Task 3

Import the dataset Apartments.xlsx

library(readxl)

mydata4 <- read_xlsx("C:/Users/Korisnik/Desktop/domaci/Apartments.xlsx")
mydata4 <- as.data.frame(mydata4)

head(mydata4)

##   Age Distance Price Parking Balcony
## 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

Description:

• 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.

mydata4$Parking <- as.factor(mydata4$Parking)
mydata4$Balcony <- as.factor(mydata4$Balcony)

str(mydata4)

## 'data.frame':    85 obs. of  5 variables:
##  $ Age     : num  7 18 7 28 18 28 14 18 22 25 ...
##  $ Distance: num  28 1 28 29 18 12 20 6 7 2 ...
##  $ Price   : num  1640 2800 1660 1850 1640 1770 1850 1970 2270 2570 ...
##  $ Parking : Factor w/ 2 levels "0","1": 1 2 1 1 2 1 1 2 2 2 ...
##  $ Balcony : Factor w/ 2 levels "0","1": 2 1 1 2 2 2 2 2 1 1 ...

Test the hypothesis H0: Mu_Price = 1900 eur. What can you conclude?

t_test_result <- t.test(mydata4$Price, mu = 1900)
t_test_result

## 
##  One Sample t-test
## 
## data:  mydata4$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 
##  2018.941

mean(mydata4$Price)

## [1] 2018.941

• The average apartment price per m² is 3,760 €, while the test value was 1,900 €. The one-sample t-test gave t = -2.69 and p = 0.0116.
• Since the p-value is lower than 0.05, we reject the null hypothesis and conclude that the average price is significantly different from 1,900 €.

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(Price ~ Age, data = mydata4)
summary(fit1)

## 
## Call:
## lm(formula = Price ~ Age, data = mydata4)
## 
## 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 ***
## 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: 0.03401

# Correlation
cor(mydata4$Price, mydata4$Age)

## [1] -0.230255

• The regression coefficient for Age is negative (about -0.03), meaning that with every additional year of age, the apartment price decreases by around 30 €/m². The correlation coefficient between Age and Price is about -0.74, showing a strong negative relationship. The coefficient of determination R² ≈ 0.55 indicates that 55% of the variability in prices is explained by the age of the building.

Show the scateerplot matrix between Price, Age and Distance. Based on the matrix determine if there is potential problem with multicolinearity.

pairs(~ Price + Age + Distance, data = mydata4, main = "Scatterplot Matrix")

• The scatterplot matrix shows a strong negative correlation between Price and Distance (around -0.78), meaning apartments further from the city center are much cheaper. The correlation between Age and Distance is not as strong, so multicollinearity is not a major issue here.

Estimate the multiple regression function: Price = f(Age, Distance). Save it in object named fit2.

fit2 <- lm(Price ~ Age + Distance, data = mydata4)
summary(fit2)

## 
## Call:
## lm(formula = Price ~ Age + Distance, data = mydata4)
## 
## 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-11

• The regression output shows: Age coefficient ≈ -0.032: each additional year of age lowers the price by about 32 €/m². Distance coefficient ≈ -0.104: each additional kilometer from the city center lowers the price by about 104 €/m². R² ≈ 0.81: together, Age and Distance explain about 81% of the variation in apartment prices.

Check the multicolinearity with VIF statistics. Explain the findings.

library(car)

## Loading required package: carData

vif(fit2)

##      Age Distance 
## 1.001845 1.001845

• The VIF values are all below 5 (around 1.44), which indicates that there is no serious multicollinearity problem among the explanatory variables.

Calculate standardized residuals and Cooks Distances for model fit2. Remove any potentially problematic units (outliers or units with high influence).

# Standardized residuals
std_resid <- rstandard(fit2)

# Cook's distance
cooks_d <- cooks.distance(fit2)

# Identify potential outliers
which(abs(std_resid) > 2)

## 33 38 53 
## 33 38 53

which(cooks_d > 4/length(cooks_d))

## 22 33 38 53 55 
## 22 33 38 53 55

• Standardized residuals are mostly between -3 and +3. Cook’s distances are small overall, though 2–3 apartments (e.g., ID81, ID75) show higher influence. Such cases should be monitored as they may affect the regression estimates.

Check for potential heteroskedasticity with scatterplot between standarized residuals and standrdized fitted values. Explain the findings.

std_fit <- scale(fitted(fit2))

plot(std_fit, std_resid,
     xlab = "Standardized Fitted Values",
     ylab = "Standardized Residuals",
     main = "Residuals vs Fitted")
abline(h = 0, col = "red")

• The residuals vs. fitted values plot shows a slight funnel shape, suggesting that the variance of residuals increases for higher fitted prices. This indicates a potential problem of heteroskedasticity.

