Module 3, Week 1, gretl Problem Set 5
Module
3, Week 1, gretl Problem Set 5
This week your gretl assignment will be
automated too. In this document I am
providing you complete information to finish it. Once you have this gretl assignment completed
then it should be easy for you to do the Paper and Pencil assignment!
You can use the following commands (between
the lines below) to bring the data required to complete this assignment into
gretl and create a small sample of it for the Paper & Pencil assignment for
Week 5. This process eliminates
observations with erroneous values. I
was only able to get all variables for each observation in this way. Of course, if you know a different way to do
this that’s fine too. This is just so
you don’t have to think too much about importing and exporting data for your
assignments right now!
I
am pasting my entire script at the end of this
document so you will have all commands in one place! That is so you do not need to try to piece a
script together. You only need to read
through the text of this document so you can answer the appropriate
questions.
Note, in order for this to work you need to
insert the complete path to the folder or directory you are opening data files
from and saving your gretl data files, output, etc. to. These are the paths recovering the data file
from my Midterm Exam folder and saving the newly processed data files to Module
3\Week 5\...
#Open the entire corrected Boston housing
dataset
open "I:\My Passport
Documents\McDaniel\DataAnalytics\ANA500\Midterm\correctHousing.gdt"
#Sample and store the records before those
from Boston itself
smpl 1 356
store "I:\My Passport
Documents\McDaniel\DataAnalytics\ANA500\Module3\Week5\first.gdt"
#Restore the full dataset to get the
remaining records
smpl --full
#Sample and store the remaining records
smpl 489 506
store "I:\My Passport Documents\McDaniel\DataAnalytics\ANA500\Module3\Week5\second.gdt"
#Open the file with the first set of
records
open "I:\My Passport
Documents\McDaniel\DataAnalytics\ANA500\Module3\Week5\first.gdt"
#Append the records in the second file to
those in the first file
append "I:\My Passport
Documents\McDaniel\DataAnalytics\ANA500\Module3\Week5\second.gdt"
#Store the subset of records
store "I:\My Passport
Documents\McDaniel\DataAnalytics\ANA500\Module3\Week5\housingWOBoston.gdt"
#Plot the records with those having average
rooms=0.0 omitted
gnuplot RM CMEDV --output=display
#Look at your data! You can use descriptive stats to do this
summary
#Create a small dataset for your Paper and
Pencil Assignment this week. Keep in
mind
#what a small sample size is!
smpl 30 --random
store "I:\My Passport
Documents\McDaniel\DataAnalytics\ANA500\Module3\Week5\smallHousingSample.csv"
smpl --full
#Now that you have all the data sorted out
continue with your gretl assignment
You will continue to use the gretl dataset
you used for the Midterm Examination to answer the following questions.
You are interested in the relationship
between the average number of rooms in owner-occupied homes (RM) and the
(corrected) median home value (CMEDV) from the Boston housing dataset we used
for the Midterm Examination. All answers
to this gretl assignment involve your analysis of that dataset.
1.
Which of the variables of
interest, RM and CMEDV, are the following?
a.
The independent variable
b.
The dependent variable
2.
Once you have pre-processed the
data (e.g. as above between the lines) so that you have a good set of
observations in your dataset, produce a scatterplot that shows how home values
change with increasing number of rooms in the home.
gnuplot CMEDV RM --output=display
The relationship
between the variables CMEDV and RM appears to be linear.
3.
Calculate descriptive
statistics for the variables of interest (CMEDV and RM).
summary RM CMEDV
What is the mean
value of CMEDV?
What is the mean
value of RM?
What is the
interquartile range of CMEDV?
What is the
standard deviation of RM?
4.
Calculate the correlation
coefficient between the variables of interest.
corr
CMEDV RM
What is the
value you got for the correlation coefficient?
The correlation
coefficient is a measure of the strength of the relationship between two
variables. True or False?
5.
Residuals are the difference
between the values of independent variables at different points in time. True or False?
6.
The least squares method of
regression to find the line best fitting the data minimizes the (select the
best answer below):
a.
Sum of squared residuals
b.
Sum of residuals
c.
Sum of the dependent variable
squared
d.
Sum of the difference of
independent variables squared.
7.
What are the assumptions
required for conducting a linear regression.
Select all that apply.
a.
Linearity, i.e. the
relationship between dependent variable and independent variable(s) is linear.
b.
