DATA 605 - Final
12/15/2019
# Import library
library(corrplot)## corrplot 0.84 loaded library(DT)
library(dplyr)##
## Attaching package: 'dplyr'## The following objects are masked from 'package:stats':
##
## filter, lag## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, union library(ggplot2)
library(knitr)
library(matrixcalc)
library(stats)Problem 1
Using R, generate a random variable X that has 10,000 random uniform numbers from 1 to N, where N can be any number of your choosing greater than or equal to 6. Then generate a random variable Y that has 10,000 random normal numbers with a mean of μ=σ=(N+1)/2.
# Random variable X
N <- round(runif(1, 6, 500))
X <- runif(10000, min=0, max=N)
#Random variable Y
Y <- rnorm(10000, (N+1)/2,(N+1)/2)Probability. Calculate as a minimum the below probabilities a through c. Assume the small letter “x” is estimated as the median of the X variable, and the small letter “y” is estimated as the 1st quartile of the Y variable. Interpret the meaning of all probabilities.
- P(X > x | X > y)
x <- median(X)
y <- quantile(Y, 0.25)
sum(X > x & X > y)/sum(X > y)## [1] 0.5948132- P(X > x, Y > y)
sum(X > x & Y > y)/10000## [1] 0.3763- P(X < x | X > y)
sum(X < x & X > y)/10000## [1] 0.3406Investigate whether P(X>x and Y>y)=P(X>x)P(Y>y) by building a table and evaluating the marginal and joint probabilities.
# We build the table!!
problemOneTable <- matrix(c(sum(X > x & Y > y), sum(X < x & Y > y), sum(X > x & Y > y) + sum(X < x & Y > y), sum(X > x & Y < y), sum(X < x & Y < y), sum(X > x & Y < y) + sum(X < x & Y < y), sum(X > x & Y > y) + sum(X > x & Y < y), sum(X < x & Y > y) + sum(X < x & Y < y), sum(X > x & Y > y) + sum(X < x & Y > y) + sum(X > x & Y < y) + sum(X < x & Y < y)), nrow=3, ncol = 3)
colnames(problemOneTable) = c("Y > y", "Y < y", "All")
rownames(problemOneTable) = c("X > x", "X < x", "All")
datatable(problemOneTable)Check to see if independence holds by using Fisher’s Exact Test and the Chi Square Test. What is the difference between the two? Which is most appropriate?
# Fisher's Exact Test
fisher.test(problemOneTable, simulate.p.value=TRUE)##
## Fisher's Exact Test for Count Data with simulated p-value (based
## on 2000 replicates)
##
## data: problemOneTable
## p-value = 0.9805
## alternative hypothesis: two.sided # Chi Square Test
chisq.test(problemOneTable, correct=TRUE)##
## Pearson's Chi-squared test
##
## data: problemOneTable
## X-squared = 0.36053, df = 4, p-value = 0.9856Problem 2
Descriptive and Inferential Statistics. Provide univariate descriptive statistics and appropriate plots for the training data set. Provide a scatterplot matrix for at least two of the independent variables and the dependent variable. Derive a correlation matrix for any three quantitative variables in the dataset. Test the hypotheses that the correlations between each pairwise set of variables is 0 and provide an 80% confidence interval. Discuss the meaning of your analysis. Would you be worried about familywise error? Why or why not?
