Assignment on econometrics PDF

Running head: ECONOMETRICS
ECONOMETRICS
Name of the Student
Name of the University
Author Note

ECONOMETRICS1
Assignment 1:
In this assignment all the variables from the ‘bears.xls’ files are first extracted and then a
prediction model(regression) for bear weight is performed with respect to other variables
chosen as independent variables. The variables are listed as follows.
MonthObs: The month in which each of bears are measured.
Gender: The gender of bears, 1 = male and 2 =female
HeadLengthInches: Head length of the bears measured in inches.
NeckInches: Circumference of the neck of bears measured in inches.
BodyLengthInches: length of the body of bears measured in inches.
ChestInches: circumference of the chest of bears measured in inches.
WeightPounds: Weight of the bears measured in pounds.
The sample of 54 bears is collected by the scientist assumed are collected in random
sampling method. From the sample the regression is performed to measure the weight of the
bears as it is practically very much difficult or impossible to measure the weight of the bears
individually. The linear regression model is performed in MATLAB which is shown below.
All the units are converted to metric form (weight of bears in Kgs and lengths or
circumference in meters).
MATLAB code:
bears = readtable('bears.xls','ReadVariableNames',true);
Meter = 1*0.0254; % inch to meter conversion unit
KGs = 1*0.45359227; % pounds to Kg conversion unit

ECONOMETRICS2
HeadLengthM = bears.HeadLengthInches.*Meter;
HeadWidthM = bears.HeadWidthInches.*Meter;
NeckM = bears.NeckInches.*Meter;
BodyLengthM = bears.BodyLengthInches.*Meter;
ChestM = bears.ChestInches.*Meter;
WeightKG = bears.WeightPounds.*KGs;
Regression = fitlm([HeadLengthM HeadWidthM NeckM BodyLengthM
ChestM],WeightKG) % Regression with linear variables
LogHeadLengthM = log(HeadLengthM);
LogHeadWidthM = log(HeadWidthM);
LogNeckM = log(NeckM);
LogBodyLengthM = log(BodyLengthM);
LogChestM = log(ChestM);
LogWeightKG = log(WeightKG);
LogRegression = fitlm([LogHeadLengthM LogHeadWidthM LogNeckM
LogBodyLengthM LogChestM],LogWeightKG) % Fitting Regression with logarithmic
variables
Output:
assign1
Regression =

ECONOMETRICS3
Linear regression model:
y ~ 1 + x1 + x2 + x3 + x4 + x5
Estimated Coefficients:
Estimate SE tStat pValue
________ ______ _______ __________
(Intercept) -111.54 14.393 -7.7493 5.2791e-10
x1 -140.18 103.78 -1.3508 0.18309
x2 50.386 89.738 0.56148 0.57708
x3 117.84 45.874 2.5687 0.013373
x4 15.99 21.178 0.75504 0.45391
x5 162.68 26.201 6.2089 1.2017e-07
Number of observations: 54, Error degrees of freedom: 48
Root Mean Squared Error: 14.3
R-squared: 0.939, Adjusted R-Squared 0.933
F-statistic vs. constant model: 148, p-value = 5.57e-28
LogRegression =
Linear regression model:
y ~ 1 + x1 + x2 + x3 + x4 + x5

ECONOMETRICS4
Estimated Coefficients:
Estimate SE tStat pValue
________ _______ ________ __________
(Intercept) 4.4229 0.42166 10.489 5.1817e-14
x1 -0.10279 0.29287 -0.35096 0.72715
x2 0.13452 0.12769 1.0535 0.29739
x3 0.49422 0.16147 3.0607 0.0036101
x4 1.1175 0.26719 4.1825 0.00012176
x5 1.4531 0.19754 7.3559 2.0979e-09
Number of observations: 54, Error degrees of freedom: 48
Root Mean Squared Error: 0.121
R-squared: 0.976, Adjusted R-Squared 0.974
F-statistic vs. constant model: 391, p-value = 1.19e-37
Here, y is bear weight in linear model and logarithmic model respectively and x1, x2...x5 are
the predictor variables as specified in the corresponding linear and logarithmic model.
Hence, the linear regression model is fitted with the sample for the variables in linear scale.
As it can be seen that the adjusted R2 value for the model is 0.933 or the variables can explain
93.30% of variation in the bear weight in kilograms. The regression model is
WeigthKG = -111.54 -140.18*HeadLengthM + 50.386*HeadWidthM + 117.84* NeckM +
15.99*BodyLengthM + 162.68*ChestM.
Similarly, the logarithmic regression model is

ECONOMETRICS5
LogWeightKG = 4.4229 + -0.10279*LogHeadLengthM + 0.13452*LogHeadWidthM +
0.49422*LogNeckM + 1.1175*LogBodyLengthM + 1.4531*LogChestM.
The model has the adjusted R^2 value of 0.974 (97.4% of variation is explained by the
model) and overall p value of 1.19e-37.
Now, backward elimination method is applied to find the most suitable models in the both
logarithmic and linear model. In this method the considered significance level is 0.05. Hence,
in each step the predictor coefficient with highest p value is removed until all the coefficients
have the p values less than 0.05. The final regression model is given below.
Regression =
Linear regression model:
y ~ 1 + x3 + x5
Estimated Coefficients:
Estimate SE tStat pValue
________ ______ _______ __________
(Intercept) -121.14 7.6844 -15.764 3.0186e-21
x3 103.05 38.267 2.693 0.0095594
x5 165.91 23.082 7.1879 2.7478e-09
Number of observations: 54, Error degrees of freedom: 51
Root Mean Squared Error: 14.2
R-squared: 0.937, Adjusted R-Squared 0.934
F-statistic vs. constant model: 377, p-value = 2.78e-31