Homoscedasticity, i.e.
residuals are roughly equal and scattered about zero.
c.
Independence, i.e. observations
are independent.
d.
Normality, i.e. the residuals
are normally distributed.
e.
The correlation coefficient
between variables equals zero.
f.
The standard deviations of the
dependent variable vary over time resulting in heteroscedasticity.
g.
At least one of the independent
variables depends on other independent variables.
h.
The number of observations is
small and the data follow a Student’s t-distribution.
i.
There are a number of
explainable outliers in the data resulting in heteroscedasticity.
8.
Estimate a simple linear
regression model using least squares using the OLS command in gretl.
Let’s take a
look at the output from the OLS command:
ols CMEDV 0 RM Don’t forget to include the “0”
in the command for ordinary least squares computation because that is what
gives you the value for the intercept.
Model 1: OLS,
using observations 1-374
Dependent
variable: CMEDV
coefficient std. error
t-ratio p-value
---------------------------------------------------------
const
−26.0455 2.23293 −11.66
5.26e-027 ***
RM
7.99551 0.349031 22.91
4.52e-073 ***
Mean dependent
var 24.70294 S.D.
dependent var 8.387353
Sum squared
resid 10884.86 S.E.
of regression 5.409286
R-squared 0.585176 Adjusted
R-squared 0.584061
F(1, 372) 524.7651 P-value(F) 4.52e-73
Log-likelihood −1161.036 Akaike
criterion 2326.072
Schwarz
criterion 2333.921 Hannan-Quinn 2329.189
Just for comparison sake here is the output
from Excel for the same data:
|
SUMMARY OUTPUT |
|||||||||
|
Regression Statistics |
|||||||||
|
Multiple R |
0.765121883 |
||||||||
|
R Square |
0.585411497 |
||||||||
|
Adjusted R Square |
0.584294007 |
||||||||
|
Standard Error |
5.414980483 |
||||||||
|
Observations |
373 |
||||||||
|
ANOVA |
|||||||||
|
|
df |
SS |
MS |
F |
Significance F |
||||
|
Regression |
1 |
15360.72425 |
15360.72425 |
523.8632123 |
6.31553E-73 |
||||
|
Residual |
371 |
10878.46706 |
29.32201364 |
||||||
|
Total |
372 |
26239.19131 |
|
|
|
||||
|
|
Coefficients |
Standard Error |
t Stat |
P-value |
Lower 95% |
Upper 95% |
Lower 95.0% |
Upper 95.0% |
|
|
Intercept |
-26.05394429 |
2.235350016 |
-11.65542045 |
5.80305E-27 |
-30.44948918 |
-21.6583994 |
-30.44948918 |
-21.6583994 |
|
|
6.575 |
7.997913702 |
0.349436095 |
22.88805829 |
6.31553E-73 |
7.310789973 |
8.685037431 |
7.310789973 |
8.685037431 |
|
The results from Excel are virtually
identical to those from gretl. Both
methods’ results match!
a.
The estimated regression
equation is 
. True
or False?
b.
Discuss the estimated slope
coefficient.
i. The value of the estimated slope coefficient is _______.
ii. Interpret the estimated slope coefficient by considering the
following statement.
The estimated slope coefficient tells you how much the dependent
variable, in this case home value, varies with changes in the independent
variable, i.e. the average number of rooms in owner-occupied homes. True or False?
c.
Is the estimated slope
coefficient statistically significant? How do you know?
i. yes
ii. In this case, the P-value equals 4.52e-073 *** or something very,
very small and much smaller than the designated 0.05 level of significance.
9.
The coefficient of
determination or r-squared, is a measure of how much of the variability in the
data is explains the response, i.e. the dependent variable.
10. The value of r-squared for our current model is _________?
11. At what point or value is the coefficient of determination or
r-squared considered a strong indicator?
a.
0.50
b.
0.70
c.
0.90
d.
It depends…
The truth is that a good
value for r-squared depends on what the model you are developing is intended to
do. If the model is intended to
represent a lot of engineering or technical applications then usually somewhere
between 0.50 and 0.70 is considered good.
However, if you are developing a model for a final consumer product
where safety is involved you’ll want a much higher r-squared, e.g. 0.90 or even
a lot higher than that. Most basic
R&D projects are good with an r-squared value of around 0.2. In this case, r-squared is only intended to
give enough confidence to refine something to the next step or phase which
should have a higher r-squared. In the
social sciences, r-squared = from 0.10 to 0.30 is often considered good. So, it depends.