trainData <- read.csv(file="/Users/suma/Desktop/CUNY SPS - Masters Data Science/DATA 605/house-prices-advanced-regression-techniques/train.csv", header=TRUE, sep=",")
# Display train data
datatable(trainData, options = list(pageLength = 5))# Summary Statistics
summary(trainData)## Id MSSubClass MSZoning LotFrontage
## Min. : 1.0 Min. : 20.0 C (all): 10 Min. : 21.00
## 1st Qu.: 365.8 1st Qu.: 20.0 FV : 65 1st Qu.: 59.00
## Median : 730.5 Median : 50.0 RH : 16 Median : 69.00
## Mean : 730.5 Mean : 56.9 RL :1151 Mean : 70.05
## 3rd Qu.:1095.2 3rd Qu.: 70.0 RM : 218 3rd Qu.: 80.00
## Max. :1460.0 Max. :190.0 Max. :313.00
## NA's :259
## LotArea Street Alley LotShape LandContour
## Min. : 1300 Grvl: 6 Grvl: 50 IR1:484 Bnk: 63
## 1st Qu.: 7554 Pave:1454 Pave: 41 IR2: 41 HLS: 50
## Median : 9478 NA's:1369 IR3: 10 Low: 36
## Mean : 10517 Reg:925 Lvl:1311
## 3rd Qu.: 11602
## Max. :215245
##
## Utilities LotConfig LandSlope Neighborhood Condition1
## AllPub:1459 Corner : 263 Gtl:1382 NAmes :225 Norm :1260
## NoSeWa: 1 CulDSac: 94 Mod: 65 CollgCr:150 Feedr : 81
## FR2 : 47 Sev: 13 OldTown:113 Artery : 48
## FR3 : 4 Edwards:100 RRAn : 26
## Inside :1052 Somerst: 86 PosN : 19
## Gilbert: 79 RRAe : 11
## (Other):707 (Other): 15
## Condition2 BldgType HouseStyle OverallQual
## Norm :1445 1Fam :1220 1Story :726 Min. : 1.000
## Feedr : 6 2fmCon: 31 2Story :445 1st Qu.: 5.000
## Artery : 2 Duplex: 52 1.5Fin :154 Median : 6.000
## PosN : 2 Twnhs : 43 SLvl : 65 Mean : 6.099
## RRNn : 2 TwnhsE: 114 SFoyer : 37 3rd Qu.: 7.000
## PosA : 1 1.5Unf : 14 Max. :10.000
## (Other): 2 (Other): 19
## OverallCond YearBuilt YearRemodAdd RoofStyle
## Min. :1.000 Min. :1872 Min. :1950 Flat : 13
## 1st Qu.:5.000 1st Qu.:1954 1st Qu.:1967 Gable :1141
## Median :5.000 Median :1973 Median :1994 Gambrel: 11
## Mean :5.575 Mean :1971 Mean :1985 Hip : 286
## 3rd Qu.:6.000 3rd Qu.:2000 3rd Qu.:2004 Mansard: 7
## Max. :9.000 Max. :2010 Max. :2010 Shed : 2
##
## RoofMatl Exterior1st Exterior2nd MasVnrType MasVnrArea
## CompShg:1434 VinylSd:515 VinylSd:504 BrkCmn : 15 Min. : 0.0
## Tar&Grv: 11 HdBoard:222 MetalSd:214 BrkFace:445 1st Qu.: 0.0
## WdShngl: 6 MetalSd:220 HdBoard:207 None :864 Median : 0.0
## WdShake: 5 Wd Sdng:206 Wd Sdng:197 Stone :128 Mean : 103.7
## ClyTile: 1 Plywood:108 Plywood:142 NA's : 8 3rd Qu.: 166.0
## Membran: 1 CemntBd: 61 CmentBd: 60 Max. :1600.0
## (Other): 2 (Other):128 (Other):136 NA's :8
## ExterQual ExterCond Foundation BsmtQual BsmtCond BsmtExposure
## Ex: 52 Ex: 3 BrkTil:146 Ex :121 Fa : 45 Av :221
## Fa: 14 Fa: 28 CBlock:634 Fa : 35 Gd : 65 Gd :134
## Gd:488 Gd: 146 PConc :647 Gd :618 Po : 2 Mn :114
## TA:906 Po: 1 Slab : 24 TA :649 TA :1311 No :953
## TA:1282 Stone : 6 NA's: 37 NA's: 37 NA's: 