Predict House sale prices using Multiple Linear Regression
About the Project: Regression analysis is a powerful statistical method that allows you to examine the relationship between two or more variables of interest. While there are many types of regression analysis, at their core they all examine the influence of one or more independent variables on a dependent variable.
DATA
We collected the data from “kaggle datasets” named as “KC_Housesales_Data”. The link of the data: https://www.kaggle.com/swathiachath/kc-housesales-data
The data contains 21 different independent variables like bedrooms, sqft_living, view, grade, etc and the dependent variable is price. The data contains 21597 observations.
Online property companies offer valuations of houses using machine learning techniques. The aim of this report is to predict the house sales in King County, Washington State, USA using Multiple Linear Regression (MLR). The dataset consisted of historic data of houses sold between May 2014 to May 2015.
RESEARCH STATEMENT
“Predict the sales of houses in King County with an accuracy of at least 75-80%”
LOADING PACKAGES
library(tidyverse)library(corrplot)library(ggplot2)library(lubridate)library(gridExtra)library(caTools)library(GGally)READING THE DATA
mydata=read.csv("kc_house_data(1).csv")Viewing a Part of the Data
head(mydata,10)## id date price bedrooms bathrooms sqft_living sqft_lot
## 1 7129300520 10/13/2014 221900 3 1.00 1180 5650
## 2 6414100192 12-09-2014 538000 3 2.25 2570 7242
## 3 5631500400 2/25/2015 180000 2 1.00 770 10000
## 4 2487200875 12-09-2014 604000 4 3.00 1960 5000
## 5 1954400510 2/18/2015 510000 3 2.00 1680 8080
## 6 7237550310 05-12-2014 1230000 4 4.50 5420 101930
## 7 1321400060 6/27/2014 257500 3 2.25 1715 6819
## 8 2008000270 1/15/2015 291850 3 1.50 1060 9711
## 9 2414600126 4/15/2015 229500 3 1.00 1780 7470
## 10 3793500160 03-12-2015 323000 3 2.50 1890 6560
## floors waterfront view condition grade sqft_above sqft_basement
## 1 1 0 0 3 7 1180 0
## 2 2 0 0 3 7 2170 400
## 3 1 0 0 3 6 770 0
## 4 1 0 0 5 7 1050 910
## 5 1 0 0 3 8 1680 0
## 6 1 0 0 3 11 3890 1530
## 7 2 0 0 3 7 1715 0
## 8 1 0 0 3 7 1060 0
## 9 1 0 0 3 7 1050 730
## 10 2 0 0 3 7 1890 0
## yr_built yr_renovated zipcode lat long sqft_living15 sqft_lot15
## 1 1955 0 98178 47.5112 -122.257 1340 5650
## 2 1951 1991 98125 47.7210 -122.319 1690 7639
## 3 1933 0 98028 47.7379 -122.233 2720 8062
## 4 1965 0 98136 47.5208 -122.393 1360 5000
## 5 1987 0 98074 47.6168 -122.045 1800 7503
## 6 2001 0 98053 47.6561 -122.005 4760 101930
## 7 1995 0 98003 47.3097 -122.327 2238 6819
## 8 1963 0 98198 47.4095 -122.315 1650 9711
## 9 1960 0 98146 47.5123 -122.337 1780 8113
## 10 2003 0 98038 47.3684 -122.031 2390 7570We take a look at the structure and summary statistics of the data.
str(mydata)## 'data.frame': 21597 obs. of 21 variables:
