605_FinalProject
library(tidyr)
library(matrixcalc)
library(corrplot)## corrplot 0.84 loadedlibrary(MASS)Problem 1
Generate random variable
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 \(\mu = \sigma = (N+1)/2\)
N <- 10
n <- 10000
X <- runif(n = n, min = 1, max = N)
Y <- rnorm(n, mean=(N+1)/2, sd = (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.
x <- median(X)
y <- quantile(Y)[2]
df <- data.frame(X, Y)a. P(X>x | X>y)
Identify first the Probability of X>y
dfXy <- subset(df, df$X>y)Identify the Probability of X>x in the above subset
n2 <- nrow(subset(dfXy, df$X>x))
n2/n## [1] 0.5b. P(X>x, Y>y)
n2 <- nrow(subset(df, df$X>x & df$Y>y))
Pxny <- n2/n
Pxny## [1] 0.3776c. P(X<x | X>y)
n2 <- nrow(subset(dfXy, df$X<x))
n2/n## [1] 0.5Marginal and joint probabilities
Investigate whether P(X>x and Y>y)=P(X>x)P(Y>y) by building a table and evaluating the marginal and joint probabilities.
Px <- nrow(subset(df, df$X>x))/n
Py <- nrow(subset(df, df$Y>y))/n
Px*Py## [1] 0.375dfPtableValues <- matrix(c( nrow( df %>% subset(df$X < x & df$Y < y) ),
nrow( df %>% subset(df$X > x & df$Y < y)),
nrow( df %>% subset(df$X < x & df$Y > y)),
nrow( df %>% subset(df$X > x & df$Y > y))
))
table <- matrix(dfPtableValues , nrow=2)
colnames(table) <- c('Y < y','Y > y')
rownames(table) <- c('X < x','X > x')
table## Y < y Y > y
## X < x 1276 3724
## X > x 1224 3776dfPtable <- c( nrow( df %>% subset(df$X < x & df$Y < y) )/n,
nrow( df %>% subset(df$X > x & df$Y < y))/n,
nrow( df %>% subset(df$X < x & df$Y > y))/n,
nrow( df %>% subset(df$X > x & df$Y > y))/n
)
table <- matrix(dfPtable , nrow=2)
colnames(table) <- c('Y < y','Y > y')
rownames(table) <- c('X < x','X > x')
table## Y < y Y > y
## X < x 0.1276 0.3724
## X > x 0.1224 0.3776Looking at the individual values, marginal and joint probabilities are almost similar i.e., 0.375 vs 0.3776
Fisher and Chi square test
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 test is used
pTable <- matrix(dfPtableValues , nrow=2)
fisher.test(pTable)##
## Fisher's Exact Test for Count Data
##
## data: pTable
## p-value = 0.2389
## alternative hypothesis: true odds ratio is not equal to 1
## 95 percent confidence interval:
## 0.964496 1.158476
## sample estimates:
## odds ratio
## 1.057068Chi Square test is used
chisq.test(pTable)##
## Pearson's Chi-squared test with Yates' continuity correction
##
## data: pTable
## X-squared = 1.3872, df = 1, p-value = 0.2389Since both tests return higher p value, they fail to reject the null hypothesis.
Problem 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?
