A Methods and Measurements Analyst for Digital Devices needs to develop a time standard for the task of assembling a computer mouse. In a preliminary study, she observed one of her employees perform this task six times with the following results:
Observation 1 2 3 4 5 6
Time (S) 46 38 40 34 42 40
a) What is the observed time (OT) for this task?
b) What is the normal time (NT) for this task if the employee worked at a 20% faster pace than is average?
c) What is the standard time (ST) for this task if the employee worked at a 20% faster pace than is average and an allowance of 25% of the workday is used? (1)
d) What is the standard time (ST) for this task if the employee worked at a 20% faster pace than is average and an allowance of 25% of job time is used?
e) How many observations should be made if she wants to be 0.8664 confident that the maximum error in the observed time is 0.5 second? Assume that the standard deviation of the task time is 4 seconds. (Value of Z =1.5 for 0.8664 confidence)

Question

13-12-2022
Computers and Technology

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A Methods and Measurements Analyst for Digital Devices needs to develop a time standard for the task of assembling a computer mouse. In a preliminary study, she observed one of her employees perform this task six times with the following results:
Observation 1 2 3 4 5 6
Time (S) 46 38 40 34 42 40
a) What is the observed time (OT) for this task?
b) What is the normal time (NT) for this task if the employee worked at a 20% faster pace than is average?
c) What is the standard time (ST) for this task if the employee worked at a 20% faster pace than is average and an allowance of 25% of the workday is used? (1)
d) What is the standard time (ST) for this task if the employee worked at a 20% faster pace than is average and an allowance of 25% of job time is used?
e) How many observations should be made if she wants to be 0.8664 confident that the maximum error in the observed time is 0.5 second? Assume that the standard deviation of the task time is 4 seconds. (Value of Z =1.5 for 0.8664 confidence)

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tutorconsortium403 tutorconsortium403 · Answer 1 · 2022-12-18T15:36:45+01:00

a. The observed time (OT) for this task is 40 seconds.

b. The normal time (NT) for this task is 48 seconds.

c. The standard time (ST) for this task is 64 seconds.

d. The standard time (ST) for this task is 60 seconds.

e. 144 observations should be made if she wants to be 0.8664 confident.

a).

Observed time for the task = Sum of the observed times / number of times task observed

= (46 + 38 + 40 + 34 + 42 + 40) / 6

= 40 seconds

b).

Observed time = 40 seconds

Expected increase = 20% faster than average observed time

Now as the employee worked at 20% faster pace than average, the normal time would be = Observed time * (100% + 20%)

= 40 * 120%

= 48 seconds

c).

Standard time = net time * allowance factor

Observed time = 40

Standard time = normal time * allowance factor

= 48 * 1 / (1 - 25%)

= 64 seconds

d).

PF = 1.2 (20% faster pace)

AFday = 1/1-.2 = 1.25 (25% allowance)

Standard time = normal time * allowance factor

= 48 * 1.25

= 60 seconds

e).

N = ((Z*STANDARD DEVIATION) / ESTIMATE MEAN)2

St. dev = 4

Confidence interval = 86.64, which gives a Z value pf 1.5

Estimated mean = 0.5

N = ((1.5 * 4) / (0.5)2

= 144

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