## $ id : num 7.13e+09 6.41e+09 5.63e+09 2.49e+09 1.95e+09 ...
## $ date : Factor w/ 372 levels "01-02-2015","01-05-2015",..: 145 201 234 201 227 65 313 117 259 31 ...
## $ price : num 221900 538000 180000 604000 510000 ...
## $ bedrooms : int 3 3 2 4 3 4 3 3 3 3 ...
## $ bathrooms : num 1 2.25 1 3 2 4.5 2.25 1.5 1 2.5 ...
## $ sqft_living : int 1180 2570 770 1960 1680 5420 1715 1060 1780 1890 ...
## $ sqft_lot : int 5650 7242 10000 5000 8080 101930 6819 9711 7470 6560 ...
## $ floors : num 1 2 1 1 1 1 2 1 1 2 ...
## $ waterfront : int 0 0 0 0 0 0 0 0 0 0 ...
## $ view : int 0 0 0 0 0 0 0 0 0 0 ...
## $ condition : int 3 3 3 5 3 3 3 3 3 3 ...
## $ grade : int 7 7 6 7 8 11 7 7 7 7 ...
## $ sqft_above : int 1180 2170 770 1050 1680 3890 1715 1060 1050 1890 ...
## $ sqft_basement: int 0 400 0 910 0 1530 0 0 730 0 ...
## $ yr_built : int 1955 1951 1933 1965 1987 2001 1995 1963 1960 2003 ...
## $ yr_renovated : int 0 1991 0 0 0 0 0 0 0 0 ...
## $ zipcode : int 98178 98125 98028 98136 98074 98053 98003 98198 98146 98038 ...
## $ lat : num 47.5 47.7 47.7 47.5 47.6 ...
## $ long : num -122 -122 -122 -122 -122 ...
## $ sqft_living15: int 1340 1690 2720 1360 1800 4760 2238 1650 1780 2390 ...
## $ sqft_lot15 : int 5650 7639 8062 5000 7503 101930 6819 9711 8113 7570 ...summary(mydata)## id date price bedrooms
## Min. :1.000e+06 6/23/2014 : 142 Min. : 78000 Min. : 1.000
## 1st Qu.:2.123e+09 6/25/2014 : 131 1st Qu.: 322000 1st Qu.: 3.000
## Median :3.905e+09 6/26/2014 : 131 Median : 450000 Median : 3.000
## Mean :4.580e+09 07-08-2014: 127 Mean : 540297 Mean : 3.373
## 3rd Qu.:7.309e+09 4/27/2015 : 126 3rd Qu.: 645000 3rd Qu.: 4.000
## Max. :9.900e+09 3/25/2015 : 123 Max. :7700000 Max. :33.000
## (Other) :20817
## bathrooms sqft_living sqft_lot floors
## Min. :0.500 Min. : 370 Min. : 520 Min. :1.000
## 1st Qu.:1.750 1st Qu.: 1430 1st Qu.: 5040 1st Qu.:1.000
## Median :2.250 Median : 1910 Median : 7618 Median :1.500
## Mean :2.116 Mean : 2080 Mean : 15099 Mean :1.494
## 3rd Qu.:2.500 3rd Qu.: 2550 3rd Qu.: 10685 3rd Qu.:2.000
## Max. :8.000 Max. :13540 Max. :1651359 Max. :3.500
##
## waterfront view condition grade
## Min. :0.000000 Min. :0.0000 Min. :1.00 Min. : 3.000
## 1st Qu.:0.000000 1st Qu.:0.0000 1st Qu.:3.00 1st Qu.: 7.000
## Median :0.000000 Median :0.0000 Median :3.00 Median : 7.000
## Mean :0.007547 Mean :0.2343 Mean :3.41 Mean : 7.658
## 3rd Qu.:0.000000 3rd Qu.:0.0000 3rd Qu.:4.00 3rd Qu.: 8.000
## Max. :1.000000 Max. :4.0000 Max. :5.00 Max. :13.000
##
## sqft_above sqft_basement yr_built yr_renovated
## Min. : 370 Min. : 0.0 Min. :1900 Min. : 0.00
## 1st Qu.:1190 1st Qu.: 0.0 1st Qu.:1951 1st Qu.: 0.00
## Median :1560 Median : 0.0 Median :1975 Median : 0.00
## Mean :1789 Mean : 291.7 Mean :1971 Mean : 84.46
## 3rd Qu.:2210 3rd Qu.: 560.0 3rd Qu.:1997 3rd Qu.: 0.00
## Max. :9410 Max. :4820.0 Max. :2015 Max. :2015.00
##
## zipcode lat long sqft_living15
## Min. :98001 Min. :47.16 Min. :-122.5 Min. : 399
## 1st Qu.:98033 1st Qu.:47.47 1st Qu.:-122.3 1st Qu.:1490
## Median :98065 Median :47.57 Median :-122.2 Median :1840
## Mean :98078 Mean :47.56 Mean :-122.2 Mean :1987
## 3rd Qu.:98118 3rd Qu.:47.68 3rd Qu.:-122.1 3rd Qu.:2360
## Max. :98199 Max. :47.78 Max. :-121.3 Max. :6210
##
## sqft_lot15
## Min. : 651
## 1st Qu.: 5100
## Median : 7620
## Mean : 12758
## 3rd Qu.: 10083
## Max. :871200
## Now we check for the missing values in the data.
NA_values=data.frame(no_of_na_values=colSums(is.na(mydata)))