#Load the training dataset
trainData <- read.csv("kaggle/train.csv")
#Show summary of the dataset
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 Utilities
## Min. : 1300 Grvl: 6 Grvl: 50 IR1:484 Bnk: 63 AllPub:1459
## 1st Qu.: 7554 Pave:1454 Pave: 41 IR2: 41 HLS: 50 NoSeWa: 1
## Median : 9478 NA's:1369 IR3: 10 Low: 36
## Mean : 10517 Reg:925 Lvl:1311
## 3rd Qu.: 11602
## Max. :215245
##
## LotConfig LandSlope Neighborhood Condition1 Condition2
## Corner : 263 Gtl:1382 NAmes :225 Norm :1260 Norm :1445
## CulDSac: 94 Mod: 65 CollgCr:150 Feedr : 81 Feedr : 6
## FR2 : 47 Sev: 13 OldTown:113 Artery : 48 Artery : 2
## FR3 : 4 Edwards:100 RRAn : 26 PosN : 2
## Inside :1052 Somerst: 86 PosN : 19 RRNn : 2
## Gilbert: 79 RRAe : 11 PosA : 1
## (Other):707 (Other): 15 (Other): 2
## BldgType HouseStyle OverallQual OverallCond YearBuilt
## 1Fam :1220 1Story :726 Min. : 1.000 Min. :1.000 Min. :1872
## 2fmCon: 31 2Story :445 1st Qu.: 5.000 1st Qu.:5.000 1st Qu.:1954
## Duplex: 52 1.5Fin :154 Median : 6.000 Median :5.000 Median :1973
## Twnhs : 43 SLvl : 65 Mean : 6.099 Mean :5.575 Mean :1971
## TwnhsE: 114 SFoyer : 37 3rd Qu.: 7.000 3rd Qu.:6.000 3rd Qu.:2000
## 1.5Unf : 14 Max. :10.000 Max. :9.000 Max. :2010
## (Other): 19
## YearRemodAdd RoofStyle RoofMatl Exterior1st Exterior2nd
## Min. :1950 Flat : 13 CompShg:1434 VinylSd:515 VinylSd:504
## 1st Qu.:1967 Gable :1141 Tar&Grv: 11 HdBoard:222 MetalSd:214
## Median :1994 Gambrel: 11 WdShngl: 6 MetalSd:220 HdBoard:207
## Mean :1985 Hip : 286 WdShake: 5 Wd Sdng:206 Wd Sdng:197
## 3rd Qu.:2004 Mansard: 7 ClyTile: 1 Plywood:108 Plywood:142
## Max. :2010 Shed : 2 Membran: 1 CemntBd: 61 CmentBd: 60
## (Other): 2 (Other):128 (Other):136
## MasVnrType MasVnrArea ExterQual ExterCond Foundation BsmtQual
## BrkCmn : 15 Min. : 0.0 Ex: 52 Ex: 3 BrkTil:146 Ex :121
## BrkFace:445 1st Qu.: 0.0 Fa: 14 Fa: 28 CBlock:634 Fa : 35
## None :864 Median : 0.0 Gd:488 Gd: 146 PConc :647 Gd :618
## Stone :128 Mean : 103.7 TA:906 Po: 1 Slab : 24 TA :649
## NA's : 8 3rd Qu.: 166.0 TA:1282 Stone : 6 NA's: 37
## Max. :1600.0 Wood : 3
## NA's :8
## BsmtCond BsmtExposure BsmtFinType1 BsmtFinSF1 BsmtFinType2
## Fa : 45 Av :221 ALQ :220 Min. : 0.0 ALQ : 19
## Gd : 65 Gd :134 BLQ :148 1st Qu.: 0.0 BLQ : 33
## Po : 2 Mn :114 GLQ :418 Median : 383.5 GLQ : 14
## TA :1311 No :953 LwQ : 74 Mean : 443.6 LwQ : 46
## NA's: 37 NA's: 38 Rec :133 3rd Qu.: 712.2 Rec : 54
## Unf :430 Max. :5644.0 Unf :1256
## NA's: 37 NA's: 38
## BsmtFinSF2 BsmtUnfSF TotalBsmtSF Heating HeatingQC
## Min. : 0.00 Min. : 0.0 Min. : 0.0 Floor: 1 Ex:741
## 1st Qu.: 0.00 1st Qu.: 223.0 1st Qu.: 795.8 GasA :1428 Fa: 49
## Median : 0.00 Median : 477.5 Median : 991.5 GasW : 18 Gd:241
## Mean : 46.55 Mean : 567.2 Mean :1057.4 Grav : 7 Po: 1
## 3rd Qu.: 0.00 3rd Qu.: 808.0 3rd Qu.:1298.2 OthW : 2 TA:428
## Max. :1474.00 Max. :2336.0 Max. :6110.0 Wall : 4
##
## CentralAir Electrical X1stFlrSF X2ndFlrSF LowQualFinSF
## N: 95 FuseA: 94 Min. : 334 Min. : 0 Min. : 0.000
## Y:1365 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 TotRmsAbvGrd
## Min. :0.0000 Min. :0.000 Min. :0.000 Ex:100 Min. : 2.000
## 1st Qu.:0.0000 1st Qu.:2.000 1st Qu.:1.000 Fa: 39 1st Qu.: 5.000
## Median :0.0000 Median :3.000 Median :1.000 Gd:586 Median : 6.000
## Mean :0.3829 Mean :2.866 Mean :1.047 TA:735 Mean : 6.518
## 3rd Qu.:1.0000 3rd Qu.:3.000 3rd Qu.:1.000 3rd Qu.: 7.000
## Max. :2.0000 Max. :8.000 Max. :3.000 Max. :14.000
##
## Functional Fireplaces FireplaceQu GarageType GarageYrBlt
## Maj1: 14 Min. :0.000 Ex : 24 2Types : 6 Min. :1900
## Maj2: 5 1st Qu.:0.000 Fa : 33 Attchd :870 1st Qu.:1961
## Min1: 31 Median :1.000 Gd :380 Basment: 19 Median :1980
## Min2: 34 Mean :0.613 Po : 20 BuiltIn: 88 Mean :1979
## Mod : 15 3rd Qu.:1.000 TA :313 CarPort: 9 3rd Qu.:2002
## Sev : 1 Max. :3.000 NA's:690 Detchd :387 Max. :2010
## Typ :1360 NA's : 81 NA's :81
## GarageFinish GarageCars GarageArea GarageQual GarageCond
## Fin :352 Min. :0.000 Min. : 0.0 Ex : 3 Ex : 2
## RFn :422 1st Qu.:1.000 1st Qu.: 334.5 Fa : 48 Fa : 35
## Unf :605 Median :2.000 Median : 480.0 Gd : 14 Gd : 9
## NA's: 81 Mean :1.767 Mean : 473.0 Po : 3 Po : 7
## 3rd Qu.:2.000 3rd Qu.: 576.0 TA :1311 TA :1326
## Max. :4.000 Max. :1418.0 NA's: 81 NA's: 81
##
## PavedDrive WoodDeckSF OpenPorchSF EnclosedPorch X3SsnPorch
## N: 90 Min. : 0.00 Min. : 0.00 Min. : 0.00 Min. : 0.00
## P: 30 1st Qu.: 0.00 1st Qu.: 0.00 1st Qu.: 0.00 1st Qu.: 0.00
## Y:1340 Median : 0.00 Median : 25.00 Median : 0.00 Median : 0.00
## Mean : 94.24 Mean : 46.66 Mean : 21.95 Mean : 3.41
## 3rd Qu.:168.00 3rd Qu.: 68.00 3rd Qu.: 0.00 3rd Qu.: 0.00
## Max. :857.00 Max. :547.00 Max. :552.00 Max. :508.00
##
## ScreenPorch PoolArea PoolQC Fence MiscFeature
## Min. : 0.00 Min. : 0.000 Ex : 2 GdPrv: 59 Gar2: 2
## 1st Qu.: 0.00 1st Qu.: 0.000 Fa : 2 GdWo : 54 Othr: 2
## Median : 0.00 Median : 0.000 Gd : 3 MnPrv: 157 Shed: 49
## Mean : 15.06 Mean : 2.759 NA's:1453 MnWw : 11 TenC: 1
## 3rd Qu.: 0.00 3rd Qu.: 0.000 NA's :1179 NA's:1406
## Max. :480.00 Max. :738.000
##
## MiscVal MoSold YrSold SaleType
## Min. : 0.00 Min. : 1.000 Min. :2006 WD :1267
## 1st Qu.: 0.00 1st Qu.: 5.000 1st Qu.:2007 New : 122
## Median : 0.00 Median : 6.000 Median :2008 COD : 43
## Mean : 43.49 Mean : 6.322 Mean :2008 ConLD : 9
## 3rd Qu.: 0.00 3rd Qu.: 8.000 3rd Qu.:2009 ConLI : 5
## Max. :15500.00 Max. :12.000 Max. :2010 ConLw : 5
## (Other): 9
## SaleCondition SalePrice
## Abnorml: 101 Min. : 34900
## AdjLand: 4 1st Qu.:129975
## Alloca : 12 Median :163000
## Family : 20 Mean :180921
## Normal :1198 3rd Qu.:214000
## Partial: 125 Max. :755000
## univariate statistics and plots
par(mfrow=c(2,3))
