take-home_exam
2025-09-20
Matej Metelko
Task 1
mydata_cars <- read.table("./car_sales_data.csv",
header=TRUE,
sep=",",
dec=";")
head(mydata_cars, 10) #Showing first 10 units## Manufacturer Model Engine.size Fuel.type Year.of.manufacture Mileage
## 1 Ford Fiesta 1.0 Petrol 2002 127300
## 2 Porsche 718 Cayman 4.0 Petrol 2016 57850
## 3 Ford Mondeo 1.6 Diesel 2014 39190
## 4 Toyota RAV4 1.8 Hybrid 1988 210814
## 5 VW Polo 1.0 Petrol 2006 127869
## 6 Ford Focus 1.4 Petrol 2018 33603
## 7 Ford Mondeo 1.8 Diesel 2010 86686
## 8 Toyota Prius 1.4 Hybrid 2015 30663
## 9 VW Polo 1.2 Petrol 2012 73470
## 10 Ford Focus 2.0 Diesel 1992 262514
## Price
## 1 3074
## 2 49704
## 3 24072
## 4 1705
## 5 4101
## 6 29204
## 7 14350
## 8 30297
## 9 9977
## 10 1049Description:
- Manufacturer: Name of the brand / manufacturer of a car
- Model: Name of the car model
- Engine.size: Engine size
- Fuel.type: Type of fuel that a car uses
- Year.of.manufacture: Year when a car was produced (in years)
- Mileage: Number of miles, that the car has driven since produced (in miles)
- Price: Price for which a car sold (in USD$)
#Because i am only interested in petrol and diesl fueld car, I want to remove all units that have different Fuel.type, and create new dataset
mydata_cars2 <- mydata_cars[mydata_cars$Fuel.type !="Hybrid",]
mydata_cars2$Fuel.type <- ifelse(mydata_cars2$Fuel.type == "Diesel", 1, 0) #Now fuel is a dummy variable 1= Diesel and 0= Petrol
head(mydata_cars2, 10) #Showing first 10 units## Manufacturer Model Engine.size Fuel.type Year.of.manufacture Mileage
## 1 Ford Fiesta 1.0 0 2002 127300
## 2 Porsche 718 Cayman 4.0 0 2016 57850
## 3 Ford Mondeo 1.6 1 2014 39190
## 5 VW Polo 1.0 0 2006 127869
## 6 Ford Focus 1.4 0 2018 33603
## 7 Ford Mondeo 1.8 1 2010 86686
## 9 VW Polo 1.2 0 2012 73470
## 10 Ford Focus 2.0 1 1992 262514
## 11 VW Golf 2.0 1 2014 83047
## 12 BMW Z4 2.0 0 1990 293666
## Price
## 1 3074
## 2 49704
## 3 24072
## 5 4101
## 6 29204
## 7 14350
## 9 9977
## 10 1049
## 11 17173
## 12 719#I would like to find the most popular/common manufacturer
tab <- table(mydata_cars2$Manufacturer)
most_common <- names(which.max(tab))
most_common## [1] "VW"# I am only interested in most common brand which is VW, now i will randomly select VW model, to analyse further
library(dplyr)##
## Attaching package: 'dplyr'## The following objects are masked from 'package:pastecs':
##
## first, last## The following object is masked from 'package:car':
##
## recode## The following objects are masked from 'package:stats':
##
## filter, lag## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, unionmydata_cars_VW <- mydata_cars2 %>%
filter(Manufacturer == "VW")
mydata_cars_VW$Manufacturer <- NULL
head(mydata_cars_VW, 5) #Showing first 5 units## Model Engine.size Fuel.type Year.of.manufacture Mileage Price
## 1 Polo 1.0 0 2006 127869 4101
## 2 Polo 1.2 0 2012 73470 9977
## 3 Golf 2.0 1 2014 83047 17173
## 4 Golf 1.2 1 2007 92697 7792
## 5 Golf 1.6 1 1989 222390 933sum(mydata_cars$Model == "Golf")## [1] 5050Within the VW brand, the Golf model was randomly selected by computer for analysis. Focusing on a single model allows me to examine relationships between price, mileage, year of manufacture, engine size and fuel type without variation caused by differences between models, ensuring clearer and more interpretable results.