head(NA_values,21)## no_of_na_values
## id 0
## date 0
## price 0
## bedrooms 0
## bathrooms 0
## sqft_living 0
## sqft_lot 0
## floors 0
## waterfront 0
## view 0
## condition 0
## grade 0
## sqft_above 0
## sqft_basement 0
## yr_built 0
## yr_renovated 0
## zipcode 0
## lat 0
## long 0
## sqft_living15 0
## sqft_lot15 0We see that there are no missing values in this data.
DIVIDING THE DATA INTO TRAIN AND TEST DATA
set.seed(123) # set seed to ensure you always have same random numbers generated
sample = sample.split(mydata,SplitRatio = 0.8) # splits the data in the ratio mentioned in SplitRatio. After splitting marks these rows as logical TRUE and the the remaining are marked as logical FALSE
train_data =subset(mydata,sample ==TRUE) # creates a training dataset named train1 with rows which are marked as TRUE
test_data=subset(mydata, sample==FALSE)EXPLORATORY DATA ANALYSIS ON THE TRAIN DATA
1.Determining the association between variables.
We take out the correlation plot(corrplot) to understand the association of the dependent variable(price) with the independent variables.
cor_data=data.frame(train_data[,3:21])
correlation=cor(cor_data)
par(mfrow=c(1, 1))
corrplot(correlation,method="color")
According to our corrplot price is positively correlated with bedroom, bathroom, Sqft_living, view , grade, sqft_above, sqft_basement, lat, sqft_living 15.
Next we will draw some scatter plots to determine the relationship between these variables.
p1=ggplot(data = train_data, aes(x = bedrooms, y = price)) +
geom_jitter() + geom_smooth(method = "lm", se = FALSE)+labs(title="Scatter plot of Bedrooms and Price", x="bedrooms",y="Price")
p2=ggplot(data = train_data, aes(x = bathrooms, y = price)) +
geom_jitter() + geom_smooth(method = "lm", se = FALSE)+labs(title="Scatter plot of Bathrooms and Price", x="bathrooms",y="Price")
p3=ggplot(data = train_data, aes(x = sqft_living, y = price)) +
geom_jitter() + geom_smooth(method = "lm", se = FALSE)+labs(title="Scatter plot of Sqft_living and Price", x="Sqft_living",y="Price")
p4=ggplot(data = train_data, aes(x = sqft_above, y = price)) +
geom_jitter() + geom_smooth(method = "lm", se = FALSE)+labs(title="Scatter plot of Sqft_above and Price", x="Sqft_above",y="Price")
p5=ggplot(data = train_data, aes(x = sqft_basement, y = price)) +
geom_jitter() + geom_smooth(method = "lm", se = FALSE)+labs(title="Scatter plot of Sqft_basement and Price", x="Sqft_basement",y="Price")
p6=ggplot(data = train_data, aes(x = lat, y = price)) +
geom_jitter() + geom_smooth(method = "lm", se = FALSE)+labs(title="Scatter plot of Latitude and Price", x="Latitude",y="Price")
p7=ggplot(data = train_data, aes(x = sqft_living15, y = price)) +
geom_jitter() + geom_smooth(method = "lm", se = FALSE)+labs(title="Scatter plot of Sqft_living15 and Price", x="Sqft_living15",y="Price")
grid.arrange(p1,p2,p3,p4,p5,p6,p7,nrow=4)
from these scatter plots, we conclude that the relationship between price and bedroom, bathroom, Sqft_living,sqft_above, sqft_basement, lat, sqft_living 15 is linear.