plot(trainData$MSZoning, main = 'Zoning')
hist(trainData$LotArea, main = 'Lot Area')
hist(trainData$TotalBsmtSF, main = 'Basement sqft')
plot(trainData$Neighborhood, main = 'Neighborhood')
hist(trainData$SalePrice, main = 'Sale price')
hist(trainData$BedroomAbvGr, main = 'Bedroom count')
Independent and response variables
#Select 2 independent variables i.e., 'GrLivArea' and 'SaleCondition', and a response variable 'SalePrice'
plot(trainData$GrLivArea, trainData$SalePrice, xlab = "Gr LivingArea", ylab = "Sale price", main='LivingArea vs Sale price')
plot(trainData$SaleCondition, trainData$SalePrice, xlab = "Sale condition", ylab = "Sale price", main='Sale condition vs Sale price')
subsetData <- trainData[c('BedroomAbvGr', 'GrLivArea', 'SalePrice')]
#Calculate the co-relations between the 3 variables
corMatrix <- cor(subsetData)
corMatrix## BedroomAbvGr GrLivArea SalePrice
## BedroomAbvGr 1.0000000 0.5212695 0.1682132
## GrLivArea 0.5212695 1.0000000 0.7086245
## SalePrice 0.1682132 0.7086245 1.0000000# Plot the co-relations
corrplot(corMatrix)
#Test the co-relations for a confidence level of 0.8
cor.test(subsetData$BedroomAbvGr, subsetData$SalePrice, conf.level = .8)##
## Pearson's product-moment correlation
##
## data: subsetData$BedroomAbvGr and subsetData$SalePrice
## t = 6.5159, df = 1458, p-value = 9.927e-11
## alternative hypothesis: true correlation is not equal to 0
## 80 percent confidence interval:
## 0.1354160 0.2006421
## sample estimates:
## cor
## 0.1682132cor.test(subsetData$GrLivArea, subsetData$SalePrice, conf.level = .8)##
## Pearson's product-moment correlation
##
## data: subsetData$GrLivArea and subsetData$SalePrice
## t = 38.348, df = 1458, p-value < 2.2e-16
## alternative hypothesis: true correlation is not equal to 0
## 80 percent confidence interval:
## 0.6915087 0.7249450
## sample estimates:
## cor
## 0.7086245cor.test(subsetData$BedroomAbvGr, subsetData$GrLivArea, conf.level = .8)##
## Pearson's product-moment correlation
##
## data: subsetData$BedroomAbvGr and subsetData$GrLivArea
## t = 23.323, df = 1458, p-value < 2.2e-16
## alternative hypothesis: true correlation is not equal to 0
## 80 percent confidence interval:
## 0.4963921 0.5452915
## sample estimates:
## cor
## 0.5212695Would you be worried about familywise error? Due to the high number of variables available, and many of which are categorical nature and as I’m focussed on quantitative variables for this analysis, it is possible that I might be making false assumptions leading to FWE.