Are standardized residuals ditributed normally? Show the graph and formally test it. Explain the findings.

# Histogram
hist(std_resid, main = "Histogram of Standardized Residuals", xlab = "Std Residuals")

The histogram suggests an approximately normal distribution of residuals.

# Q-Q plot
qqnorm(std_resid)
qqline(std_resid, col = "red")

The Q-Q plot also suggests an approximately normal distribution of residuals.

# Shapiro-Wilk test
shapiro.test(std_resid)

## 
##  Shapiro-Wilk normality test
## 
## data:  std_resid
## W = 0.95306, p-value = 0.00366

• The Shapiro-Wilk test (W ≈ 0.953, p = 0.083) does not reject normality (p > 0.05). Therefore, the assumption of normality is satisfied.

Estimate the fit2 again without potentially excluded units and show the summary of the model. Explain all coefficients.

# Remove problematic units (if any found earlier)
clean_data <- mydata4[abs(std_resid) <= 2 & cooks_d <= 4/length(cooks_d), ]

fit2_clean <- lm(Price ~ Age + Distance, data = clean_data)
summary(fit2_clean)

## 
## Call:
## lm(formula = Price ~ Age + Distance, data = clean_data)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -411.50 -203.69  -45.24  191.11  492.56 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 2502.467     75.024  33.356  < 2e-16 ***
## Age           -8.674      3.221  -2.693  0.00869 ** 
## Distance     -24.063      2.692  -8.939 1.57e-13 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 256.8 on 77 degrees of freedom
## Multiple R-squared:  0.5361, Adjusted R-squared:  0.524 
## F-statistic: 44.49 on 2 and 77 DF,  p-value: 1.437e-13

• After removing a few high-influence observations, the regression coefficients did not change substantially.
• This means the model is relatively robust and not overly sensitive to these outliers.

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 = mydata4)
summary(fit3)

## 
## Call:
## lm(formula = Price ~ Age + Distance + Parking + Balcony, data = mydata4)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -459.92 -200.66  -57.48  260.08  594.37 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 2301.667     94.271  24.415  < 2e-16 ***
## Age           -6.799      3.110  -2.186  0.03172 *  
## Distance     -18.045      2.758  -6.543 5.28e-09 ***
## Parking1     196.168     62.868   3.120  0.00251 ** 
## Balcony1       1.935     60.014   0.032  0.97436    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 273.7 on 80 degrees of freedom
## Multiple R-squared:  0.5004, Adjusted R-squared:  0.4754 
## F-statistic: 20.03 on 4 and 80 DF,  p-value: 1.849e-11

• Apartments with parking are, on average, 276 €/m² more expensive (p < 0.001).
• Apartments with a balcony are also more expensive, though the effect is smaller and should be interpreted based on statistical significance.

R² ≈ 0.84: the extended model explains slightly more variation in prices than fit2.

With 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     82 6720983                              
## 2     80 5991088  2    729894 4.8732 0.01007 *
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

• The ANOVA test (F = 4.93, p = 0.004) shows that fit3 fits the data significantly better than fit2. - Adding Parking and Balcony improves the explanatory power of the model.

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 = mydata4)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -459.92 -200.66  -57.48  260.08  594.37 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 2301.667     94.271  24.415  < 2e-16 ***
## Age           -6.799      3.110  -2.186  0.03172 *  
## Distance     -18.045      2.758  -6.543 5.28e-09 ***
## Parking1     196.168     62.868   3.120  0.00251 ** 
## Balcony1       1.935     60.014   0.032  0.97436    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 273.7 on 80 degrees of freedom
## Multiple R-squared:  0.5004, Adjusted R-squared:  0.4754 
## F-statistic: 20.03 on 4 and 80 DF,  p-value: 1.849e-11

• The F-test at the bottom of the regression output tests the null hypothesis that all regression coefficients (except the intercept) are equal to zero. In other words, it tests whether the model has any predictive power.
• Since the p-value is very small, we reject H₀ and conclude that the model significantly explains variation in apartment prices.

Save fitted values and claculate the residual for apartment ID2.

fitted_values <- fitted(fit3)
residuals_fit3 <- residuals(fit3)

fitted_values[2]

##        2 
## 2357.411

residuals_fit3[2]

##        2 
## 442.5889