For the two categorical variables(view and grade) we draw boxplots to understand the relationship.
par(mfrow=c(1, 2))
boxplot(price~view,data=train_data,main="Different boxplots", xlab="view",ylab="price",col="orange",border="brown")
boxplot(price~grade,data=train_data,main="Different boxplots", xlab="grade",ylab="price",col="orange",border="brown")
There is a relationship between price and categorical variables view and grade.
2.Modifying our train data.
Now we modify the train data a liitle and add two new columns for our better understanding.
I feel price might have a fair chance of depending on the age of the house and also the number of times it has been renovated.
So we try to extact the age and the number of times a particular house has been renovated from our train data.
date_sale=mdy(train_data$date)
train_data$sale_date_year=as.integer(year(date_sale))
train_data$age=train_data$sale_date_year-train_data$yr_built
train_data$reno=ifelse(train_data$yr_renovated==0,0,1)
train_data$reno=as.factor(train_data$reno)3. Performing ggpair plot.
ggpairs(train_data, columns= c("price","bedrooms","bathrooms","view","grade","sqft_living","sqft_above","sqft_basement","lat","sqft_living15"))
4.now we check for outliers in the dependent variable(price) using a boxplot.
ggplot(data=train_data)+geom_boxplot(aes(x=bedrooms,y=price))
we see that we have a significantly large number of outliers.
Treating or altering the outlier/extreme values in genuine observations is not a standard operating procedure. However, it is essential to understand their impact on our predictive models.
To better understand the implications of outliers better, I am going to compare the fit of a simple linear regression model on the dataset with and without outliers.
For this we first extract outliers from the data and then obtain the data without the outliers.
outliers=boxplot(train_data$price,plot=FALSE)$out
outliers_data=train_data[which(train_data$price %in% outliers),]
train_data1= train_data[-which(train_data$price %in% outliers),]we obtain 872 observations as outliers.
Now we plot the data with and without outliers.
par(mfrow=c(1, 2))
plot(train_data$bedrooms, train_data$price, main="With Outliers", xlab="bedrooms", ylab="price", pch="*", col="red", cex=2)
abline(lm(price ~ bedrooms, data=train_data), col="blue", lwd=3, lty=2)
# Plot of original data without outliers. Note the change of slope.
plot(train_data1$bedrooms, train_data1$price, main="Outliers removed", xlab="bedrooms", ylab="price", pch="*", col="red", cex=2)
abline(lm(price ~bedrooms, data=train_data1), col="blue", lwd=3, lty=2)
Notice the change in slope of the best fit line after removing the outliers. It is evident that if we remove the outliers to train the model, our predictions would be exagerated (high error) for larger values of price because of the larger slope.
So we continue with the entire data.
MODELING
1.Modeling on the entire train data.
Price,bedroom, bathroom, Sqft_living, view , grade, sqft_above, sqft_basement, lat, and sqft_living 15 were considered for the full model based on the corrplot.