38
## Wood : 3
##
## BsmtFinType1 BsmtFinSF1 BsmtFinType2 BsmtFinSF2
## ALQ :220 Min. : 0.0 ALQ : 19 Min. : 0.00
## BLQ :148 1st Qu.: 0.0 BLQ : 33 1st Qu.: 0.00
## GLQ :418 Median : 383.5 GLQ : 14 Median : 0.00
## LwQ : 74 Mean : 443.6 LwQ : 46 Mean : 46.55
## Rec :133 3rd Qu.: 712.2 Rec : 54 3rd Qu.: 0.00
## Unf :430 Max. :5644.0 Unf :1256 Max. :1474.00
## NA's: 37 NA's: 38
## BsmtUnfSF TotalBsmtSF Heating HeatingQC CentralAir
## Min. : 0.0 Min. : 0.0 Floor: 1 Ex:741 N: 95
## 1st Qu.: 223.0 1st Qu.: 795.8 GasA :1428 Fa: 49 Y:1365
## Median : 477.5 Median : 991.5 GasW : 18 Gd:241
## Mean : 567.2 Mean :1057.4 Grav : 7 Po: 1
## 3rd Qu.: 808.0 3rd Qu.:1298.2 OthW : 2 TA:428
## Max. :2336.0 Max. :6110.0 Wall : 4
##
## Electrical X1stFlrSF X2ndFlrSF LowQualFinSF
## FuseA: 94 Min. : 334 Min. : 0 Min. : 0.000
## FuseF: 27 1st Qu.: 882 1st Qu.: 0 1st Qu.: 0.000
## FuseP: 3 Median :1087 Median : 0 Median : 0.000
## Mix : 1 Mean :1163 Mean : 347 Mean : 5.845
## SBrkr:1334 3rd Qu.:1391 3rd Qu.: 728 3rd Qu.: 0.000
## NA's : 1 Max. :4692 Max. :2065 Max. :572.000
##
## GrLivArea BsmtFullBath BsmtHalfBath FullBath
## Min. : 334 Min. :0.0000 Min. :0.00000 Min. :0.000
## 1st Qu.:1130 1st Qu.:0.0000 1st Qu.:0.00000 1st Qu.:1.000
## Median :1464 Median :0.0000 Median :0.00000 Median :2.000
## Mean :1515 Mean :0.4253 Mean :0.05753 Mean :1.565
## 3rd Qu.:1777 3rd Qu.:1.0000 3rd Qu.:0.00000 3rd Qu.:2.000
## Max. :5642 Max. :3.0000 Max. :2.00000 Max. :3.000
##
## HalfBath BedroomAbvGr KitchenAbvGr KitchenQual
## Min. :0.0000 Min. :0.000 Min. :0.000 Ex:100
## 1st Qu.:0.0000 1st Qu.:2.000 1st Qu.:1.000 Fa: 39
## Median :0.0000 Median :3.000 Median :1.000 Gd:586
## Mean :0.3829 Mean :2.866 Mean :1.047 TA:735
## 3rd Qu.:1.0000 3rd Qu.:3.000 3rd Qu.:1.000
## Max. :2.0000 Max. :8.000 Max. :3.000
##
## TotRmsAbvGrd Functional Fireplaces FireplaceQu GarageType
## Min. : 2.000 Maj1: 14 Min. :0.000 Ex : 24 2Types : 6
## 1st Qu.: 5.000 Maj2: 5 1st Qu.:0.000 Fa : 33 Attchd :870
## Median : 6.000 Min1: 31 Median :1.000 Gd :380 Basment: 19
## Mean : 6.518 Min2: 34 Mean :0.613 Po : 20 BuiltIn: 88
## 3rd Qu.: 7.000 Mod : 15 3rd Qu.:1.000 TA :313 CarPort: 9
## Max. :14.000 Sev : 1 Max. :3.000 NA's:690 Detchd :387
## Typ :1360 NA's : 81
## GarageYrBlt GarageFinish GarageCars GarageArea GarageQual
## Min. :1900 Fin :352 Min. :0.000 Min. : 0.0 Ex : 3
## 1st Qu.:1961 RFn :422 1st Qu.:1.000 1st Qu.: 334.5 Fa : 48
## Median :1980 Unf :605 Median :2.000 Median : 480.0 Gd : 14
## Mean :1979 NA's: 81 Mean :1.767 Mean : 473.0 Po : 3
## 3rd Qu.:2002 3rd Qu.:2.000 3rd Qu.: 576.0 TA :1311
## Max. :2010 Max. :4.000 Max. :1418.0 NA's: 81
## NA's :81
## GarageCond PavedDrive WoodDeckSF OpenPorchSF EnclosedPorch
## Ex : 2 N: 90 Min. : 0.00 