Question

Learning curves are used in production operations to estimate the time required to complete a repetitive task as an operator gains experience. Suppose a production manager has compiled 30 time values (in minutes) for a particular operator as she progressed down the learning curve during the first 100 units. A portion of this data is shown in the accompanying table.

Time Per Unit	Unit Number
18.30	3
17.50	5
12.80	8
11.30	12
10.00	17
8.50	21
8.90	24
8.70	27
8.10	30
8.20	32
8.30	37
7.60	39
6.90	41
7.30	44
7.20	48
7.00	52
7.10	54
6.30	58
6.60	60
6.50	64
6.80	67
6.90	69
6.20	75
6.10	78
6.00	82
5.70	87
5.90	90
5.80	92
5.70	96
5.60	100

b. Estimate a simple linear regression model and a logarithmic regression model with time per unit as the response variable and unit number as the explanatory variable. (Negative values should be indicated by a minus sign. Round answers to 2 decimal places.)

Timeˆ=Time^=	+  Unit
Timeˆ=Time^=	+  ln(Unit)

c. Based on R², use the best-fitting model to predict the time that was required for the operator to build Unit 50. (Round coefficient estimates to at least 4 decimal places and final answer to 2 decimal places.)

Timeˆ=Time^=

minutes

Answer 1

Answer:

The first linear regression model that we have to estimate is

$Time=\alpha+\beta \text{unit} +\epsilon$

where Time is the time taken per unit

$\alpha$ is the intercept

$\beta$ is the slope coefficient for unit number

$\epsilon \sim \mathcal{N}(0,\sigma^2)$ is the random distirbance which is iid and normally distributed

Now we run regression in excel using data-->data analysis-->regression as

Regression Input Input Y Range: $B$1:$B$31 $A$1:$A$31 ОК Cancel Input X Range: Help Labels Confidence Level: Constant is Zero 95 % I A B Time per Unit Unit 1 Number (minutes) 18.30 17.50 12.80 11.30 10.00 8.50 8.90 8.70 8.10 8.20 8.30 7.60 6.90 $E$1 Output options Output Range: New Worksheet Ply: New Workbook Residuals Residuals Standardized Residuals Residual Plots Line Fit Plots Normal Probability Normal Probability Plots 7.30 7.20

We get the following output:

From this, looking at the coefficients table our estimated regression equation is

$\widehat{Time}=12.43-0.09\times unit$

Next we want to estimate the following regression model with natural lo of unit number as explanatory variable

$Time=\alpha+\beta\times \ln\text{unit} +\epsilon$

where $\ln\text{unit}$ is the natural log of unit number

The data with the new column is given below

Unit Number	Time per Unit (minutes)	ln (unit number)
3	18.3	=LN(A2)
5	17.5	=LN(A3)
8	12.8	=LN(A4)
12	11.3	=LN(A5)
17	10	=LN(A6)
21	8.5	=LN(A7)
24	8.9	=LN(A8)
27	8.7	=LN(A9)
30	8.1	=LN(A10)
32	8.2	=LN(A11)
37	8.3	=LN(A12)
39	7.6	=LN(A13)
41	6.9	=LN(A14)
44	7.3	=LN(A15)
48	7.2	=LN(A16)
52	7	=LN(A17)
54	7.1	=LN(A18)
58	6.3	=LN(A19)
60	6.6	=LN(A20)
64	6.5	=LN(A21)
67	6.8	=LN(A22)
69	6.9	=LN(A23)
75	6.2	=LN(A24)
78	6.1	=LN(A25)
82	6	=LN(A26)
87	5.7	=LN(A27)
90	5.9	=LN(A28)
92	5.8	=LN(A29)
96	5.7	=LN(A30)
100	5.6	=LN(A31)

and the data is

Unit Number	Time per Unit (minutes)	ln (unit number)
3	18.30	1.09861
5	17.50	1.60944
8	12.80	2.07944
12	11.30	2.48491
17	10.00	2.83321
21	8.50	3.04452
24	8.90	3.17805
27	8.70	3.29584
30	8.10	3.4012
32	8.20	3.46574
37	8.30	3.61092
39	7.60	3.66356
41	6.90	3.71357
44	7.30	3.78419
48	7.20	3.8712
52	7.00	3.95124
54	7.10	3.98898
58	6.30	4.06044
60	6.60	4.09434
64	6.50	4.15888
67	6.80	4.20469
69	6.90	4.23411
75	6.20	4.31749
78	6.10	4.35671
82	6.00	4.40672
87	5.70	4.46591
90	5.90	4.49981
92	5.80	4.52179
96	5.70	4.56435
100	5.60	4.60517

Next we run the regression as:

Regression Input Input Y Range: OK $8$1:$8$31 Cancel Input X Range: $C$1:$C$31 Help — Labels Confidence Level: Constant is Zero 95 % A B C Time per In (unit Unit Unit number 1 Number (minutes) 18.30 1.09861 17.50 1.60944 12.80 2.07944 11.30 2.48491 10.00 2.83321 8.50 3.04452 8.90 3.17805 8.70 3.29584 8.10 3.4012 8.20 3.46574 8.30 3.61092 7.60 3.66356 6.90 3.71357 7.30 3.78419 7.20 3.8712 $E$11 Output options Qutput Range: New Worksheet Ply: New Workbook Residuals Residuals Standardized Residuals Residual Plots Line Fit Plots Normal Probability Normal Probability Plots

The output is:

SUMMARY OUTPUT Regression Statistics Multiple F 0.964405 R Square 0.930076 Adjustedi 0.927579 Standard I 0.845084 Observati ANOVA df F gnificance F 372.437 1.03E-17 Regressio Residual Total SSM S 1 265.982 265.982 28 19.99666 0.714166 29 285.9787 Coefficientandard Err Stat P-value Lower 95%Upper 95%ower 95.0%pper 95.09 Intercept 20.56234 0.662595 31.03302 3.13E-23 19.20507 21.9196 19.20507 21.9196 In (unit nu -3.40501 0.176438 -19.2986 1.03E-17 -3.76643 -3.04359 -3.76643 -3.04359

From the coefficient table above the estimated regression equation is

$\widehat{Time}=20.56-3.41\times \ln \text{unit}$

c) We look at the R2 values for the 2 outputs from the above tables

For the first regression,  $R^2=0.6256$

For the second regression, $R^2=0.9301$

Since the second regression has a better R2, we can say that the second regression where log of unit number is the explanatory variable is the best fitting model.

We want to predict the time taken per unit when the unit number is 50. Using excel =LN(50) is 3.912

That is

$\begin{align*} \widehat{Time}&=20.56-3.41\times \ln \text{unit}\\ &=20.56-3.41\times \ln(50)\\ &= 20.56-3.41\times 3.912\\ &=7.22\end{align*}$

the time taken is 7.22 minutes

NOTE: I HOPE YOU HAPPY WITH MY ANSWER PLS RATE.