Background
This is a real dataset of house prices sold in Seattle, WA, USA between August and December 2022.
Data Description
The dataset consists of a dataframe with 1665 observations with the following 6 variables:
1.beds = Number of bedrooms in property 2. region = based on zip code, north, central or south Seattle. 3. baths = Number of bathrooms in property. Note 0.5 corresponds to a half-bath which has a sink and toilet but no tub or shower 4. size = Total floor area of property in square feet 5. lot_size = Total area of the land where the property is located on in square feet 6. price = Price the property was sold for (US dollars) ##THIS IS THE RESPONSE VARIABLE
Source: Kaggle
Load the data
No changes needed
# reading the dataset in to RStudio
# Call the data table price_data
#https://drive.google.com/file/d/1LGxmER8s0PiGtrxoJA0fLCGU5IvCB4mh/view?usp=sharing
# Set up the actual file ID from the share link
file_id <- "1LGxmER8s0PiGtrxoJA0fLCGU5IvCB4mh"
data_url <- paste0("https://docs.google.com/uc?export=download&id=", file_id)
#Read the data into R
price_data <- read.csv(data_url, header = TRUE)
# Show the first few rows of data
head(price_data)## beds Region baths size lot_size price
## 1 1 Central 1 704 500 645000
## 2 3 Central 3 1524 513 850000
## 3 3 Central 3 1524 513 875000
## 4 3 North 3 1450 525 750000
## 5 1 Central 1 480 560 275000
## 6 2 North 1 800 560 690000
# Exploratory Data Analysis
``` r
# Compute univariate numeric summary statistics
summary(price_data)## beds Region baths size
## Min. : 1.000 Length:1665 Min. :0.500 Min. : 376
## 1st Qu.: 2.000 Class :character 1st Qu.:1.500 1st Qu.: 1260
## Median : 3.000 Mode :character Median :2.000 Median : 1720
## Mean : 3.126 Mean :2.298 Mean : 1896
## 3rd Qu.: 4.000 3rd Qu.:3.000 3rd Qu.: 2360
## Max. :15.000 Max. :9.000 Max. :11010
## lot_size price
## Min. : 500 Min. : 159488
## 1st Qu.: 2734 1st Qu.: 680000
## Median : 5000 Median : 865000
## Mean : 9673 Mean :1010483
## 3rd Qu.: 7350 3rd Qu.:1175000
## Max. :400752 Max. :6250000# Univariate Charts--histograms for the quantitative variables
# use hist(price_data$VARNAME)
par(mfrow = c(3, 2))
hist(price_data$beds)
hist(price_data$baths)
hist(price_data$size)
hist(price_data$lot_size)
hist(price_data$price)
Compute the correlation coefficient for each quantitative predicting variables against the response
#Create the correlation matrix for the quantitative variables CHANGE THE DATA TABLE NAME
round(cor(price_data [,-2]),2)## beds baths size lot_size price
## beds 1.00 0.59 0.73 -0.13 0.46
## baths 0.59 1.00 0.62 -0.08 0.54
## size 0.73 0.62 1.00 -0.07 0.74
## lot_size -0.13 -0.08 -0.07 1.00 -0.09
## price 0.46 0.54 0.74 -0.09 1.00Describe the strength and direction of the top 3 predicting variables that have the strongest linear relationships with the response.
Answer: The three predictors that have the strongest linear relationships with price are size, baths, and beds.
Create a boxplot of the qualitative predicting variable versus the response variable.
# Box plot CHANGE THE VARIABLES AND DATA TABLE
boxplot(price_data$price ~ price_data$Region)
Explain the relationship between the qualitative predicting variable versus the response variable.
Answer: The boxplot of price by region shows that average house prices are not the same across the three regions.