mydata_golf <- mydata_cars_VW[mydata_cars_VW$Model == "Golf",]
head(mydata_golf, 5) #Showing first 5 units## Model Engine.size Fuel.type Year.of.manufacture Mileage Price
## 3 Golf 2.0 1 2014 83047 17173
## 4 Golf 1.2 1 2007 92697 7792
## 5 Golf 1.6 1 1989 222390 933
## 7 Golf 2.0 0 2020 18985 36387
## 20 Golf 1.2 0 2002 139884 3713mydata_golf$Engine.size <- factor(mydata_golf$Engine.size)
levels(mydata_golf$Engine.size)## [1] "1.2" "1.4" "1.6" "1.8" "2.0"round(stat.desc(mydata_golf[ , -c(1,2,3)]), 2)## Year.of.manufacture Mileage Price
## nbr.val 4537.00 4.537000e+03 4537.00
## nbr.null 0.00 0.000000e+00 0.00
## nbr.na 0.00 0.000000e+00 0.00
## min 1984.00 6.300000e+02 135.00
## max 2022.00 4.219790e+05 46830.00
## range 38.00 4.213490e+05 46695.00
## sum 9093222.00 5.092727e+08 45983700.00
## median 2004.00 1.021160e+05 6472.00
## mean 2004.24 1.122488e+05 10135.27
## SE.mean 0.14 1.054210e+03 142.41
## CI.mean.0.95 0.28 2.066770e+03 279.20
## var 92.47 5.042236e+09 92016939.14
## std.dev 9.62 7.100871e+04 9592.55
## coef.var 0.00 6.300000e-01 0.95table(mydata_golf$Fuel.type)##
## 0 1
## 3002 1535table(mydata_golf$Engine.size)##
## 1.2 1.4 1.6 1.8 2.0
## 912 952 885 917 871Description:
- Year of manufacture: From descriptive statistics we can see, that the oldest car taht is in our sample of Golfs was made in 1984, and the newest one was made in 2022. Thats why range(=range is a value that we get when we deduct the minimum value from the maximum value of our sample for a certain variable) is 38 years. On avrage the car is from 2004, showing a roughly symmetric distribution.
- Mileage: Car with lowest milage was sold at only 630 miles, while the most was at a 421,979 miles. Which gives us a large range. From the values of median (= median is a value, where 50% of units have lower or equal value to median, and 50% have a higher value than median) and mean we can assume that the graph will be skewed to the right. Due to sold cars being used, high mileage is very common.
- Price: Here we can also see a high range, because the cheapest car sold for only 135$ and the most expensive for 46,830$. With mean (=mean is the sum of all values devided by the number of values) > median, we can again assume that the graph will be skewed to the right.
- Fuel Type: I have also included frequency of two different fuel types, where we can see there was mor petrol cars sold.
- Engine size: I have also included frequency of two different fuel types, where we can see that most sold Golfs had 1.4 engine and the ferwest had 1.6 engine.
ggplot(mydata_golf, aes(x = Price)) +
geom_histogram(binwidth = 1000, fill = "skyblue", color = "black") +
labs(title = "Distribution of used VW Golf Prices", x = "Price (€)", y = "Count")+
theme_minimal()
ggplot(mydata_golf, aes(x = Mileage)) +
geom_histogram(binwidth = 10000, fill = "lightgreen", color = "black") +
labs(title = "Distribution of used VW Golf Mileage", x = "Mileage (km)", y = "Count")+
theme_minimal()
ggplot(mydata_golf, aes(x = Year.of.manufacture)) +
geom_histogram(binwidth = 1, fill = "lightcoral", color = "black") +
labs(title = "Distribution of Year of Manufacture", x = "Year", y = "Count")+
theme_minimal()
ggplot(mydata_golf, aes(x = factor(Fuel.type))) +
geom_bar(fill = "lightblue") +
scale_x_discrete(labels = c("Petrol", "Diesel")) +
labs(title = "Distribution of Fuel Type", x = "Fuel Type", y = "Count")+
theme_minimal()
ggplot(mydata_golf, aes(x = Engine.size)) +
geom_bar(fill = "pink") +
labs(title = "Distribution of Engine Size", x = "Engine Size (L)", y = "Count")+
theme_minimal()
Explination:
As assumed both price and mileage distribution is skiewed to the right.Where we can still see long tails due to few cars being sold at very high mileage and high prices for a used car. While in contrastz year of make is roughly normally distributed.
