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IE 322 CASE STUDY

Group #:__28_____

Name :___Andre Fernandes________ Class #:__28______

Name: ____Lukas Gurrera___________ Class #:__40______

Name:_____Justin Rees________________ Class#:_72_______

Please SUBMIT THIS FILE as GROUP_#.pdf to ANGEL DROP BOX

Part I: Correlation and Covariance Task 1:

Scatter Plot #1: Y(Market Share) vs. X1 (Absolute Unit Price) (5 Points)

Scatter Plot #2: Y(Market Share) vs. X2 (Relative Unit Price) (5 Points)

R² = 0.4502

0.15

0.2

0.25

0.3

36 42 48 54 60 66 72 78 84 90 96

SCATTER PLOT 1

SCATTER PLOT 1

Linear (SCATTER PLOT1)

R² = 0.6334

0.15

0.2

0.25

0.3

0.2 0.4 0.6 0.8

SCATTER PLOT 2

SCATTER PLOT 2

Linear (SCATTER PLOT2)

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Task 2: (12 Points)

COV[X1, Y] = -.13699 COV[X2,Y]= .001002

CORR COEFF[X1, Y] = -.67098 CORR COEFF[X2, Y] = .795854

Show Sample Calculations Below for COV[X1, Y]; COV[X2,Y]; CORR COEFF[X1, Y];

CORR COEFF[X2, Y] (8 Points):

COV(X,Y) = E[XY] – E[X]E[Y]

E[XY] = ∑(Xi *Yi) / 120

E[X] = ∑ Xi / 120

E[Y] = ∑ Yi / 120

COV(X1,Y) = 11.08048 – 48.48889*0.231341

COV(X1,Y) = -0.13699

COV(X2,Y) = 0.115627 – 0.495481*0.231341

COV(X2,Y) = 0.115627

CORR(X,Y) = COV(X,Y) / (σx σy)

σx = √ (∑(Xi – Xaverage)2 / (n-1) )

σy = √ (∑(Yi – Yaverage)2 / (n-1) )

CORR(X1,Y) = -0.13699/10.88615 = -0.67098

CORR(X2,Y) = 0.115627/0.06714 = 0.795854

Built in Excel functions were used to readily calculate average and standard deviation of

variables. Functions used included:

E(x) ‘=average (array of x)’

V(x) ‘=STDEV.P(array of x)’

Built in Excel functions were used to validate covariance and correlation coefficient

COV(X,Y) ‘=COVAR(array of x, array of y)’

CORR(X,Y) ‘=CORREL(array of x, array of y)’

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Task 3: Answer the Following Questions (12 Points)

(1) Comment on the scatter plots from Task #1. Explain what you can tell about

the relationship between Y and X1 and Y and X2 based on the scatter plots. You

may want to use the linear trendline and resulting R2 value to help you with this.

(4 Points)

We can observe that the trend line of the X1 and Y plot yields a negative

slope, representing the negative correlation between both variables. In other words,

as X1 increase, Y will decrease. The trend line on the X2 and Y plot yielded a

positive slope, meaning that there is a positive relation between both variables and

as X2 increases, so does the Y value .The R2 value represents the strength of the

relationship between both variables since it is the square of the correlation

coefficient, indicating that the higher the R2 value, the stronger the correlation

coefficient. In other words, the higher the R2 value, a stronger change in Y is

observed with a change in X variables.

(2) Comment on the Correlation Coefficients calculated and what they tell you

about the relationship between Y and X1 and Y and X2. Is there a relationship? IF

yes, is it positive or negative? Is it slight or strong? (4 Points)

The correlation coefficient of X1 came out to be -.67098 and X2 equaled

.795854 meaning they both have a relationship. X1 has a strong negative

relationship because it is closer to -1 than 0, likewise X2 has a strong positive

relationship because its value is closer to 1 than 0.

(3) Do you feel as though the relationships you discussed in (1) and (2) can be used

to accurately predict the market share (i.e. the dependent variable). In other words,

do you think the changes in the independent variables (X1, X2) can accurately

predict the dependent variable (Y) ? (4 Points)

The relationship allows us to predict a trend line but cannot accurately

predict the market share because the values are based on the average of the trend

line.

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Part II: Regression Analysis

Task 1: (10 Points)

Paste the Minitab Output below for Y vs. X1, X2

Regression Analysis: y versus x1, x2 The regression equation is

y = 0.179 - 0.00105 x1 + 0.208 x2

Predictor Coef SE Coef T P

Constant 0.179129 0.000521 343.67 0.000

x1 -0.00104546 0.00000532 -196.48 0.000

x2 0.207687 0.000863 240.72 0.000

S = 0.000632182 R-Sq = 99.9% R-Sq(adj) = 99.9%

Analysis of Variance

Source DF SS MS F P

Regression 2 0.042162 0.021081 52748.26 0.000

Residual Error 117 0.000047 0.000000

Total 119 0.042209

Source DF Seq SS

x1 1 0.019003

x2 1 0.023159

Unusual Observations

Obs x1 y Fit SE Fit Residual St Resid

28 92.2 0.194436 0.193747 0.000245 0.000689 1.18 X

61 46.0 0.281676 0.280911 0.000203 0.000764 1.28 X

73 84.9 0.195084 0.195358 0.000203 -0.000273 -0.46 X

106 88.8 0.203899 0.203395 0.000235 0.000504 0.86 X

111 86.0 0.178107 0.177150 0.000212 0.000957 1.61 X

X denotes an observation whose X value gives it large leverage.

