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ASSIGNMENT 4

SAM

OCTOBER 20, 2015

RIZWAN ASIM

2017-01-0037

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ASSIGNMENT 4

Assignment 4

Q1:

A) Scatter Plot

Figure 1 Figure 2

From above plots it is clear that R 2 has best possible value when we use quadratic function of age

and ln(length).

B) Is there a linear relationship between age and length?

As explained by the scatter plots in question that there is no linear relation between age and length.The R square is better in case of quadratic relationship. R square increases to 0.744 when quadratic

relation is considered using ln transformation.

Table 1

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ASSIGNMENT 4

C) Propose a best fit model to estimate a length of bluegill given the age is equal to 2 years.

Best fir model is given by the following equation.

Ln (length) = -0.064 (age)2 + 0.601(age) + 3.661

At Age =2 Average Length of bluegill = 100.18 mm Table 2

D) On average, at what age a bluegill reaches its maximum length?

On average at age of 4.7, length will be maximum

Q 2

A) Run a scatter plot to find out a possible relationship between temperature and yield.

From scatter plot it is clear that, best possible relation between temp and yield is quadratic. There

is a sudden increase in value of R 2 when we change it from linear to quadratic. (From 0.092 to

0.673)

Figure 3

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ASSIGNMENT 4

B) Propose a best fit model.

Table 3

The best fit model explained by regression is given below

yield = 0.001(Temp)2 - 0.154(Temp) + 7.960

C) Write a detailed interpretation of the model. (hint: coefficient, R2, significance of the model, model

assumptions).

Table 4

62% of variation in yield is explained by independent variable temp in this model. Yield is decreasing at a

constant decreasing rate at 2.0 it reaches its lowest value at a temp of 77 after which its start increasing.

The coefficients of independent variables are statistically significant and have expected signs.

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ASSIGNMENT 4

Evaluating Assumptions

Residuals follow normal probability distribution.

Fig 1

It’s very much clear from figure I that residuals follow normal distribution.

Variation of residual along regression

Fig 2

It’s clear from fig 2 that no heteroscedasticity with is model.

No Auto Correlation.

Model’s Durbin Watson is 2.284 which lies in the acceptable range, therefore in this case NullHypothesis i.e. there is no autocorrelation.

Multicollinearity

From table 3, it’s clear that VIF value is greater than 10, but this collinearity is because of temp 2

directly depends upon value of temp. so there is no multi-collinearity.

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ASSIGNMENT 4

Linear relation

Scatter plot shows a very weak linear relationship.

Q: 3

A) Run a scatter plot to find out a possible relationship between speed and economy?

Best relation is explained by quadratic model. R square for quadratic is 0.965.

Table 5

Model Summary

Model R R Square Adjusted R

Square

Std. Error of the

Estimate

1 .982a .965 .957 .5139

a. Predictors: (Constant), Speed, Speed2

Fig 3

B) Propose a best fit model.

Coefficients a

Model Unstandardized Coefficients Standardized

Coefficients

t Sig.

B Std. Error Beta

1

(Constant) 13.472 .936 14.396 .000

Speed2 -.008 .001 -5.168 -14.568 .000

Speed .746 .049 5.446 15.350 .000

a. Dependent Variable: Economy

Fuel economy = -0.008(Speed)2 + 0.746(Speed) + 13.47

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C) Find the speed that maximizes a Honda car’s fuel economy.

Y= Fuel economy

X = Speed

Y=-0.008x2 +0.746x + 13.472

Taking derivative

2*-0.008x + 0.746 = 0

X = 46.625

The exact speed that maximizes the economy is 46.625 mile/hour

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ASSIGNMENT 5

SAM

OCTOBER 20, 2015

MUHAMMAD RIZWAN ASIM

2017 – 01 - 0037

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Assignment 5

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Assignment 5

A) Has advertisement impacted the sales positively?

There is a strong positive correlation between sales and advertisements, which is clear from

Correlation matrix and scatter plot.

Table 1

Correlations

Sales (1000s) Advertising (1000s)

Sales (1000s)

Pearson Correlation 1 .957**

Sig. (2-tailed) .000

N 36 36

Advertising (1000s)

Pearson Correlation .957** 1

Sig. (2-tailed) .000

N 36 36

**. Correlation is significant at the 0.01 level (2-tailed).

Figure 1

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Assignment 5

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Regression Model:

Table 2

Model Summary c

Model R R Square Adjusted R Square Std. Error of the

Estimate

Durbin-Watson

1 .963a .926 .920 4.1491

2 .998b .995 .992 1.3255 2.165

a. Predictors: (Constant), Y2014, Y2013, Advertising (1000s)

b. Predictors: (Constant), Y2014, Y2013, Advertising (1000s), May, Jun, Sep, Jul, Mar, Oct, Apr, Aug, Dec,

Feb, Nov

c. Dependent Variable: Sales (1000s)

Table 3

ANOVA a

Model Sum of Squares df Mean Square F Sig.