While on last two graphs we can see the distribution of frequency for Engine size and Fuel type. We can see, that there was more petrol cars sold, while engine size is quite evenly distributed.
ggplot(mydata_golf, aes(x = Engine.size, y = Price)) +
geom_boxplot(fill = "pink") +
labs(title = "Price by Engine Size", x = "Engine Size (L)", y = "Price (€)")+
theme_minimal()
ggplot(mydata_golf, aes(x = factor(Fuel.type), y = Price)) +
geom_boxplot(fill = "lightblue") +
scale_x_discrete(labels = c("Petrol", "Diesel")) +
labs(title = "Price by Fuel Type", x = "Fuel Type", y = "Price (€)")+
theme_minimal()
library(car)
scatterplotMatrix(mydata_golf[ , c(4, 5, 6)], #Scatterplot matrix
smooth = FALSE,
regLine = list(col = "red"))
Explination:
Scatterplot matrix (year, mileage and price):
- we can see that, taht negative correlation between year of manufacture and mileage, which means the older the car more mileage it has
- Price has a positive correlation wiht year of manufacture, meaning newer the car, higher the price
- While the price and mileage have negative correlation, meaning if mileage increases, the price decreases. Higher mileage means lower market price for a car
Boxplot: Price by engine size:
- We can see that the median price increases with larger engines, however the engine size has a small effect on price of used Golfs, according to Boxplot
- We can assume that more expensive models have bigger engines, which we can see from number of outliers and their spread
Boxplot: Price by fuel type:
- We can say fuel type does not strongly affect the median price of used Golfs
lm(Price ~ Year.of.manufacture + Mileage + Fuel.type + Engine.size, data = mydata_golf)##
## Call:
## lm(formula = Price ~ Year.of.manufacture + Mileage + Fuel.type +
## Engine.size, data = mydata_golf)
##
## Coefficients:
## (Intercept) Year.of.manufacture Mileage
## -1.428e+06 7.186e+02 -2.918e-02
## Fuel.type Engine.size1.4 Engine.size1.6
## 3.010e+02 6.929e+02 1.432e+03
## Engine.size1.8 Engine.size2.0
## 1.982e+03 2.722e+03Explination of fit of multiple regression:
- On avrage for each additional year of use, the car depreciates by 718.6$, assuming every thing else stays the same
- On avrage each additional mile decreases the price by about 0.03$, assuming every thing else stays the same
- Fuel type is set as Dummy variable. We can see that Diesel (coded = 1) on avrage increases the price by 301$, assuming every thing else stays the same
- From last variable we can see that engine size increases the price as the engine size increases, assuming every thing else stays the same
Task 2
library(readxl)
mydata_MBA <- read_excel("D:/IMB/R/R Take Home Exam 2025/Business School.xlsx")
head(mydata_MBA, 10) #Showing first 10 units## # A tibble: 10 × 9
## `Student ID` `Undergrad Degree` `Undergrad Grade` `MBA Grade`
## <dbl> <chr> <dbl> <dbl>
## 1 1 Business 68.4 90.2
## 2 2 Computer Science 70.2 68.7
## 3 3 Finance 76.4 83.3
## 4 4 Business 82.6 88.7
## 5 5 Finance 76.9 75.4
## 6 6 Computer Science 83.3 82.1
## 7 7 Engineering 76 66.9
## 8 8 Engineering 82.8 76.8
## 9 9 Business 76 72.3
## 10 10 Finance 76.9 72.4
## # ℹ 5 more variables: `Work Experience` <chr>, `Employability (Before)` <dbl>,
## # `Employability (After)` <dbl>, Status <chr>, `Annual Salary` <dbl>ggplot(mydata_MBA, aes(x = `Undergrad Degree`))+
geom_bar(fill ="lightblue")+
labs(title = "Distribution of undergraduate degrees",
x = "Undergraduate degree",
y = "Count")+
theme_minimal()
Explination:
We can assume from the graph, that the Business degree is the most common Undergraduate degree.
round(stat.desc(mydata_MBA$`Annual Salary`))## nbr.val nbr.null nbr.na min max range
## 100 0 0 20000 340000 320000
## sum median mean SE.mean CI.mean.0.95 var
## 10905800 103500 109058 4150 8235 1722373475
## std.dev coef.var
## 41501 0Explination:
- we can see that the sample contains 100 units, from which for variable Annual Salary have value 0 or are missing value.
- We can also see that the minimum Annual Salary is 20,000 and maximum is 340,000, which gives us a big range of 340,000.