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Task 2: Provide Answers to the Following Questions Based on the Minitab Output Above:

(24 Points)

(1) Looking at the R2 value (R-sq) in the output for the regression analysis- please

write a one sentence interpretation of this value in the context of this problem.

(6 Points)

The R^2 value equaled 99.9% indicating that the model is very accurate in

determining the data variance.

(2) From the output above, would you say that the Absolute Price (X1) is a good

predictor of the Market Share (Y)? Why or Why Not? Please make sure you try to

answer this using the p value for the Absolute Price shown. (6 Points)

Yes X1 is a good predictor of the market share. The regression analysis

yielded a p value of 0.0 meaning that the model is very significant in determining

the market share for X1 variable. With p equaling 0.0 we can say with a greater

than 95% certainty that the model correctly predicts the market share.

(3) From the output above, would you say that the Relative Price (X2) is a good

predictor of the Market Share (Y)? Why or Why Not? Please make sure you try to

answer this using the p value for the Relative Price shown. (6 Points)

Yes X2 is a good predictor of the market share. The regression analysis

yielded a p value of 0.0 meaning that the model is very significant in determining

the market share for X2 variable. With p equaling 0.0 we can say with a greater

than 95% certainty that the model correctly predicts the market share.

(4) Do You feel as though the linear regression model (i.e. the regression equation)

does a good job of estimating (Y)? Explain. (6 Points)

Yes the linear regression model does a good job of closely estimating Y values. By

plugging different X values it is observed that the calculated Y value is very close to the actual

market share value.

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Part III: Matching Distributions

Task 1: Paste Your (3) Minitab Histograms Below (X1, X2, Y): (6 Points)

90807060504030

50

40

30

20

10

0

x1

Fre

qu

en

cy

Mean 48.49

StDev 10.93

N 120

Histogram of x1Normal

0.700.650.600.550.500.450.400.35

20

15

10

5

0

x2

Fre

qu

en

cy

Mean 0.4955

StDev 0.06742

N 120

Histogram of x2Normal

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0.280.260.240.220.200.18

30

25

20

15

10

5

0

y

Fre

qu

en

cy

Mean 0.2313

StDev 0.01883

N 120

Histogram of yNormal

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Task 2: Paste Your (9) Empirical CDF Plots Below (Normal, Exponential, Gamma): (9

Points)

NORMAL

1009080706050403020

100

80

60

40

20

0

x1

Pe

rce

nt

Mean 48.49

StDev 10.93

N 120

Empirical CDF of x1Normal

0.70.60.50.40.3

100

80

60

40

20

0

x2

Pe

rce

nt

Mean 0.4955

StDev 0.06742

N 120

Empirical CDF of x2Normal

9

0.280.260.240.220.200.18

100

80

60

40

20

0

y

Pe

rce

nt

Mean 0.2313

StDev 0.01883

N 120

Empirical CDF of yNormal

EXPONENTIAL

250200150100500

100

80

60

40

20

0

x1

Pe

rce

nt

Mean 48.49

N 120

Empirical CDF of x1Exponential

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2.52.01.51.00.50.0

100

80

60

40

20

0

x2

Pe

rce

nt

Mean 0.4955

N 120

Empirical CDF of x2Exponential

1.21.00.80.60.40.20.0

100

80

60

40

20

0

y

Pe

rce

nt

Mean 0.2313

N 120

Empirical CDF of yExponential

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GAMMA

1009080706050403020

100

80

60

40

20

0

x1

Pe

rce

nt

Shape 24.23

Scale 2.001

N 120

Empirical CDF of x1Gamma

0.70.60.50.40.3

100

80

60

40

20

0

x2

Pe

rce

nt

Shape 55.46

Scale 0.008934

N 120

Empirical CDF of x2Gamma

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0.280.260.240.220.200.18

100

80

60

40

20

0

y

Pe

rce

nt

Shape 149.6

Scale 0.001546

N 120

Empirical CDF of yGamma

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Task 3: Provide Answers to the Following Questions: (9 Points)

(1) From your work in Task 1 & Task 2 above, Which of the distributions do you feel as through the Relative Market Share (Y) follows? (3 Points) Normal

(2) From your work in Task 1 & Task 2 above, which of the distributions do you

feel as though the Absolute Unit Price (X1) follows? (3 Points)

Gamma

(3) From your work in Task 1 & Task 2 above, which of the distributions do you

feel as though the Relative Unit Price (X2) follows? (3 Points)

Gamma

*All results were obtained by first looking at how close of a fit to the Empirical CDF lines were

and narrowing it down by observing the normal distribution trends.

GRADE: ________/ 100