**Z-Value for given confidence level is 2.58. Each year, Lord & Taylor, Ltd., se
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Question
**Z-Value for given confidence level is 2.58.
Each year, Lord & Taylor, Ltd., sets up a gift-wrapping station to assist its customers with holiday shopping. Preliminary observations of one worker at the station produced the following sample time (in minutes per package): 34, 3.1, 4.2, 3.6, 3.9. Assume a 99% confidence level and an accuracy of 5%. Round all intermediate calculations to two decimal places before proceeding with further calculations.) Based on this small sample and the given confidence level and accuracy level, the number of observations that would be necessary to determine the true cycle time observations (round your response up to the next whole number).Explanation / Answer
First calculate the mean and standard deviation
Observations
Time (Minutes)
1
3.4
2
3.1
3
4.2
4
3.6
5
3.9
Mean (x-bar)
3.64
Standard Deviation (s)
0.428
Number of observations that are required for a 99% confidence level within 5% accuracy
Number of observations needed, n = (z *s / h*x-bar) ^2
Where,
z = number of std. dev. needed for desired confidence level of 99%= 2.58
s = sample Standard Deviation = 0.428
h = desired accuracy = 5% or 0.05
x –bar = sample mean =3.64 minutes
Therefore
n = (2.58*0.428 / 0.05* 3.64) ^2
= (6.064) ^2
= 36.77 or 37 observations
Observations
Time (Minutes)
1
3.4
2
3.1
3
4.2
4
3.6
5
3.9
Mean (x-bar)
3.64
Standard Deviation (s)
0.428
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