- Total value of variable Annual Salary for our sample is 10,905,800
- Median tells us, that 50% of units have lower or equal Annual Salary to 103,500 and the other 50% of units have a salary higher than 103,500
- Mean tells us the avarage Annual Salary for our sample, which is equal to 109,058
- SE.
ggplot(mydata_MBA, aes(x = `Annual Salary`)) +
geom_histogram(binwidth = 5000, colour = "black", fill = "lightblue") +
labs(
title = "Annual Salary distribution",
x = "Annual Salary",
y = "Frequency")+
theme_minimal()
#for easier description of distribution
summary(mydata_MBA$`Annual Salary`)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 20000 87125 103500 109058 124000 340000Description:
From the histogram above we can see that the graph is skiewed to the right. There is few units that have Annual salary above 200,000, and they strech the range to the right, these units are outsliers. Mean is higher than median, which causes the skiew to the right. With the help of function summary we can also say that 75% of people have a salary equal or less to 124,000, while the maximum Annual salary is 340,000.
Two sided t-test for hypotesis:
𝐻0:𝜇MBA Grade = 74
t.test(mydata_MBA$`MBA Grade`, mu = 74)##
## One Sample t-test
##
## data: mydata_MBA$`MBA Grade`
## t = 2.6587, df = 99, p-value = 0.00915
## alternative hypothesis: true mean is not equal to 74
## 95 percent confidence interval:
## 74.51764 77.56346
## sample estimates:
## mean of x
## 76.04055#install.packages("effectsize")
library(effectsize)
cohens_d(mydata_MBA$`MBA Grade`, mu = 74)## Cohen's d | 95% CI
## ------------------------
## 0.27 | [0.07, 0.46]
##
## - Deviation from a difference of 74.mean(mydata_MBA$`MBA Grade`)## [1] 76.04055We can reject the H0, at α = 5%. Mean of variable MBA Grade is not 74. Even though we rejected H0, Cohen’s d test showed us that effect size is small / weak.
Task 3
Import the dataset Apartments.xlsx
library(readxl)
Apartments <- read_excel("D:/IMB/R/R Take Home Exam 2025/Apartments.xlsx")
head(Apartments, 10) #Showing first 10 units## # A tibble: 10 × 5
## Age Distance Price Parking Balcony
## <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 7 28 1640 0 1
## 2 18 1 2800 1 0
## 3 7 28 1660 0 0
## 4 28 29 1850 0 1
## 5 18 18 1640 1 1
## 6 28 12 1770 0 1
## 7 14 20 1850 0 1
## 8 18 6 1970 1 1
## 9 22 7 2270 1 0
## 10 25 2 2570 1 0Description:
- Age: Age of an apartment in years
- Distance: The distance from city center in km
- Price: Price per m2
- Parking: 0-No, 1-Yes
- Balcony: 0-No, 1-Yes
Change categorical variables into factors.
Apartments$Parking <- factor(Apartments$Parking,
levels = c(0, 1),
labels = c("No parking", "Has parking"))
Apartments$Balcony <- factor(Apartments$Balcony,
levels = c(0, 1),
labels = c("No balcony", "Has balcony"))
head(Apartments, 10) #Showing first 10 units## # A tibble: 10 × 5
## Age Distance Price Parking Balcony
## <dbl> <dbl> <dbl> <fct> <fct>
## 1 7 28 1640 No parking Has balcony
## 2 18 1 2800 Has parking No balcony
## 3 7 28 1660 No parking No balcony
## 4 28 29 1850 No parking Has balcony
## 5 18 18 1640 Has parking Has balcony
## 6 28 12 1770 No parking Has balcony
## 7 14 20 1850 No parking Has balcony
## 8 18 6 1970 Has parking Has balcony
## 9 22 7 2270 Has parking No balcony
## 10 25 2 2570 Has parking No balconyTest the hypothesis H0: Mu_Price = 1900 eur. What can you conclude?
H0: Mu_Price = 1900 eur
t.test(Apartments$Price, mu = 1900)##
## One Sample t-test
##
## data: Apartments$Price
## t = 2.9022, df = 84, p-value = 0.004731
## alternative hypothesis: true mean is not equal to 1900
## 95 percent confidence interval:
## 1937.443 2100.440
## sample estimates:
## mean of x
## 2018.941We can conclude, that we can reject H0 at α = 5%, mean is not 1900eur. We can also see, that the 95% confidence interval assumes that the mean for price of the apartment lies on the interval 1937.443eur and 2100.440eur. While the sample estimate of the mean for price is 2018.941eur.