1 Regression 6943.944 3 2314.648 134.453 .000b

Residual 550.888 32 17.215

Total 7494.832 35

2 Regression 7457.938 14 532.710 303.219 .000c

Residual 36.894 21 1.757

Total 7494.832 35

a. Dependent Variable: Sales (1000s)

b. Predictors: (Constant), Y2014, Y2013, Advertising (1000s)

c. Predictors: (Constant), Y2014, Y2013, Advertising (1000s), May, Jun, Sep, Jul, Mar, Oct, Apr, Aug, Dec,

Feb, Nov

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Assignment 5

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Table 4

Coefficients a

Model Unstandardized

Coefficients

Standardize

d

Coefficients

t Sig. Collinearity

Statistics

B Std. Error Beta Toleran

ce

VIF

1 (Constant) 36.934 9.773 3.779 .001

Advertising

(1000s)

12.577 1.548 1.061 8.126 .000 .135 7.418

Y2013 1.294 2.557 .041 .506 .616 .345 2.901

Y2014 -3.591 4.512 -.120 -.796 .432 .102 9.821

2 (Constant) 49.331 5.731 8.608 .000

Advertising

(1000s)

8.727 1.014 .736 8.602 .000 .032 31.226

Y2013 6.295 1.379 .201 4.565 .000 .121 8.267

Y2014 6.572 2.777 .219 2.367 .028 .027 36.459

Feb 10.046 1.095 .192 9.171 .000 .532 1.878

Mar 11.049 1.124 .212 9.828 .000 .505 1.978

Apr 15.030 1.108 .288 13.568 .000 .521 1.921

May 12.931 1.242 .248 10.414 .000 .414 2.413

Jun 11.510 1.333 .220 8.636 .000 .360 2.780

Jul 12.213 1.416 .234 8.625 .000 .319 3.138

Aug 15.149 1.507 .290 10.055 .000 .281 3.553

Sep 15.270 1.394 .293 10.952 .000 .329 3.043

Oct 14.040 1.483 .269 9.465 .000 .290 3.444

Nov 8.280 1.680 .159 4.928 .000 .226 4.418

Dec 15.275 1.485 .293 10.286 .000 .290 3.452


Q2: The management believes that the month of March has been the most successful month in terms of

sales generation. Is it correct?

From Table 4, it is clear that with reference to January sales for month of march are 11.049 thousand

more when sales of other months are constant. But this is not highest value, so sales for month of march

are not most successful.

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Assignment 5

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Q3: What are the most successful and the unsuccessful months for the sales?

From table 4, it is clear that with reference to January sales of Dec are 15.275 thousand are more when

all other months’ sales are constant and its most successful month.

From table 4, it’s clear that with reference to January sales of Nov are 8.28 thousand are more when all

other months’ sales are constant and its most unsuccessful month.

Q4: What is the annual sales growth? What is the quarterly growth?

Annual Turnover:

From Table 4, it is clear that with reference to 2012 sales of 2013 6.295 more while all other factors are

constant. Sales of 2014 are 6.572 more than 2012.

Quarterly:

Table 5

Coefficients a

Model Unstandardized Coefficients Standardized

Coefficients

t Sig.

B Std. Error Beta

1 (Constant) 33.344 13.942 2.392 .023

Y2012 3.591 4.512 .117 .796 .432

Y2013 4.885 2.871 .156 1.701 .099

Advertising (1000s) 12.577 1.548 1.061 8.126 .000

2 (Constant) 66.439 19.393 3.426 .002

Y2012 -7.829 6.318 -.256 -1.239 .225

Y2013 -1.299 3.599 -.041 -.361 .721

Advertising (1000s) 8.337 2.295 .703 3.632 .001

Q2 6.273 1.913 .188 3.278 .003

Q3 7.482 2.469 .225 3.030 .005

Q4 5.810 2.738 .174 2.122 .043


Sales for quarter 2 are 6.273 more than comparing to quarter1Sales of quarter 3 are 7.482 more comparing to quarter 1.

Sales of quarter 4 are 5.810 more comparing to quarter 1.

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Assignment 5

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Q5: Has the year 2014 the better sales turnover?

Annual turnaround of 2014 are better than turnaround for 2012 and 2013 when all other factors

remain constant.

Q6: How confident the management should be about its inferences? Hint: Check the R-square of the

model generated; check the OLS assumptions.

From table 2, it is clear that value of R2 is 0.995, which shows that 99.5% variation is explained by this

mode.

Assumption 1:

Figure 1

From Fig 1 it is clear that residual follow normal distribution.

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Assumption 2:

From figure it is clear that there is no hetroscadicity in model.

Assumption 3:

Figure 2

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Assignment 5

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From figure 2, it is clear that there is a strong multi collinearity year sales and advertisement.

Assumption 4:

Table 6

ANOVAa

Model Sum of Squares df Mean Square F Sig.

1

Regression 6943.944 3 2314.648 134.453 .000b

Residual 550.888 32 17.215

Total 7494.832 35

2

Regression 7457.938 14 532.710 303.219 .000c

Residual 36.894 21 1.757

Total 7494.832 35


b. Predictors: (Constant), Advertising (1000s), Y2013, Y2012

c. Predictors: (Constant), Advertising (1000s), Y2013, Y2012, May, Jun, Sep, Jul, Mar, Oct, Apr,

Aug, Dec, Feb, Jan

F value from anova table is 303.219 which lies in no auto correlation region of Durbin Watson table.

17010037 - a4 & 5.pdf

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