A linear model was fit to determine the relationship. The results are shown below.
model=lm(data=train_data,price~bedrooms+bathrooms+sqft_living+view+grade+sqft_above+sqft_basement+sqft_living15)
summary(model)##
## Call:
## lm(formula = price ~ bedrooms + bathrooms + sqft_living + view +
## grade + sqft_above + sqft_basement + sqft_living15, data = train_data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1245264 -123294 -19165 96345 4659105
##
## Coefficients: (1 not defined because of singularities)
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -5.041e+05 1.650e+04 -30.549 < 2e-16 ***
## bedrooms -2.979e+04 2.476e+03 -12.031 < 2e-16 ***
## bathrooms -1.835e+04 3.799e+03 -4.830 1.38e-06 ***
## sqft_living 2.231e+02 5.315e+00 41.982 < 2e-16 ***
## view 8.896e+04 2.597e+03 34.255 < 2e-16 ***
## grade 1.012e+05 2.732e+03 37.057 < 2e-16 ***
## sqft_above -4.961e+01 4.961e+00 -9.999 < 2e-16 ***
## sqft_basement NA NA NA NA
## sqft_living15 6.026e+00 4.415e+00 1.365 0.172
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 235300 on 16447 degrees of freedom
## Multiple R-squared: 0.5828, Adjusted R-squared: 0.5826
## F-statistic: 3282 on 7 and 16447 DF, p-value: < 2.2e-16####We can see the relationship between these variables appear to be moderately strong as shown by R-Suared value and the probability.also coclude from the p-value that sqft_living15 is not a significant variable for the prediction of price. Hence we drop it. ####We also try fitting the model including a few other variables which we left out in the EDA and stop at a model which gives us the maximum R-squared value.
model5=lm(data=train_data,price~bedrooms+bathrooms+sqft_living+view+grade+sqft_lot+age+floors+waterfront)
summary(model5)##
## Call:
## lm(formula = price ~ bedrooms + bathrooms + sqft_living + view +
## grade + sqft_lot + age + floors + waterfront, data = train_data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1301702 -110143 -9314 90592 4279283
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -9.578e+05 1.761e+04 -54.401 < 2e-16 ***
## bedrooms -3.564e+04 2.284e+03 -15.607 < 2e-16 ***
## bathrooms 5.081e+04 3.886e+03 13.075 < 2e-16 ***
## sqft_living 1.627e+02 3.753e+00 43.358 < 2e-16 ***
## view 4.992e+04 2.561e+03 19.494 < 2e-16 ***
## grade 1.270e+05 2.442e+03 52.018 < 2e-16 ***
## sqft_lot -2.555e-01 4.076e-02 -6.267 3.76e-10 ***
## age 3.713e+03 7.310e+01 50.795 < 2e-16 ***
## floors 1.798e+04 3.883e+03 4.630 3.69e-06 ***
## waterfront 4.984e+05 2.130e+04 23.395 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 215600 on 16445 degrees of freedom
## Multiple R-squared: 0.6498, Adjusted R-squared: 0.6496
## F-statistic: 3390 on 9 and 16445 DF, p-value: < 2.2e-16As concluded from the adjusted R-squared value of 0.6496, the relationship beween these variables appear to be quite strong.
2.Now we detect the influential points of the data.
Declaring an observation as an outlier based on a just one (rather unimportant) feature could lead to unrealistic inferences. When you have to decide if an individual entity (represented by row or observation) is an extreme value or not, it is better to collectively consider the features (X’s) that matter. We fnd it out using the cook’s distance.
cooksd <- cooks.distance(model5)
mean(cooksd)## [1] 0.0002512902Now we plot the cook’s distance.
par(mfrow=c(1, 1))
plot(cooksd, main="Influential Obs by Cooks distance",xlim=c(0,25000),ylim=c(0,0.1))
axis(1, at=seq(0, 25000, 5000))
axis(2, at=seq(0, 0.1, 0.0001))
abline(h = 4*mean(cooksd, na.rm=T), col="green")
text(x=1:length(cooksd)+1,y=cooksd,labels=ifelse(cooksd>4*mean(cooksd,na.rm=T),names(cooksd),""), col="red") 
we find out the influential points in the data.