Estimate the simple regression function: Price = f(Age). Save results in object fit1 and explain the estimate of regression coefficient, coefficient of correlation and coefficient of determination.
fit1 <- lm(Apartments$Price ~ Apartments$Age, data = Apartments)
summary(fit1)##
## Call:
## lm(formula = Apartments$Price ~ Apartments$Age, data = Apartments)
##
## Residuals:
## Min 1Q Median 3Q Max
## -623.9 -278.0 -69.8 243.5 776.1
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 2185.455 87.043 25.108 <2e-16 ***
## Apartments$Age -8.975 4.164 -2.156 0.034 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 369.9 on 83 degrees of freedom
## Multiple R-squared: 0.05302, Adjusted R-squared: 0.04161
## F-statistic: 4.647 on 1 and 83 DF, p-value: 0.03401cor(Apartments$Price, Apartments$Age)## [1] -0.230255- Intercept (β₀ = 2185.46) => tells us that the expectede price of a new apartment is 2185eur.
- β₁ = -8.98 => this value tells us the slope, and means that for on year increase in age, the price is expected to decrease on avrage by 8.98eur.
- Multiple R-squared: 0.05302 => this is coefficient of determination, and it tells us, that we can explain only 5.3% variation of price can be explained by linear effect of age
- We can also calculate the correlation coefficient: r = -0.23, which tells us there is weak negative correlation between variables age and price
Show the scateerplot matrix between Price, Age and Distance. Based on the matrix determine if there is potential problem with multicolinearity.
scatterplotMatrix(Apartments[ , c(1, 2, 3)], #Scatterplot matrix
smooth = FALSE,
regLine = list(col = "cyan3"))
From the scatter plot matrix we can see that there is no multicollinearity between age and distance variables. They are not strongly correlated, which means we can include both into a regression model without any concerns.
Estimate the multiple regression function: Price = f(Age, Distance). Save it in object named fit2.
fit2 <- lm(Price ~ Age + Distance, data = Apartments)
summary(fit2)##
## Call:
## lm(formula = Price ~ Age + Distance, data = Apartments)
##
## Residuals:
## Min 1Q Median 3Q Max
## -603.23 -219.94 -85.68 211.31 689.58
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 2460.101 76.632 32.10 < 2e-16 ***
## Age -7.934 3.225 -2.46 0.016 *
## Distance -20.667 2.748 -7.52 6.18e-11 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 286.3 on 82 degrees of freedom
## Multiple R-squared: 0.4396, Adjusted R-squared: 0.4259
## F-statistic: 32.16 on 2 and 82 DF, p-value: 4.896e-11Chech the multicolinearity with VIF statistics. Explain the findings.
library(car)
vif(fit2)## Age Distance
## 1.001845 1.001845VIF statistics confirm our assumption about multicollinearity between age and distance. Because both values are almost 1, they are almost completley independent of each other. Both can be included in our regression model.
Calculate standardized residuals and Cooks Distances for model fit2. Remove any potentially problematic units (outliers or units with high influence).
