineer will be able to assert with 98% confidence that the error is at most 0.3 minutes? (9 points)

Explanation / Answer

Part a

One can assert with 98% confidence that the maximum error would be only 2% if Xbar = 12.50 is used as a point estimate for the true average time it takes to assemble the computer component.

We are given

Xbar = 12.50

Variance = 4,

So SD = S = sqrt(4) = 2, and

Sample size = n = 64,

df = n – 1 = 63

Confidence level = 98%

Critical t value = 2.3870

Standard error = SD/sqrt(n) = 2/sqrt(64) = 2/8 = ¼ = 0.25

Margin of error = t*Standard error = 2.3870*0.25 = 0.5968

With 98% confidence, one can asset that maximum error would be 0.5968.

Part b

We are given

Xbar = 12.50

Variance = 4,

So SD = S = sqrt(4) = 2, and

Sample size = n = 64,

df = n – 1 = 63

Confidence level = 98%

Critical t value = 2.3870

Confidence interval = Xbar -/+ t*S/sqrt(n)

Confidence interval = 12.5 -/+ 2.3870*2/sqrt(64)

Confidence interval = 12.5 -/+ 0.5968

Lower limit = 12.5 - 0.5968 = 11.9032

Upper limit = 12.5 + 0.5968 = 13.0968

Confidence interval = (11.9032, 13.0968)

Part c

We are given

C = 98%

E = 0.3

= 2

Z = 2.3263 (by using z-table or excel)

Sample size = n = (Z*/E)2

n = (2.3263*2/0.3)^2 = 240.5187

n = 241

Required sample size = 241

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