Homework 2
Stenroos
October 5, 2017
House Prices
This data set includes prices and characteristics of 128 houses. Some of the variables are Price, Size, Bedrooms, Bathrooms, and Neighborhood.
HP <- read.csv("~/Business Analytics/HousePrices.csv")
plot(Price~SqFt,data=HP)
plot(Price~Bedrooms,data=HP)
plot(Price~Bathrooms,data=HP)
plot(Price~Offers,data=HP)
plot(Price~Brick,data=HP)
plot(Price~Neighborhood,data=HP)
Above are all of the XY plots of our data set. These plots all look good and none show any signs of the X variables being colinear.
The correlation matrix supports our assumption of X variables being independent. Now to look at our regression output.
m1=lm(Price~.,data=HP)
summary(m1)##
## Call:
## lm(formula = Price ~ ., data = HP)
##
## Residuals:
## Min 1Q Median 3Q Max
## -27897.8 -6074.8 -48.7 5551.8 27536.4
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 308.114 9605.692 0.032 0.974465
## HomeID -11.456 25.387 -0.451 0.652616
## SqFt 53.634 5.926 9.051 3.30e-15 ***
## Bedrooms 4136.461 1621.775 2.551 0.012023 *
## Bathrooms 7975.157 2133.831 3.737 0.000287 ***
## Offers -8350.128 1103.693 -7.566 8.96e-12 ***
## BrickYes 17313.540 1988.548 8.707 2.12e-14 ***
## NeighborhoodNorth 1729.613 2433.756 0.711 0.478675
## NeighborhoodWest 22264.319 2540.699 8.763 1.56e-14 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 10050 on 119 degrees of freedom
## Multiple R-squared: 0.8688, Adjusted R-squared: 0.86
## F-statistic: 98.54 on 8 and 119 DF, p-value: < 2.2e-16Our regression output gives us an adjusted R^2 of .86 which tells us that our model is a good fit for our data set. We can also see that Sqft, bedrooms, bathrooms, offers, and bricks are all significant.
The equation for our line is:
Price=308.11-11.46HomeID+53.63SqFt+4136.46Bedrooms+7975.16Bathrooms-8350.13Offers+17313.54BrickYes+1729.613NeighborhoodNorth+22264.32NeighborhoodWest
Next, I will take 90 random cases and create a model with them and use this model to predict a test set of 38 cases. Then I will examine the error.
set.seed(1)
n=length(HP$Price)
n1=90
n2=n-n1
train=sample(1:n,n1)
m1=lm(Price~.,data=HP[train,])
summary(m1)##
## Call:
## lm(formula = Price ~ ., data = HP[train, ])
##
## Residuals:
## Min 1Q Median 3Q Max
## -29267 -4859 -1429 5576 28896
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1782.970 11103.095 0.161 0.87282
## HomeID -18.470 29.611 -0.624 0.53453
## SqFt 51.466 6.655 7.734 2.51e-11 ***
## Bedrooms 4692.374 1854.417 2.530 0.01333 *
## Bathrooms 7309.842 2387.949 3.061 0.00299 **
## Offers -7737.062 1229.031 -6.295 1.49e-08 ***
## BrickYes 17915.620 2434.550 7.359 1.36e-10 ***
## NeighborhoodNorth 3941.988 3001.474 1.313 0.19277
## NeighborhoodWest 22785.492 2919.501 7.805 1.82e-11 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 9586 on 81 degrees of freedom
## Multiple R-squared: 0.8777, Adjusted R-squared: 0.8656
## F-statistic: 72.68 on 8 and 81 DF, p-value: < 2.2e-16pred=predict(m1,newdat=HP[-train,])
obs=HP$Price[-train]
diff=obs-pred
percdiff=abs(diff)/obs
me=mean(diff)
rmse=sqrt(sum(diff**2)/n2)
mape=100*(mean(percdiff))
me # mean error## [1] 1904.989rmse # root mean square error## [1] 11328.64mape # mean absolute percent error## [1] 6.882381With our mean absolute percent error at roughly 7% we can conclude our training set was a good predictor for our test set which tell us our data is a good set of data to make models from and should translate to more data of the same nature.