Apartments$StdResid <- round(rstandard(fit2), 3) #Standardized residuals
Apartments$CooksD <- round(cooks.distance(fit2), 3) #Cooks distances
hist(Apartments$StdResid,
xlab = "Standardized residuals",
ylab = "Frequency",
main = "Histogram of standardized residuals")
hist(Apartments$CooksD,
xlab = "Cooks distance",
ylab = "Frequency",
main = "Histogram of Cooks distances")
Apartments$ID <- 1:nrow(Apartments)
head(Apartments, 10)## # A tibble: 10 × 8
## Age Distance Price Parking Balcony StdResid CooksD ID
## <dbl> <dbl> <dbl> <fct> <fct> <dbl> <dbl> <int>
## 1 7 28 1640 No parking Has balcony -0.665 0.007 1
## 2 18 1 2800 Has parking No balcony 1.78 0.03 2
## 3 7 28 1660 No parking No balcony -0.594 0.006 3
## 4 28 29 1850 No parking Has balcony 0.754 0.008 4
## 5 18 18 1640 Has parking Has balcony -1.07 0.005 5
## 6 28 12 1770 No parking Has balcony -0.778 0.005 6
## 7 14 20 1850 No parking Has balcony -0.302 0.001 7
## 8 18 6 1970 Has parking Has balcony -0.787 0.004 8
## 9 22 7 2270 Has parking No balcony 0.455 0.001 9
## 10 25 2 2570 Has parking No balcony 1.24 0.017 10head(Apartments[order(Apartments$StdResid),], 5) #Three units with lowest value of stand. residuals## # A tibble: 5 × 8
## Age Distance Price Parking Balcony StdResid CooksD ID
## <dbl> <dbl> <dbl> <fct> <fct> <dbl> <dbl> <int>
## 1 7 2 1760 No parking Has balcony -2.15 0.066 53
## 2 12 14 1650 No parking Has balcony -1.50 0.013 13
## 3 12 14 1650 No parking No balcony -1.50 0.013 72
## 4 13 8 1800 No parking No balcony -1.38 0.012 20
## 5 14 16 1660 No parking Has balcony -1.26 0.008 35head(Apartments[order(-Apartments$CooksD),], 6) #Six units with highest value of Cooks distance## # A tibble: 6 × 8
## Age Distance Price Parking Balcony StdResid CooksD ID
## <dbl> <dbl> <dbl> <fct> <fct> <dbl> <dbl> <int>
## 1 5 45 2180 Has parking Has balcony 2.58 0.32 38
## 2 43 37 1740 No parking No balcony 1.44 0.104 55
## 3 2 11 2790 Has parking No balcony 2.05 0.069 33
## 4 7 2 1760 No parking Has balcony -2.15 0.066 53
## 5 37 3 2540 Has parking Has balcony 1.58 0.061 22
## 6 40 2 2400 No parking Has balcony 1.09 0.038 39Looking at standard residuals only units 53, 38 and 33 stand out as potential outlier, but with standrard residual values under 3. While looking at Cooks distances, with rule Cooks D there shouldnt be any gaps in the histogram, thats why we need to remove only unit 38.
library(dplyr)
Apartments <- Apartments %>%
filter(!ID == "38") #Removing 38th unitCheck for potential heteroskedasticity with scatterplot between standarized residuals and standrdized fitted values. Explain the findings.
fit2 <- lm(Price ~ Age + Distance, data = Apartments)
Apartments$StdFitted <- scale(fit2$fitted.values)
library(car)
scatterplot(y = Apartments$StdResid, x = Apartments$StdFitted,
ylab = "Standardized residuals",
xlab = "Standardized fitted values",
boxplots = FALSE,
regLine = FALSE,
smooth = FALSE)
library(olsrr)##
## Attaching package: 'olsrr'## The following object is masked from 'package:datasets':
##
## riversols_test_breusch_pagan(fit2)##
## Breusch Pagan Test for Heteroskedasticity
## -----------------------------------------
## Ho: the variance is constant
## Ha: the variance is not constant
##
## Data
## ---------------------------------
## Response : Price
## Variables: fitted values of Price
##
## Test Summary
## -----------------------------
## DF = 1
## Chi2 = 2.927455
## Prob > Chi2 = 0.08708469The scatterplot of standardized residuals versus fitted values does not show any clear funnel shape. With Breusch Pagan Test for Heteroskedasticity we get a result of p=0.087 > α = 5% , therefore we can not reject the H0.Our findings support the validity of standard regression assumptions.
Are standardized residuals ditributed normally? Show the graph and formally test it. Explain the findings.
H0: standardized residuals are ditributed normally H1: standardized residuals are not ditributed normally
hist(Apartments$StdResid,
xlab = "Standardized residuals",
ylab = "Frequency",
main = "Histogram of standardized residuals")
shapiro.test(Apartments$StdResid)##
## Shapiro-Wilk normality test
##
## data: Apartments$StdResid
## W = 0.94879, p-value = 0.002187We can see from the histogram, that there is a small deviation from normal distribution which is also confirmed by Shapiro-Wilk normality test, which results came back at w=0.956 and p=0.0063, we can reject the H0 at p < α, we conclude that the standardized residuals are not distributed normally.