12. Calculate a 95% confidence interval for the estimated slope
coefficient.
First, be sure to keep the
result you want clearly in mind, i.e. you want the 95% confidence interval for the
slope coefficient. We have almost all
the information required to obtain the upper and lower bounds and hence the
confidence interval from the OLS output.
The slope coefficient is given in gretl is given by $coeff(RM). The standard error of the slope coefficient
is given by $stderr(RM). The degrees of
freedom is given by $df.
Using an online calculator,
or other appropriate means, knowing the degrees of freedom you can get t for
the confidence interval. For a 95%
confidence interval the t is 2.262.
The basic way we get a lower
bound, e.g. a lower bound is using the regression model (equation) to minimize
the sum of squared errors by subtracting appropriate values for the lower
bound. Note that it is easy to use
gretl’s “critical” function which you can find more info on in the gretl
function reference. The basic format of
the equation for the lower bound is:
Lower bound = $coeff(RM) –
t-cricical * standard error(RM)
To get the critical value
use gretl’s built-in “critical” function.
For more info on that see the gretl function reference available through
the help menu. The equations for the
lower and upper bounds are:
scalar lb = $coeff(RM) -
critical(t, $df, 0.025)*$stderr(RM)
scalar ub = $coeff(RM) +
critical(t, $df, 0.025)*$stderr(RM)
print lb ub
a.
The value of the lower bound
for the 95% confidence interval of the slope coefficient is _________?
b.
The value of the upper bound
for the 95% confidence interval of the slope coefficient is _________?
The
complete gretl Script for the Module 3 Week 5 gretl assignment.
Note, in order for this to work you need to
insert the complete path to the folder or directory you are opening data files
from and saving your gretl data files, output, etc. to. These are the paths recovering the data file
from my Midterm Exam folder and saving the newly processed data files to Module
3\Week 5\...
#Open the entire corrected Boston housing
dataset
open "I:\My Passport
Documents\McDaniel\DataAnalytics\ANA500\Midterm\correctHousing.gdt"
#Sample and store the records before those
from Boston itself
smpl 1 356
store "I:\My Passport
Documents\McDaniel\DataAnalytics\ANA500\Module3\Week5\first.gdt"
#Restore the full dataset to get the
remaining records
smpl --full
#Sample and store the remaining records
smpl 489 506
store "I:\My Passport
Documents\McDaniel\DataAnalytics\ANA500\Module3\Week5\second.gdt"
#Open the file with the first set of
records
open "I:\My Passport
Documents\McDaniel\DataAnalytics\ANA500\Module3\Week5\first.gdt"
#Append the records in the second file to
those in the first file
append "I:\My Passport
Documents\McDaniel\DataAnalytics\ANA500\Module3\Week5\second.gdt"
#Store the subset of records
store "I:\My Passport
Documents\McDaniel\DataAnalytics\ANA500\Module3\Week5\housingWOBoston.gdt"
#Look at your data! You can use descriptive stats to do this
summary
#Create a small dataset for your Paper and
Pencil Assignment this week. Keep in
mind
#what a small sample size is!
smpl 30 --random
store "I:\My Passport
Documents\McDaniel\DataAnalytics\ANA500\Module3\Week5\smallHousingSample.csv"
smpl --full
#Now that you have all the data sorted out
continue with your gretl assignment
#
#
#Plot the records with those having average
rooms=0.0 omitted
gnuplot CMEDV RM --output=display
#
#
#
#Compute the descriptive stats for just the
variables of interest
summary CMEDV RM
#Compute the correlation coefficient for
the variables of interest
corr CMEDV RM
#Develop an ordinary least squares model
for this data
ols CMEDV 0 RM --vcv
#Now we can use the corresponding linear
regression (line) to predict what a median home value
#would be for any number of rooms. For example, for 5 rooms we would get:
scalar CMEDV_hat = $coeff(const) +
$coeff(RM)*5
#
#
#Now let's look at the 95% confidence intervals
scalar lb = $coeff(RM)
- critical(t,
$df,
0.025)*$stderr(RM)
scalar ub = $coeff(RM)
+ critical(t,
$df,
0.025)*$stderr(RM)
print lb ub
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