Linear 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.
inverseCorMatrix <- solve(corMatrix)
x <- round(corMatrix %*% inverseCorMatrix, 1)
x## BedroomAbvGr GrLivArea SalePrice
## BedroomAbvGr 1 0 0
## GrLivArea 0 1 0
## SalePrice 0 0 1y <- round(inverseCorMatrix %*% corMatrix, 1)
y## BedroomAbvGr GrLivArea SalePrice
## BedroomAbvGr 1 0 0
## GrLivArea 0 1 0
## SalePrice 0 0 1lu.decomposition(x)## $L
## [,1] [,2] [,3]
## [1,] 1 0 0
## [2,] 0 1 0
## [3,] 0 0 1
##
## $U
## [,1] [,2] [,3]
## [1,] 1 0 0
## [2,] 0 1 0
## [3,] 0 0 1lu.decomposition(y)## $L
## [,1] [,2] [,3]
## [1,] 1 0 0
## [2,] 0 1 0
## [3,] 0 0 1
##
## $U
## [,1] [,2] [,3]
## [1,] 1 0 0
## [2,] 0 1 0
## [3,] 0 0 1The decomposition of both x and y matrix resolve to same value.
Calculus-Based Probability & Statistics
Find the optimal value of \(\lambda\) for this distribution, and then take 1000 samples from this exponential distribution using this value (e.g., rexp(1000, \(\lambda\))). 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.
a <- fitdistr(x=trainData$SalePrice, densfun = "exponential")
value <- a$estimate
data <- rexp(1000, value)
quantile(data, c(.05,.95))## 5% 95%
## 10542.15 547971.98#Plot graphs
par(mfrow=c(1,2))
hist(trainData$SalePrice,main = "Actual train data")
#hist(trainData$SalePrice_log,main = "Skew adjusted train data")
hist(data,main = "distrbution data")
5th and 95th percentiles using the cumulative distribution function (CDF)
quantile(data, c(.05,.95))## 5% 95%
## 10542.15 547971.98Empirical 5th percentile and 95th percentile of the data
quantile(trainData$SalePrice, c(.05,.95))## 5% 95%
## 88000 326100data <- rnorm(length(trainData$SalePrice), mean = mean(trainData$SalePrice), sd=sd(trainData$SalePrice))
hist(data, main = "Normalized data")
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.
I observed there are high statistical outliers in the data in certain cases and calculated the quantile data for each variable I’m considering for this regression model. Based on my observation, I have mostly selected 98 percentile data and removed first 5 percentile data for a particular variable (yearbuild)
#Calculate and remove observations with outliers for variables 'GrLivArea','TotalBsmtSF','TotRmsAbvGrd','GarageArea','YearBuilt'
# Showcasing quantile data for Living area
quantile(trainData$GrLivArea, c(1:100)/100)## 1% 2% 3% 4% 5% 6% 7% 8% 9% 10%
## 692.18 768.00 796.00 828.80 848.00 864.00 864.00 885.44 895.55 912.00
## 11% 12% 13% 14% 15% 16% 17% 18% 19% 20%
## 924.49 948.00 960.00 980.00 988.00 1004.00 1034.00 1040.00 1054.00 1066.60
## 21% 22% 23% 24% 25% 26% 27% 28% 29% 30%
## 1078.78 1092.00 1109.57 1120.00 1129.50 1144.00 1154.93 1178.00 1196.22 1208.00
## 31% 32% 33% 34% 35% 36% 37% 38% 39% 40%
## 1217.29 1224.00 1236.00 1251.06 1262.00 1279.96 1300.49 1311.68 1324.00 1339.00
## 41% 42% 43% 44% 45% 