Min. : 0.00 Min. : 0.00
## Fa : 35 P: 30 1st Qu.: 0.00 1st Qu.: 0.00 1st Qu.: 0.00
## Gd : 9 Y:1340 Median : 0.00 Median : 25.00 Median : 0.00
## Po : 7 Mean : 94.24 Mean : 46.66 Mean : 21.95
## TA :1326 3rd Qu.:168.00 3rd Qu.: 68.00 3rd Qu.: 0.00
## NA's: 81 Max. :857.00 Max. :547.00 Max. :552.00
##
## X3SsnPorch ScreenPorch PoolArea PoolQC
## Min. : 0.00 Min. : 0.00 Min. : 0.000 Ex : 2
## 1st Qu.: 0.00 1st Qu.: 0.00 1st Qu.: 0.000 Fa : 2
## Median : 0.00 Median : 0.00 Median : 0.000 Gd : 3
## Mean : 3.41 Mean : 15.06 Mean : 2.759 NA's:1453
## 3rd Qu.: 0.00 3rd Qu.: 0.00 3rd Qu.: 0.000
## Max. :508.00 Max. :480.00 Max. :738.000
##
## Fence MiscFeature MiscVal MoSold
## GdPrv: 59 Gar2: 2 Min. : 0.00 Min. : 1.000
## GdWo : 54 Othr: 2 1st Qu.: 0.00 1st Qu.: 5.000
## MnPrv: 157 Shed: 49 Median : 0.00 Median : 6.000
## MnWw : 11 TenC: 1 Mean : 43.49 Mean : 6.322
## NA's :1179 NA's:1406 3rd Qu.: 0.00 3rd Qu.: 8.000
## Max. :15500.00 Max. :12.000
##
## YrSold SaleType SaleCondition SalePrice
## Min. :2006 WD :1267 Abnorml: 101 Min. : 34900
## 1st Qu.:2007 New : 122 AdjLand: 4 1st Qu.:129975
## Median :2008 COD : 43 Alloca : 12 Median :163000
## Mean :2008 ConLD : 9 Family : 20 Mean :180921
## 3rd Qu.:2009 ConLI : 5 Normal :1198 3rd Qu.:214000
## Max. :2010 ConLw : 5 Partial: 125 Max. :755000
## (Other): 9# Appropriate plots
# Boxplot
ggplot(trainData, aes(x=YearBuilt, y=OverallCond, fill=YearBuilt, group=YearBuilt)) + geom_boxplot() + labs(title="Overall Condition vs YearBuilt", x="Year Built", y="Overall Condition")
# Histograms
ggplot(trainData, aes(x=SalePrice)) + geom_histogram(fill="blue") + labs(title="Sale Price", x="Sale Price")## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
ggplot(trainData, aes(x=OverallCond, color=LotShape)) +
geom_histogram(fill="white", alpha=0.5, position="identity") + labs(title = "Overall Condition by LotShape", x = "Overall Condition")## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
# Scatterplot
ggplot(trainData, aes(x=YearBuilt, y=SalePrice)) + geom_point() + labs(title="Sale Price vs Year Built Scatter Plot", x="Year Built", y="Sale Price")
# 2 Independent Variables: Lot Frontage and Garage Area, and Dependent Variable Sale Price
pairs(~SalePrice+LotFrontage+GarageArea,data=trainData, main="Scatterplot Matrix")
Correlation Matrix:
trainCorrMatrix <- cor(dplyr::select(trainData, GarageArea, BsmtUnfSF, GrLivArea, SalePrice))
trainCorrMatrix## GarageArea BsmtUnfSF GrLivArea SalePrice
## GarageArea 1.0000000 0.1833027 0.4689975 0.6234314
## BsmtUnfSF 0.1833027 1.0000000 0.2402573 0.2144791
## GrLivArea 0.4689975 0.2402573 1.0000000 0.7086245
## SalePrice 0.6234314 0.2144791 0.7086245 1.0000000 # Correlation Plot - two useful ways to view
# With Numbers
corrplot(trainCorrMatrix, method="number")