Create boxplots of the qualitative predicting variable versus the quantitative PREDICTOR variables
# Box plots by vehicle type CHANGE THE VARIABLES AND DATA TABLE
par(mfrow = c(2, 2))
boxplot(price_data$beds ~ price_data$Region)
boxplot(price_data$baths ~ price_data$Region)
boxplot(price_data$size ~ price_data$Region)
boxplot(price_data$lot_size ~ price_data$Region)
Explain the relationship between the variables.
Answer: The boxplots of beds, baths, size, and lot_size by region show that the characteristics of homes differ by region.
Create scatterplots of the response variable against each quantitative predicting variable Describe the general trend of each plot.
## Explore the relationships between all predictor variables and the response
## CHANGE THE VARIABLES AND DATA TABLE AND THE XLAB AND YLAB
par(mfrow = c(2, 3))
plot(price_data$beds,price_data$price,xlab = "beds", ylab = "price")
plot(price_data$baths,price_data$price,xlab = "baths", ylab = "price")
plot(price_data$size,price_data$price,xlab = "size", ylab = "price")
plot(price_data$lot_size,price_data$price,xlab = "lot_size", ylab = "price")
plot(price_data$price,price_data$price,xlab = "price", ylab = "price")
Describe the general trend of each plot.
Answer: For price vs beds, there is a upward trend. For price vs baths there is also an upward trend. For price vs size, there is a clear positive relationship. For price vs lot_size, there is a positive trend.
Regression Model Fitting and Interpretation
Fit a multiple linear regression model called model1 using with the response and the top 2 predicting variables with the strongest relationship with the response variable
# Fit a linear regression model CHANGE THE VARIABLES AND DATA TABLE
model1 <- lm(price ~ size + baths, data = price_data )
summary(model1)##
## Call:
## lm(formula = price ~ size + baths, data = price_data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1512769 -184967 -14583 139168 4682870
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 54645.49 24309.14 2.248 0.0247 *
## size 412.82 12.88 32.043 < 2e-16 ***
## baths 75416.75 11711.01 6.440 1.56e-10 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 377700 on 1662 degrees of freedom
## Multiple R-squared: 0.5643, Adjusted R-squared: 0.5637
## F-statistic: 1076 on 2 and 1662 DF, p-value: < 2.2e-16Using alpha = 0.05, which of the estimated coefficients are statistically significant?
Answer: In the model, both the size coefficient and the baths coefficient have p-values below 0.05, so they are statistically significant predictors of price.
Is the overall regression significant? And why or why not?
Answer: Yes, the overall regression is significant.
What is the estimated coefficient for the intercept? Interpret this coefficient in the context of the dataset.
Answer: The estimated intercept is the value of the price when both size = 0 and baths = 0. In context, this does not have a realistic meaning because a house cannot have 0 square feet and 0 baths.
Answer:
Goodness of Fit
# Extract the standardized residuals
#NO CHANGES NEEDED
resids = rstandard(model1)
fits = model1$fitted
# Constant Variance Assumption
plot(fits, resids,
xlab="Fitted Values",
ylab="Residuals",
main="")
abline(0, 0, lty=2, lwd=2)
Answer: Looking at the plot of standardized residuals versus fitted values for Model 1, the residuals are roughly spread around zero without a strong pattern. There may be some slight widening of the spread at higher fitted values, but there is no extreme shape.
Fit a linear regression model called model2 with response variable and the qualitataive variable as the predicting variable
# CHANGE THE VARIABLES AND DATA TABLE
model2 <- lm (price ~ Region, price_data )
summary(model2)##
## Call:
## lm(formula = price ~ Region, data = price_data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -976039 -305368 -122368 149942 5023961
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1226039 26408 46.426 < 2e-16 ***
## RegionNorth -193671 33439 -5.792 8.31e-09 ***
## RegionSouth -425981 35779 -11.906 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 548900 on 1662 degrees of freedom
## Multiple R-squared: 0.07961, Adjusted R-squared: 0.0785
## F-statistic: 71.87 on 2 and 1662 DF, p-value: < 2.2e-16What is the model for each of the values of the qualitataive variable?