46% 47% 48% 49% 50%
## 1348.00 1360.00 1368.00 1382.00 1392.55 1414.00 1427.46 1437.96 1452.91 1464.00
## 51% 52% 53% 54% 55% 56% 57% 58% 59% 60%
## 1474.09 1483.36 1494.00 1501.86 1509.45 1525.00 1540.26 1557.22 1571.00 1578.00
## 61% 62% 63% 64% 65% 66% 67% 68% 69% 70%
## 1599.95 1612.74 1626.51 1639.76 1651.35 1660.00 1668.00 1683.12 1694.00 1709.30
## 71% 72% 73% 74% 75% 76% 77% 78% 79% 80%
## 1717.00 1728.00 1739.07 1765.32 1776.75 1792.00 1803.43 1836.00 1849.22 1869.00
## 81% 82% 83% 84% 85% 86% 87% 88% 89% 90%
## 1907.37 1928.00 1949.97 1966.24 1987.30 2020.74 2058.66 2090.00 2120.02 2158.30
## 91% 92% 93% 94% 95% 96% 97% 98% 99% 100%
## 2221.14 2264.12 2328.35 2385.52 2466.10 2545.72 2633.23 2782.38 3123.48 5642.00trainData_nonOutlier <- trainData[c('SalePrice','GrLivArea','TotalBsmtSF','TotRmsAbvGrd','GarageArea','YearBuilt')]
trainData_nonOutlier <- subset(trainData_nonOutlier, trainData_nonOutlier$GrLivArea<quantile(trainData_nonOutlier$GrLivArea, c(1:100)/100)[98][1])
trainData_nonOutlier <- subset(trainData_nonOutlier, trainData_nonOutlier$TotalBsmtSF<quantile(trainData_nonOutlier$TotalBsmtSF, c(1:100)/100)[98][1])
trainData_nonOutlier <- subset(trainData_nonOutlier, trainData_nonOutlier$TotRmsAbvGrd<quantile(trainData_nonOutlier$TotRmsAbvGrd, c(1:100)/100)[98][1])
trainData_nonOutlier <- subset(trainData_nonOutlier, trainData_nonOutlier$GarageArea<quantile(trainData_nonOutlier$GarageArea, c(1:100)/100)[98][1])
trainData_nonOutlier <- subset(trainData_nonOutlier, trainData_nonOutlier$YearBuilt>quantile(trainData_nonOutlier$YearBuilt, c(1:100)/100)[5][1])m2 <- lm(SalePrice ~ ., data=trainData_nonOutlier)
summary(m2)##
## Call:
## lm(formula = SalePrice ~ ., data = trainData_nonOutlier)
##
## Residuals:
## Min 1Q Median 3Q Max
## -118635 -15917 -1968 14031 179803
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -1.162e+06 7.416e+04 -15.674 < 2e-16 ***
## GrLivArea 7.944e+01 3.635e+00 21.851 < 2e-16 ***
## TotalBsmtSF 4.537e+01 2.612e+00 17.370 < 2e-16 ***
## TotRmsAbvGrd -3.449e+03 1.052e+03 -3.280 0.00107 **
## GarageArea 5.171e+01 5.846e+00 8.846 < 2e-16 ***
## YearBuilt 5.940e+02 3.835e+01 15.491 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 29430 on 1248 degrees of freedom
## Multiple R-squared: 0.7678, Adjusted R-squared: 0.7669
## F-statistic: 825.6 on 5 and 1248 DF, p-value: < 2.2e-16plot(m2$residuals, main='Residuals')
res <- resid(m2)
qqnorm(res)
qqline(res)
testData <- read.csv("kaggle/test.csv")
pred <- predict(m2, testData)kaggleResult <- data.frame( Id = testData[,"Id"], SalePrice =pred)
kaggleResult[kaggleResult<0] <- 0
kaggleResult <- replace(kaggleResult,is.na(kaggleResult),0)
write.csv(kaggleResult, file="kaggleResult.csv", row.names = FALSE)Kaggle score
My username is Cheruku and I scored .49038 in first attempt. Will making few more tries to improve my score.