# With Colors
corrplot(trainCorrMatrix, method="color")
Test the hypothesis
We see that all p-values for all three quantitative variables are consistently very small small, under 0.05. Thus, we reject the null hypothesis. Generally though, we know that p-values alone are valid enough to provide accurate predictions. Family wise error <= 1 - (1-alpha)^c. I think Family wise error will be high because there are not enough series of tests done - the more tests that are done, the lower the error.
80% confidence that there is correlation with these two variables between 0.60 and 0.64
cor.test(trainData$SalePrice,trainData$GarageArea, conf.level = 0.8)##
## Pearson's product-moment correlation
##
## data: trainData$SalePrice and trainData$GarageArea
## t = 30.446, df = 1458, p-value < 2.2e-16
## alternative hypothesis: true correlation is not equal to 0
## 80 percent confidence interval:
## 0.6024756 0.6435283
## sample estimates:
## cor
## 0.623431480% confidence that there is correlation with these two variables between 0.18 and 0.24
cor.test(trainData$SalePrice,trainData$BsmtUnfSF, conf.level = 0.8)##
## Pearson's product-moment correlation
##
## data: trainData$SalePrice and trainData$BsmtUnfSF
## t = 8.3847, df = 1458, p-value < 2.2e-16
## alternative hypothesis: true correlation is not equal to 0
## 80 percent confidence interval:
## 0.1822292 0.2462680
## sample estimates:
## cor
## 0.214479180% confidence that there is correlation with these two variables between 0.69 and 0.72
cor.test(trainData$SalePrice,trainData$GrLivArea, conf.level = 0.8)##
## Pearson's product-moment correlation
##
## data: trainData$SalePrice and trainData$GrLivArea
## t = 38.348, df = 1458, p-value < 2.2e-16
## alternative hypothesis: true correlation is not equal to 0
## 80 percent confidence interval:
## 0.6915087 0.7249450
## sample estimates:
## cor
## 0.7086245Linear Algebra and Correlation. Invert your correlation matrix from above. (This is known as the precision matrix and contains variance inflation factors on the diagonal.) Multiply the correlation matrix by the precision matrix, and then multiply the precision matrix by the correlation matrix. Conduct LU decomposition on the matrix.
# Precision Matrix - Invert correlation matrix
trainPrecisionMatrix <- solve(trainCorrMatrix)
# Multiply correlation matrix by the precision matrix
round(trainCorrMatrix %*% trainPrecisionMatrix)## GarageArea BsmtUnfSF GrLivArea SalePrice
## GarageArea 1 0 0 0
## BsmtUnfSF 0 1 0 0
## GrLivArea 0 0 1 0
## SalePrice 0 0 0 1 # Multiply the precisio matrix by the correlation matrix
round(trainPrecisionMatrix %*% trainCorrMatrix)## GarageArea BsmtUnfSF GrLivArea SalePrice
