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Estimating production time.A widely used technique for estimating the length of time it takes workers to produce a product is the time study. In a time study, the task to be studied is divided into measurable parts, and each is timed with a stopwatch or filmed for later analysis. For each worker, this process is repeated many times for each subtask. Then the average and standard deviation of the time required to complete each subtask are computed for each worker. A worker鈥檚 overall time to complete the task under study is then determined by adding his or her subtask-time averages (Gaither and Frazier, Operations Management, 2001). The data (in minutes) given in the table are the result of a time study of a production operation involving two subtasks.


Worker AWorker B

Repetition

Subtask 1

Subtask 2

Subtask 1

Subtask 2

1

30

2

31

7

2

28

4

30

2

3

31

3

32

6

4

38

3

30

5

5

25

2

29

4

6

29

4

30

1

7

30

3

31

4

a.Find the overall time it took each worker to complete the manufacturing operation under study.

b.For each worker, find the standard deviation of the seven times for subtask 1.

c.In the context of this problem, what are the standard deviations you computed in part bmeasuring?

d.Repeat part b for subtask 2.

e.If you could choose workers similar to A or workers similar to B to perform subtasks 1 and 2, which type would you assign to each subtask? Explain your decisions on the basis of your answers to parts a鈥揹.

Short Answer

Expert verified

The overall time taken by B is more than A.

Step by step solution

01

Finding the overall time taken to complete the work

Worker A

Meanforsubtask1=30+28+31+38+25+29+307=2117=30.14Meanforsubtask2=2+4+3+3+2+4+37=217=3Overalltimetaken=Meanforsubtask1+Meanforsubtask2=30.14+3=33.14

Therefore, the overall time taken by Worker A to complete the work under study is 33.14 mins.

Worker B

Meanforsubtask1=31+30+32+30+29+30+317=2137=30.43Meanforsubtask2=7+2+6+5+4+1+47=297=4.14Overalltimetaken=Meanforsubtask1+Meanforsubtask2=30.43+4.14=34.57

Therefore, the overall time taken by Worker A to complete the work under study is 34.57 mins.

02

Calculating the standard deviation for each worker for subtask 1

Worker AWorker B

x

(x-x)

(x-x)2

y

(y-y)

(y-y)2

30

-0.14

0.0196

31

0.57

0.3249

28

-2.14

4.5796

30

-0.43

0.1849

31

0.86

0.7396

32

1.57

2.4649

38

7.86

61.7796

30

-0.43

0.1849

25

-5.14

26.4196

29

-1.43

2.0449

29

-1.14

1.2996

30

-0.43

0.1849

30

-0.14

0.0196

31

0.57

0.3249

SUM

0

94.8572

SUM

0

5.7143


localid="1668432137286" VarianceforworkerA=(x-x)2n=94.85727=13.55Standarddeviation=Variance=13.55=3.68VarianceforworkerB=(x-x)2n=5.71437=0.816Standarddeviation=Variance=0.816=0.9

03

Standard deviations computed in part b measuring

Worker A's standard deviation for completing subtask 1 is 3.68 minutes.

The standard deviation for completing subtask 1 for Worker B is 0.9 minutes.

04

Computing the standard deviation for each worker for subtask 2

Worker AWorker B

x

(x-x)

(x-x)2

y

(y-y)

(y-y)2

2

-1

1

7

2.86

8.1796

4

1

1

2

-2.14

4.5796

3

0

0

6

1.86

3.4596

3

0

0

5

0.86

0.7396

2

-1

1

4

-0.14

0.0196

4

1

1

1

-3.14

9.8596

3

0

0

4

-0.14

0.0196

SUM

0

4

SUM

0

26.8527

localid="1668431443790" VarianceforworkerA=(x-x)2n=47=0.57Standarddeviation=Variance=0.57=0.75VarianceforworkerB=(x-x)2n=26.85277=3.8361Standarddeviation=Variance=3.8361=1.96

Worker A's standard deviation for completing subtask 2 is 0.75 minutes.

The standard deviation for completing subtask 2 for Worker B is 1.96 minutes.

05

Determining the distribution of work

The overall time taken by B is more than A.

The standard deviation for subtask 1 is more significant for Worker A than B. Greater standard deviation implies that A's time to complete subtask 1 varies more, and hence there is no certainty as to when the work will get done. Therefore, type B workers should be given subtask 1.

Type A workers should be given subtask 2 because their standard deviation is smaller than worker Bs, implying that there will be a certainty that the work will get done quickly and the time taken will not vary much.

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Most popular questions from this chapter

Question: The purpose of this exercise is to compare the variability of with the variability of .

a. Suppose the first sample is selected from a population with mean and variance . Within what range should the sample mean vary about of the time in repeated samples of measurements from this distribution? That is, construct an interval extending standard deviations of on each side of .

b. Suppose the second sample is selected independently of the first from a second population with mean and variance . Within what range should the sample mean vary about the time in repeated samples of measurements from this distribution? That is, construct an interval extending standard deviations on each side .

c. Now consider the difference between the two sample means . What are the mean and standard deviation of the sampling distribution ?

d. Within what range should the difference in sample means vary about the time in repeated independent samples of measurements each from the two populations?

e. What, in general, can be said about the variability of the difference between independent sample means relative to the variability of the individual sample means?

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d. Form a 90 confidence interval for the difference between the actual redemption rates. Give a practical interpretation of the output.

e. Explain the meaning of the expression 鈥90 confident鈥 in your answer to part d.

f. Grounded on the interval, part d, is there a 鈥 statistically.鈥 A significant difference between the redemption rates? (Recall that a result is 鈥 statistically鈥 significant if there is substantiation to show that the true difference in proportions isn't 0.)

g. Assume the true difference between redemption rates must exceed.01 ( i.e., 1) for the experimenters to consider the difference 鈥 virtually鈥 significant. Based on the interval, part c, is there a 鈥渧irtually鈥 significant difference between the redemption rates?

Given the following values of x, s, and n, form a 90% confidence interval for2

a. x=21,s=2.5,n=50

b. x=1.3,s=0.02,n=15

c. x=167,s=31,n=22

d.x=9.4,s=1.5,n=5


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