Estimate the fit2 again without potentially excluded units and show the summary of the model. Explain all coefficients.
summary(fit2)##
## Call:
## lm(formula = Price ~ Age + Distance, data = Apartments)
##
## Residuals:
## Min 1Q Median 3Q Max
## -604.92 -229.63 -56.49 192.97 599.35
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 2456.076 73.931 33.221 < 2e-16 ***
## Age -6.464 3.159 -2.046 0.044 *
## Distance -22.955 2.786 -8.240 2.52e-12 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 276.1 on 81 degrees of freedom
## Multiple R-squared: 0.4838, Adjusted R-squared: 0.4711
## F-statistic: 37.96 on 2 and 81 DF, p-value: 2.339e-12- Explinationt:
- intercept =>is price estimate when age and distance both equal 0
- Age intercept tells us, that for each additional yearof apartments age the price on avrage decreases by 6.46eur, if everything else stays the same
- Distance estimate tells us that for each additional unit of distance from the city center, the price on avrage decreases by 22.96eur, assuming everything else stays the same.
- From residual standard error we can see, that it decreased in comparison to the model before removal, which tells us the new fit2 fits better
- We can also see, that the multiple r-squared value has increased, which tells us, that now after removing the unit we can explain more variance
- F-statistics tells us, that the newer model is slightly stronger after removing the unit, hence the higher value
Estimate the linear regression function Price = f(Age, Distance, Parking and Balcony). Be careful to correctly include categorical variables. Save the object named fit3.
fit3 <- lm(Price ~ Age + Distance + Parking + Balcony, data = Apartments)
summary(fit3)##
## Call:
## lm(formula = Price ~ Age + Distance + Parking + Balcony, data = Apartments)
##
## Residuals:
## Min 1Q Median 3Q Max
## -473.21 -192.37 -28.89 204.17 558.77
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 2329.724 93.066 25.033 < 2e-16 ***
## Age -5.821 3.074 -1.894 0.06190 .
## Distance -20.279 2.886 -7.026 6.66e-10 ***
## ParkingHas parking 167.531 62.864 2.665 0.00933 **
## BalconyHas balcony -15.207 59.201 -0.257 0.79795
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 267.5 on 79 degrees of freedom
## Multiple R-squared: 0.5275, Adjusted R-squared: 0.5035
## F-statistic: 22.04 on 4 and 79 DF, p-value: 3.018e-12With function anova check if model fit3 fits data better than model fit2.
anova(fit2, fit3)## Analysis of Variance Table
##
## Model 1: Price ~ Age + Distance
## Model 2: Price ~ Age + Distance + Parking + Balcony
## Res.Df RSS Df Sum of Sq F Pr(>F)
## 1 81 6176767
## 2 79 5654480 2 522287 3.6485 0.03051 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Based on ANOVA result the fit3 fits data better than model fit2.
Show the results of fit3 and explain regression coefficient for both categorical variables. Can you write down the hypothesis which is being tested with F-statistics, shown at the bottom of the output?
summary(fit3)##
## Call:
## lm(formula = Price ~ Age + Distance + Parking + Balcony, data = Apartments)
##
## Residuals:
## Min 1Q Median 3Q Max
## -473.21 -192.37 -28.89 204.17 558.77
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 2329.724 93.066 25.033 < 2e-16 ***
## Age -5.821 3.074 -1.894 0.06190 .
## Distance -20.279 2.886 -7.026 6.66e-10 ***
## ParkingHas parking 167.531 62.864 2.665 0.00933 **
## BalconyHas balcony -15.207 59.201 -0.257 0.79795
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 267.5 on 79 degrees of freedom
## Multiple R-squared: 0.5275, Adjusted R-squared: 0.5035
## F-statistic: 22.04 on 4 and 79 DF, p-value: 3.018e-12- Explination:
- apartments with balcony are expected to be on average by 15.21eur cheaper, then those without the balcony, assuming all other variables stay the same. Looking at p-value=0.798 which is very high, we can say this estimate is not statistically significant
- apartments with parking are on average 167.53eur more expensive by 167.53eur than those without parking, assuming everything else stays the same. This estimate is statistically significant, because of p-value=0.009
- F-statistics tests the overall significance of the regression model
- H0: All regression coefficients are equal to zero
- H1: At least one coefficient is not zero
Save fitted values and claculate the residual for apartment ID2.
Apartments$fitted_price <- fitted(fit3)
Apartments$Residuals <- residuals(fit3)
Apartments$residual <- Apartments$Price - Apartments$fitted_price
residual_ID2 <- Apartments$Price[Apartments$ID == 2] -
fitted(fit3)[Apartments$ID == 2]
residual_ID2## 2
## 427.8029