## GarageArea 1 0 0 0
## BsmtUnfSF 0 1 0 0
## GrLivArea 0 0 1 0
## SalePrice 0 0 0 1LU decomposition on the matrix.
lu.decomposition(trainCorrMatrix)$U## [,1] [,2] [,3] [,4]
## [1,] 1 0.1833027 0.4689975 0.6234314
## [2,] 0 0.9664001 0.1542888 0.1002024
## [3,] 0 0.0000000 0.7554087 0.4002391
## [4,] 0 0.0000000 0.0000000 0.3888845 lu.decomposition(trainCorrMatrix)$L## [,1] [,2] [,3] [,4]
## [1,] 1.0000000 0.0000000 0.0000000 0
## [2,] 0.1833027 1.0000000 0.0000000 0
## [3,] 0.4689975 0.1596531 1.0000000 0
## [4,] 0.6234314 0.1036863 0.5298312 1 lu.decomposition(trainPrecisionMatrix)$U## [,1] [,2] [,3] [,4]
## [1,] 1.645523 -0.07849925 -0.07604617 -0.95514586
## [2,] 0.000000 1.06570325 -0.18895578 -0.09467239
## [3,] 0.000000 0.00000000 2.00863169 -1.42336558
## [4,] 0.000000 0.00000000 0.00000000 1.00000000 lu.decomposition(trainPrecisionMatrix)$L## [,1] [,2] [,3] [,4]
## [1,] 1.00000000 0.0000000 0.0000000 0
## [2,] -0.04770475 1.0000000 0.0000000 0
## [3,] -0.04621399 -0.1773062 1.0000000 0
## [4,] -0.58045140 -0.0888356 -0.7086245 1Calculus-Based Probability & Statistics. Many times, it makes sense to fit a closed form distribution to data. Select a variable in the Kaggle.com training dataset that is skewed to the right, shift it so that the minimum value is absolutely above zero if necessary. Then load the MASS package and run fitdistr to fit an exponential probability density function. (See https://stat.ethz.ch/R-manual/R-devel/library/MASS/html/fitdistr.html ). Find the optimal value of λ for this distribution, and then take 1000 samples from this exponential distribution using this value (e.g., rexp(1000, λ)). Plot a histogram and compare it with a histogram of your original variable. Using the exponential pdf, find the 5th and 95th percentiles using the cumulative distribution function (CDF). Also generate a 95% confidence interval from the empirical data, assuming normality. Finally, provide the empirical 5th percentile and 95th percentile of the data. Discuss.
The samples histogram shows a right skewed distribution. Similarly, the GarageArea variable histogram somewhat resembles a right skewed distribution.
library(MASS)##
## Attaching package: 'MASS'## The following object is masked from 'package:dplyr':
##
## select library(spatstat)## Loading required package: spatstat.data## Loading required package: nlme##
## Attaching package: 'nlme'## The following object is masked from 'package:dplyr':
##
## collapse## Loading required package: rpart##
## spatstat 1.62-2 (nickname: 'Shape-shifting lizard')
## For an introduction to spatstat, type 'beginner'##
## Attaching package: 'spatstat'## The following object is masked from 'package:MASS':
##
## area # Select a variable in the Kaggle.com training dataset that is skewed to the right, shift it so that the minimum value is absolutely above zero if necessary.
min <- min(trainData$GarageArea)
min## [1] 0 # run fitdistr to fit an exponential probability density function
expPDF <- fitdistr(trainData$GarageArea, densfun = "exponential")
# optimal value
lambda <- expPDF$estimate
# samples histogram
hist(rexp(1000, lambda), breaks=100, main="Simulation PDF - 1000 Samples")
# variable histogram
hist(trainData$GarageArea, breaks=100, main="Garage Area Variable")
# Using the exponential pdf, find the 5th and 95th percentiles using the cumulative distribution function (CDF)
quantile(lambda, probs=c(0.05, 0.95))## 5% 95%
## 0.002114254 0.002114254 # generate a 95% confidence interval from the empirical data, assuming normality
qnorm(0.95, mean(trainData$GarageArea), sd(trainData$GarageArea))## [1] 824.6578 # provide the empirical 5th percentile and 95th percentile of the data
quantile(trainData$GarageArea, probs=c(0.05, 0.95))## 5% 95%
## 0.0 850.1Discuss:
The histograms tend to skew to the right and there is not much of a normal distribution amongst the mean. The simulation somewhat resembles the our variable data’s distribution more and less of a normal distribution. The empirical data percentile and normal distribution percentile were very close.