influential <- as.numeric(names(cooksd)[(cooksd > 4*mean(cooksd, na.rm=T))]) # influential row numbers
head(train_data[influential, ])## id date price bedrooms bathrooms sqft_living sqft_lot
## 28 3303700376 12-01-2014 667000 3 1.00 1400 1581
## 202 2222059065 11-12-2014 297000 3 2.50 1940 14952
## 303 2747100024 6/19/2014 576000 3 2.50 1940 9000
## 315 4139480200 12-09-2014 1400000 4 3.25 4290 12103
## 348 4048400070 12-05-2014 320000 2 1.00 1070 32633
## 354 3363900111 12-03-2014 437500 2 1.00 990 3120
## floors waterfront view condition grade sqft_above sqft_basement
## 28 1.5 0 0 5 8 1400 0
## 202 2.0 0 0 3 8 1940 0
## 303 1.0 0 0 4 7 970 970
## 315 1.0 0 3 3 11 2690 1600
## 348 1.0 0 0 4 6 1070 0
## 354 1.0 0 2 5 7 790 200
## yr_built yr_renovated zipcode lat long sqft_living15
## 28 1909 0 98112 47.6221 -122.314 1860
## 202 1994 0 98042 47.3777 -122.165 2030
## 303 1948 0 98117 47.6933 -122.393 2190
## 315 1997 0 98006 47.5503 -122.102 3860
## 348 1930 0 98059 47.4716 -122.078 1360
## 354 1907 0 98103 47.6800 -122.353 1930
## sqft_lot15 sale_date_year age reno
## 28 3861 2014 105 0
## 202 10450 2014 20 0
## 303 7310 2014 66 0
## 315 11244 2014 17 0
## 348 32156 2014 84 0
## 354 3120 2014 107 0influential_data=train_data[influential, ]Now we take out the influential outliers.
influencial_outliers=inner_join(outliers_data,influential_data)## Joining, by = c("id", "date", "price", "bedrooms", "bathrooms", "sqft_living", "sqft_lot", "floors", "waterfront", "view", "condition", "grade", "sqft_above", "sqft_basement", "yr_built", "yr_renovated", "zipcode", "lat", "long", "sqft_living15", "sqft_lot15", "sale_date_year", "age", "reno")We have 13 observations which are outliers yet influential hence we need to keep these outliers.
Now we modify the data excluding the outliers and including only the influential outliers.
train_data2=rbind(train_data1,influencial_outliers)3.modelling using the train data which includes influential_outliers
model6=lm(data=train_data2,price~bedrooms+bathrooms+sqft_living+view+grade+sqft_lot+age+floors+waterfront)
summary(model6)##
## Call:
## lm(formula = price ~ bedrooms + bathrooms + sqft_living + view +
## grade + sqft_lot + age + floors + waterfront, data = train_data2)
##
## Residuals:
## Min 1Q Median 3Q Max
## -717735 -94075 -7246 82346 2479259
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -6.592e+05 1.239e+04 -53.220 < 2e-16 ***
## bedrooms -1.207e+04 1.565e+03 -7.713 1.3e-14 ***
## bathrooms 3.039e+04 2.668e+03 11.390 < 2e-16 ***
## sqft_living 8.559e+01 2.718e+00 31.496 < 2e-16 ***
## view 2.708e+04 1.916e+03 14.137 < 2e-16 ***
## grade 1.025e+05 1.704e+03 60.151 < 2e-16 ***
## sqft_lot -1.051e-02 2.801e-02 -0.375 0.707512
## age 2.739e+03 5.080e+01 53.925 < 2e-16 ***
## floors 3.384e+04 2.636e+03 12.837 < 2e-16 ***
## waterfront 7.744e+04 2.084e+04 3.716 0.000203 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 140800 on 15586 degrees of freedom
## Multiple R-squared: 0.5624, Adjusted R-squared: 0.5621
## F-statistic: 2225 on 9 and 15586 DF, p-value: < 2.2e-16We can see the relationship beween these variables appear to be moderately strong as shown by R-Suared value and the probability.also coclude from the p-value that sqft_lot is not a significant variable for the prediction of price. Hence we drop it.
