Chapter 07 – Work Design and Measurement
We must estimate the sample standard deviation for Element A:
Data
Step 1
Compute
Mean
Step 2
Compute
Differences
Step 3
Square Differences &
Sum the Squares
1.40
1.40 – 1.40 = 0.00
0.002 = 0.000
1.42
1.42 – 1.40 = 0.02
0.022 = 0.0004
1.39
1.39 – 1.40 = –0.01
-0.012 = 0.0001
1.38
1.38 – 1.40 = –0.02
-0.022 = 0.0004
1.41
1.41 – 1.40 = 0.01
0.012 = 0.0001
Mean = 1.40
Sum = 0.001
Step 4: Divide the sum of the squared differences by n – 1 (n = 5):
0.001 / (5–1) = 0.00025
Step 5: Find the square root of the value from Step 4:
(round to 4 decimals)
Number of Observations Needed:
confidence.
s = Sample standard deviation (0.0158 from above)
a = Desired accuracy % (4% = .04 from above)
Sample mean (1.40 from above)
Conclusion: 1 observation is needed.
c. Given: Desired accuracy for mean time of Element C = within 0.10 minutes of the true value
with a confidence of 90%.
[use this formula when desired accuracy is a time]
where:
z = Number of normal standard deviations for desired confidence
6
1.20
1.17
1.16
1.22
1.24
1.15
1.19
2
0.83
0.87
0.78
0.82
0.85
—
0.83
Number of Observations Needed:
Using z = 1.64:
(round up to 2)
Using z = 1.65:
(round up to 2)
where:
z = Number of normal standard deviations for desired confidence
Using App. B Table A, we look for .90/2 = .45; .45 is midway between .4495 (z = 1.64) &
.4505 (z = 1.65).
s = Sample standard deviation (0.0791 from above)
e = Maximum acceptable amount of time error (0.10 from above)
Conclusion: 2 observations are needed.
10. Given: The data in the table below represent time study observations for an operation.
a. Given: Allowance = 15% of job time.
OBSERVATIONS (minutes per cycle)
Element
PR
1
2
3
4
5
6
1
1.10
1.20
1.17
1.16
1.22
1.24
1.15
2
1.15
0.83
0.87
0.78
0.82
0.85
—
3
1.05
0.58
0.53
0.52
0.59
0.60
0.54
Standard Time for Operation = 1.51 + 1.09 + 0.68 = 3.28 minutes
b. Given: Desired accuracy for mean time of Element 2 = within 1% of the true value with a
confidence of 95.5%.
Element
NT
AFjob
ST
1
1.31
1.15
1.51
2
0.95
1.15
1.09
3
0.59
1.15
0.68
7-15
Education.
0.0046 / (5–1) = 0.00115
Step 5: Find the square root of the value from Step 4:
(round to 4 decimals)
Number of Observations Needed:
2.01 is used.
Chapter 07 – Work Design and Measurement
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c. Given: Desired accuracy for mean time of Element 2 = within 0.01 minutes of the true value
with a confidence of 95.5%.
[use this formula when desired accuracy is a time]
where:
z = Number of normal standard deviations for desired confidence (95.5%)
s = Sample standard deviation
e = Maximum acceptable amount of time error
We must estimate the sample standard deviation for Element 2:
From part b: s = 0.0339
Using z = 2.01:
(round up to 47)
where:
Conclusion: 46 observations are needed if z = 2.00 is used; 47 observations are needed if z =
2.01 is used.
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11. Given: An operation has a standard deviation of 1.5 minutes per piece. The goal is to estimate the
mean time per piece to within 0.4 minutes of its true value with a confidence of 95.5%.
[use this formula when desired accuracy is a time]
where:
z = Number of normal standard deviations for desired confidence
s = Sample standard deviation
e = Maximum acceptable amount of time error
Number of Observations Needed:
e = Maximum acceptable amount of time error (0.4 from above)
Chapter 07 – Work Design and Measurement
7-18
Education.
12. Given: An operation yielded times of 5.2, 5.5, 5.8, 5.3, 5.5, and 5.1. The goal is to estimate the
mean time to within 2% of its true value with a confidence of 99%. The standard deviation is
given already as 0.253 minutes per cycle.
Mean Time = (5.2 + 5.5 + 5.8 + 5.3 + 5.5 + 5.1) / 6 = 5.4 minutes per cycle
Number of Observations Needed:
Using z = 2.57:
where:
z = Number of normal standard deviations for desired confidence (99%)
Using App. B Table A, we look for .99/2 = .495; .495 is midway between .4949 (z = 2.57) &
.4951 (z = 2.58).
s = Sample standard deviation (0.253 from above)
a = Desired accuracy % (2% = .02 from above)
Sample mean (5.4 from above)