HousePrices Exploratory Analysis

Filip Dragicevic

2022-03-16

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> setwd("C:")
> houses <- read.csv("C:/Users/filip/OneDrive/Desktop/MSCI 3230/HousePrices.csv")

> houses <- 
+   read.table("C:/Users/filip/OneDrive/Desktop/MSCI 3230/HousePrices.csv", 
+   header=TRUE, stringsAsFactors=TRUE, sep=",", na.strings="NA", dec=".", 
+   strip.white=TRUE)

> library(abind, pos=17)

> with(houses, Hist(Price, scale="frequency", breaks="Sturges", 
+   col="darkgray"))

> Boxplot( ~ Price, data=houses, id=list(method="y"))

[1] "104"

From the prices histogram and boxplot, we can see that price looks to be relatively normally distributed (with a possibility of being right-skewed.)

We can further confirm this later using methods like shapiro-wilk normality test

There is also only one major outlier in house prices, shown by the 104.

> normalityTest(~Price, test="shapiro.test", data=houses)


    Shapiro-Wilk normality test

data:  Price
W = 0.98023, p-value = 0.05836

From the shapiro-wilk analysis, we can see that the p-value is greater than 0.05.

This tells us that we fail to reject the null hypothesis, which confirms that it is normally distributed and can be used for linear regression

> with(houses, Hist(Price, groups=Neighborhood, scale="frequency", 
+   breaks="Sturges", col="darkgray"))

## By using neighbourhood as a variable to group by, we are able to see the price distribution for each neighbourhood ## All are normally distributed, however West seems to be slightly right skewed

> with(houses, Hist(Price, groups=Brick, scale="frequency", breaks="Sturges", 
+   col="darkgray"))

## The same can be said when using brick as a grouping variable. Houses with and without brick are normally distributed

> local({
+   .Table <- xtabs(~Neighborhood+Brick, data=houses)
+   cat("\nFrequency table:\n")
+   print(.Table)
+   .Test <- chisq.test(.Table, correct=FALSE)
+   print(.Test)
+ })


Frequency table:
            Brick
Neighborhood No Yes
       East  26  19
       North 37   7
       West  23  16

    Pearson's Chi-squared test

data:  .Table
X-squared = 8.7033, df = 2, p-value = 0.01289

We can also create a two-way contingency table to observe the number of observations regarding ‘neighbourhood’ and ‘brick’

This checks whether these variables are independent or not, and with a p-value less than 0.05, we reject the null hypothesis and confirm that there is no relationship

> scatterplot(Price~Bathrooms, regLine=FALSE, smooth=FALSE, boxplots=FALSE, 
+   data=houses)

## Using scatterplots to compare price with another variable can be a good visualization of how price is impacted ## Houses with 2 bathrooms are normally distributed with a mean price lower than houses with 3. There is only one instance of a house with 4 bathrooms.

> scatterplot(Price~Bedrooms, regLine=FALSE, smooth=FALSE, boxplots=FALSE, 
+   data=houses)

## The same can be said for bathrooms, as number of bathrooms increases so does price

> scatterplot(Price~SqFt, regLine=FALSE, smooth=FALSE, boxplots=FALSE, 
+   data=houses)

## This is also a trend for square footage, as it has a line of best fit with a positive slope.

> scatterplot(Price~Offers, regLine=FALSE, smooth=FALSE, boxplots=FALSE, 
+   data=houses)

## By comparing offers and price, we can see that the more offers, the lower the price

> cor(houses[,c("Bathrooms","Bedrooms","Price","SqFt")], use="complete")

          Bathrooms  Bedrooms     Price      SqFt
Bathrooms 1.0000000 0.4145560 0.5232578 0.5227453
Bedrooms  0.4145560 1.0000000 0.5259261 0.4838071
Price     0.5232578 0.5259261 1.0000000 0.5529822
SqFt      0.5227453 0.4838071 0.5529822 1.0000000