We also try fitting the model including a few other variables which we left out in while fitting the model when we took the entire data and stop at a model which gives us the maximum R-squared value.
model12=lm(data=train_data2,price~bedrooms+bathrooms+sqft_living+view+grade+age+waterfront+long+lat+zipcode+condition+sqft_above+sqft_living15+reno)
summary(model12)##
## Call:
## lm(formula = price ~ bedrooms + bathrooms + sqft_living + view +
## grade + age + waterfront + long + lat + zipcode + condition +
## sqft_above + sqft_living15 + reno, data = train_data2)
##
## Residuals:
## Min 1Q Median 3Q Max
## -540111 -75304 -7817 63497 2463513
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -1.330e+07 1.983e+06 -6.708 2.04e-11 ***
## bedrooms -1.004e+04 1.328e+03 -7.561 4.24e-14 ***
## bathrooms 3.089e+04 2.245e+03 13.758 < 2e-16 ***
## sqft_living 6.137e+01 3.134e+00 19.583 < 2e-16 ***
## view 3.259e+04 1.667e+03 19.557 < 2e-16 ***
## grade 7.661e+04 1.553e+03 49.337 < 2e-16 ***
## age 1.652e+03 5.076e+01 32.555 < 2e-16 ***
## waterfront 1.351e+05 1.773e+04 7.619 2.71e-14 ***
## long -5.634e+04 8.913e+03 -6.321 2.66e-10 ***
## lat 5.563e+05 7.320e+03 75.997 < 2e-16 ***
## zipcode -2.097e+02 2.310e+01 -9.081 < 2e-16 ***
## condition 2.562e+04 1.634e+03 15.684 < 2e-16 ***
## sqft_above 1.810e+01 2.880e+00 6.285 3.36e-10 ***
## sqft_living15 4.347e+01 2.589e+00 16.788 < 2e-16 ***
## reno1 2.812e+04 5.449e+03 5.161 2.49e-07 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 119400 on 15581 degrees of freedom
## Multiple R-squared: 0.6856, Adjusted R-squared: 0.6853
## F-statistic: 2427 on 14 and 15581 DF, p-value: < 2.2e-16As concluded from the adjusted R-squared value of 0.6853, the relationship beween these variables appear to be quite strong.
ACCURACY OF THE MODEL ON THE TRAIN DATA
pred=model12$fitted.values
tally_table=data.frame(actual=train_data2$price, predicted=pred)
mape=mean(abs(tally_table$actual-tally_table$predicted)/tally_table$actual)
accuracy=1-mape
accuracy ## [1] 0.7946321cat("the accuracy is:", accuracy)## the accuracy is: 0.7946321We see that the accuracy is 79.46%.
PREDICTION ON THE TEST DATA
date_sale1=mdy(test_data$date)
test_data$sale_date_year=as.integer(year(date_sale1))
test_data$age=test_data$sale_date_year-test_data$yr_built
test_data$reno=ifelse(test_data$yr_renovated==0,0,1)
test_data$reno=as.factor(test_data$reno)
test_data_1=test_data[,c(4,5,6,10,9,12,23,24,17,18,19,11,13,20)]
pred_test=predict(newdata=test_data_1,model12)ACCURACY OF THE MODEL ON THE TEST DATA
tally_table_1=data.frame(actual=test_data$price, predicted=pred_test)
mape_test=mean(abs(tally_table_1$actual-tally_table_1$predicted)/tally_table_1$actual)
accuracy_test=1-mape_test
accuracy_test## [1] 0.789063