Answer: Each region’s mean price is the intercept plus the corresponding region coefficient
Goodness of Fit
# Extract the standardized residuals
#NO CHANGES NEEDED
resids2 = rstandard(model2)
fits2 = model2$fitted
# Constant Variance Assumption
plot(fits2, resids2,
xlab="Fitted Values",
ylab="Residuals",
main="")
abline(0, 0, lty=2, lwd=2)
Does the fitted values vs. standardized residuals plot indicate constant variance?
Answer: The residual vs fitted plot shows the residuals scattered around zero with no strong increasing or decreasing spread as fitted values change.
Fit a multiple linear regression model called model3 using the response variable and your choice of predictor variables.
# CHANGE THE VARIABLES AND DATA TABLE
model3 <- lm (price ~ Region + size + baths, data = price_data)
summary(model3)##
## Call:
## lm(formula = price ~ Region + size + baths, data = price_data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1478315 -174273 -25602 134080 4555225
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 237727.73 28270.85 8.409 < 2e-16 ***
## RegionNorth -127937.34 21863.00 -5.852 5.85e-09 ***
## RegionSouth -319851.40 23423.98 -13.655 < 2e-16 ***
## size 405.70 12.24 33.140 < 2e-16 ***
## baths 68780.22 11126.49 6.182 7.97e-10 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 357700 on 1660 degrees of freedom
## Multiple R-squared: 0.6097, Adjusted R-squared: 0.6087
## F-statistic: 648.2 on 4 and 1660 DF, p-value: < 2.2e-16Using alpha = 0.05, which of the estimated coefficients arestatistically significant in model2?
Answer: The coefficients that are statistically significant are those with p-values < 0.05 which include size and baths.
Interpret the estimated coefficient for TypeSUV in the context of the dataset.
Answer: The coefficient for north region represents the difference in average price between houses in the central region.
Is the overall regression (model3) significant at an alpha-level of 0.05? Explain how you determined the answer.
Answer: Yes, Model 3 is significant at an alpha level of 0.05.
Goodness of Fit
# Extract the standardized residuals
#NO CHANGES NEEDED
resids3 = rstandard(model3)
fits3 = model3$fitted
# Constant Variance Assumption
plot(fits3, resids3,
xlab="Fitted Values",
ylab="Residuals",
main="")
abline(0, 0, lty=2, lwd=2)
Does the fitted values vs. standardized residuals plot indicate constant variance?
Answer: The residual vs fitted plot for Model 3 shows the residuals scattered around zero with no strong pattern. There might be some variation in the spread, but there is no clear shape so the constant variance assumption appears to be satisfied for Model 3
What are the Adjusted R^2 values for model 1 and model 3?
# Return the R^2 values for model 1 and model 3
#NO CHANGES NEEDED
paste("Model 1 adjusted R^2:", round(summary(model1)$adj.r.squared,2))## [1] "Model 1 adjusted R^2: 0.56"paste("Model 3 adjusted R^2:",round(summary(model3)$adj.r.squared,2))## [1] "Model 3 adjusted R^2: 0.61"How would you describe the performance of model 1 and model 3 based on the R^2 value.
Prediction
Using model3, predict the mean response for 2 sets of characteristics:
CHANGE THE VARIABLES AND THE VALUES TO MATCH YOUR DATA SET
Type=“Sedan”, MPG=25, Weight=3000, Horsepower=250
Type=“SUV”, MPG=25, Weight=3000, Horsepower=250,
#new data
# CHANGE THE VARIABLES AND DATA TABLE AND VALUES
newvals1 <- data.frame(Region="North", size=704, baths=2)
newvals2 <- data.frame(Region="Central", size=1010, baths=3)
# Confidence Interval for the response variable
predict(model3,newvals1,interval='confidence',level=.95)## fit lwr upr
## 1 532963.6 497329.5 568597.8predict(model3,newvals2,interval='confidence',level=.95)## fit lwr upr
## 1 853825.4 805391 902259.9
Comment on whether the plot shows constant variance.