Gender Discrimination
This data set includes data on gender, work experience, and salary.
GD <- read.csv("~/Business Analytics/GenderDiscrimination.csv")
plot(Salary~Gender,data=GD)
plot(Salary~Experience,data=GD)
Above are all of the XY plots of our data set. These plots all look good and none show any signs of the X variables being colinear.
m2=lm(Salary~.,data=GD)
summary(m2)##
## Call:
## lm(formula = Salary ~ ., data = GD)
##
## Residuals:
## Min 1Q Median 3Q Max
## -52779 -9806 -121 8347 60913
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 53260.0 2416.6 22.039 < 2e-16 ***
## GenderMale 17020.6 2499.6 6.809 1.06e-10 ***
## Experience 1744.6 160.7 10.858 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 16910 on 205 degrees of freedom
## Multiple R-squared: 0.4413, Adjusted R-squared: 0.4359
## F-statistic: 80.98 on 2 and 205 DF, p-value: < 2.2e-16Our regression output gives us an adjusted R^2 of .44 which tells us that our model is not a good fit for our data set and less than 50% of the variation is explained by our model. We can also see that all variables are significant.
The equation for our line is:
Salary= 53260+17020.6GenderMale+1744.6Experience
Direct Marketing
This data set includes data from a direct marketer who sells his products via mail.
DM <- read.csv("~/Business Analytics/DirectMarketing.csv")
plot(AmountSpent~Age,data=DM)
plot(AmountSpent~Gender,data=DM)
plot(AmountSpent~OwnHome,data=DM)
plot(AmountSpent~Married,data=DM)
plot(AmountSpent~Location,data=DM)
plot(AmountSpent~Children,data=DM)
plot(AmountSpent~History,data=DM)
plot(AmountSpent~Catalogs,data=DM)
plot(AmountSpent~Salary,data=DM)
Above are all of the XY plots of our data set. These plots all look good and none show any signs of the X variables being colinear.
m3=lm(AmountSpent~.,data=DM)
summary(m3)##
## Call:
## lm(formula = AmountSpent ~ ., data = DM)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1648.11 -286.72 -12.63 218.21 2771.25
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -2.496e+02 1.340e+02 -1.862 0.06302 .
## AgeOld 4.139e+01 5.276e+01 0.784 0.43311
## AgeYoung 8.965e+01 5.874e+01 1.526 0.12740
## GenderMale -5.370e+01 3.802e+01 -1.413 0.15823
## OwnHomeRent -1.829e+01 4.151e+01 -0.441 0.65967
## MarriedSingle 1.950e+01 4.981e+01 0.392 0.69553
## LocationFar 6.090e+02 4.399e+01 13.845 < 2e-16 ***
## Salary 1.883e-02 1.245e-03 15.124 < 2e-16 ***
## Children -2.683e+02 2.502e+01 -10.723 < 2e-16 ***
## HistoryLow -2.675e+02 8.862e+01 -3.019 0.00263 **
## HistoryMedium -3.446e+02 5.996e+01 -5.746 1.38e-08 ***
## Catalogs 4.052e+01 2.868e+00 14.128 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 463.5 on 685 degrees of freedom
## (303 observations deleted due to missingness)
## Multiple R-squared: 0.7887, Adjusted R-squared: 0.7853
## F-statistic: 232.5 on 11 and 685 DF, p-value: < 2.2e-16Our regression output gives us an adjusted R^2 of .785 which tells us that our model is a good fit for our data set. We can also see that LocationFar, Salary, Children, History Low, HistoryMedium, and catalogs are all significant.
The equation for our line is:
AmountSpent=-249.6+41.39AgeOld+89.65AgeYoung-53.7GenderMale-18.29OwnHomeRent+19.5MarriedSingle+609LocationFar+.0188Salary-268.3Children-267.5HistoryLow-344.6HistoryMedium+40.52*Catalogs