Modeling. Build some type of multiple regression model and submit your model to the competition board. Provide your complete model summary and results with analysis. Report your Kaggle.com user name and score.
Linear Model:
linearModel <- lm(trainData$SalePrice ~ OverallCond + OverallQual + GarageArea + LotArea + YearBuilt, data=trainData)
linearModel##
## Call:
## lm(formula = trainData$SalePrice ~ OverallCond + OverallQual +
## GarageArea + LotArea + YearBuilt, data = trainData)
##
## Coefficients:
## (Intercept) OverallCond OverallQual GarageArea LotArea
## -5.417e+05 3.146e+03 3.480e+04 8.183e+01 1.266e+00
## YearBuilt
## 2.237e+02Examine the summary: If the line is a good fit, the residuals should be normally distributed arounda mean of zero. Here we have have a median of -24492. A good model would have a median near zero and Min and Max of roughly the same magnitude. Below we have roughly similar magnitudes of Min -307954 and Max 390339. So far this model looks ok. The R^2 value explains 69.94% of the data’s variation. The adjusted R^2 value takes into account the number predictors used in the model, and this value will always be smaller than the R^2 value.
summary(linearModel)##
## Call:
## lm(formula = trainData$SalePrice ~ OverallCond + OverallQual +
## GarageArea + LotArea + YearBuilt, data = trainData)
##
## Residuals:
## Min 1Q Median 3Q Max
## -307954 -24492 -4342 16485 390339
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -5.417e+05 1.006e+05 -5.383 8.54e-08 ***
## OverallCond 3.146e+03 1.123e+03 2.801 0.00516 **
## OverallQual 3.480e+04 1.115e+03 31.216 < 2e-16 ***
## GarageArea 8.183e+01 6.735e+00 12.150 < 2e-16 ***
## LotArea 1.266e+00 1.169e-01 10.831 < 2e-16 ***
## YearBuilt 2.237e+02 5.150e+01 4.343 1.50e-05 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 43630 on 1454 degrees of freedom
## Multiple R-squared: 0.6994, Adjusted R-squared: 0.6984
## F-statistic: 676.7 on 5 and 1454 DF, p-value: < 2.2e-16Residual Analysis:
Pairwise comparison of the data:
pairs(trainData$SalePrice ~ OverallCond + OverallQual + GarageArea + LotArea + YearBuilt, data=trainData, gap=0.5)
The Normal Q-Q plot displays the residuals overall following closely to the line, which supports for a good model - but the ends diverge, with heavy tails. The histogram displays a normal looking (Gaussian) distribution centered at 0! This provides supports that our model is valid.
plot(linearModel, col=c("blue"))
hist(linearModel$residuals)
Kaggle Submission:
Username: kiwi #2 Score: 0.79716
knitr::include_graphics("/Users/suma/Desktop/KaggleUsernameScore.png")
testData <- read.csv(file="/Users/suma/Desktop/CUNY SPS - Masters Data Science/DATA 605/house-prices-advanced-regression-techniques/test.csv", header=TRUE, sep=",")
p <- predict(linearModel, testData)
df <- data.frame(cbind(testData$Id, p))
colnames(df) <- c("Id", "SalePrice")
df <- replace(df, is.na(df),0)
write.csv(df, file="model.csv", row.names = FALSE)