HW 3

Stella Maria Lourdes Kusumawardhani

1. Import the dataset Apartments.xlsx

library(readxl)
mydata <- read_xlsx("./Apartments.xlsx")
mydata <- as.data.frame(mydata)

head(mydata)

##   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

2.What could be a possible research question given the data you analyze? (1 p)

Does the age of an apartment in years affect its price?

3. Change categorical variables into factors. (0.5 p)

mydata$ParkingF <- factor(mydata$Parking,
                          levels = c(0, 1),
                          labels = c("No Parking", "Parking"))

mydata$BalconyF <- factor(mydata$Balcony,
                          levels = c(0, 1),
                          labels = c("No Balcony", "Balcony"))

4. Test the hypothesis H0: Mu_Price = 1900 eur. What can you conclude? (1 p)

library(psych)
describe(mydata$Price)

##    vars  n    mean     sd median trimmed    mad  min  max range skew kurtosis    se
## X1    1 85 2018.94 377.84   1950 1990.29 429.95 1400 2820  1420 0.54    -0.69 40.98

H0: On average the price of apartments is equal to 1900 euro

H1: On average the price of apartments is not equal to 1900 euro

Normality Assumption Test

library(ggplot2)

## 
## Attaching package: 'ggplot2'

## The following objects are masked from 'package:psych':
## 
##     %+%, alpha

ggplot(mydata, aes(x = Price)) +
  geom_histogram(binwidth = 250, colour = "black") +
  ylab("EUR") +
  xlab("Price of Apartments")

shapiro.test(mydata$Price)

## 
##  Shapiro-Wilk normality test
## 
## data:  mydata$Price
## W = 0.94017, p-value = 0.0006513

H0: Price of apartments is normally distributed

H1: Price of apartments is not normally distributed

We reject H0 at p < 0.001, meaning the price of apartments is not normally distributed
The hypothesis testing should proceed with an alternative non-parametric test

Wilcoxon Signed Rank Test

median(mydata$Price)

## [1] 1950

wilcox.test(mydata$Price,
            mu = 1900,
            correct = FALSE)

## 
##  Wilcoxon signed rank test
## 
## data:  mydata$Price
## V = 2328, p-value = 0.02828
## alternative hypothesis: true location is not equal to 1900

H0: Median of apartment price is equal to 1900 euro

H1: Median of apartment price is not equal to 1900 euro

We reject null hypothesis at p = 0.03, meaning the median of apartment price is not equal to 1900 euro

Effect Size

library(psych)
library(effectsize)

## 
## Attaching package: 'effectsize'

## The following object is masked from 'package:psych':
## 
##     phi

effectsize(wilcox.test(mydata$Price,
                       mu = 1900,
                       correct = FALSE))

## r (rank biserial) |       95% CI
## --------------------------------
## 0.27              | [0.04, 0.48]
## 
## - Deviation from a difference of 1900.

Conclusion

Based on the sample data, we found that the median price of apartments is not equal to 1900 euro (p = 0.03, r = 0.27 - medium effect). The median was 1950 euro.

5. 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. (1 p)

fit1 <- lm(Price ~ Age,
           data = mydata)

summary(fit1)

## 
## Call:
## lm(formula = Price ~ Age, data = mydata)
## 
## 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

Interpretation

The estimate of regression coefficient is -8.975, meaning that if the age of apartment increases by 1 year, the price on average decreases by 8.975 euro (p = 0.034)
The correlation coefficient is 0.23, indicating a weak relationship between apartment age and prices
The coefficient of determination is 0.053, meaning that 5.3% of the variability in apartment prices is explained by the linear effect of age

6. Show the scateerplot matrix between Price, Age and Distance. Based on the matrix determine if there is potential problem with multicolinearity. (0.5 p)

library(car)

## Loading required package: carData

## 
## Attaching package: 'car'

## The following object is masked from 'package:psych':
## 
##     logit

scatterplotMatrix(mydata[c("Price", "Age", "Distance")], smooth = FALSE)

based on the scatterplot between age and distance, the regression line does not show a strong linear trend, indicating that age and distance are not strongly correlated
there is no potential issue of multicolinearity

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

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

summary(fit2)

## 
## Call:
## lm(formula = Price ~ Age + Distance, data = mydata)
## 
## 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

Interpretation

If apartment age is increased by 1 year, the apartment price on average decreases by 7.934 euro, holding other variable constant (p = 0.016)
If the distance between apartment and city center increases by 1km, the apartment price on average decreases by 20.667 euro, holding other variable constant (p < 0.001)
The correlation coefficient is 0.663, meaning that the linear relationship between price, age and distance is semi strong
43.96% of the variability in apartment prices is explained by the linear effect of age and distance

8. Check the multicolinearity with VIF statistics. Explain the findings. (0.5 p)

library(car)
vif(fit2)

##      Age Distance 
## 1.001845 1.001845

The VIF statistic for age and distance of apartment is 1.001
Since this value is well below 5, it indicates that there is no problem with too strong multicolinearity

9. Calculate standardized residuals and Cooks Distances for model fit2. Remove any potentially problematic units (outliers or units with high influence). (1 p)

mydata$StdResid <- round(rstandard(fit2), 3)
mydata$CooksD <- round(cooks.distance(fit2), 3)

hist(mydata$StdResid,
     xlab = "Standardized residuals",
     ylab = "Frequency",
     main = "Histogram of standardized residuals")

shapiro.test(mydata$StdResid)

## 
##  Shapiro-Wilk normality test
## 
## data:  mydata$StdResid
## W = 0.95303, p-value = 0.003645

H0: error is normally distributed H1: error is not normally distributed

we reject H0 at p = 0.004, meaning that error is not normally distributed

hist(mydata$CooksD,
     xlab = "Cooks distance",
     ylab = "Frequency",
     main = "Histogram of Cooks distances")

mydata$ID <- seq(1, nrow(mydata))

head(mydata[order(mydata$StdResid), c("ID", "StdResid")], 3)

##    ID StdResid
## 53 53   -2.152
## 13 13   -1.499
## 72 72   -1.499

head(mydata[order(-mydata$StdResid), c("ID", "StdResid")], 3)

##    ID StdResid
## 38 38    2.577
## 33 33    2.051
## 2   2    1.783

head(mydata[order(-mydata$CooksD), c("ID", "CooksD")], 6)

##    ID CooksD
## 38 38  0.320
## 55 55  0.104
## 33 33  0.069
## 53 53  0.066
## 22 22  0.061
## 39 39  0.038

library(dplyr)

## 
## Attaching package: 'dplyr'

## 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, union

library(car)
mydata <- mydata %>%
  filter(!ID %in% c(38, 55))

10. Check for potential heteroskedasticity with scatterplot between standarized residuals and standrdized fitted values. Explain the findings. (0.5 p)

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

mydata$StdFitted <- scale(fit2$fitted.values)

library(car)
scatterplot(y = mydata$StdResid, x = mydata$StdFitted,
            ylab = "Standardized residuals",
            xlab = "Standardized fitted values",
            boxplots = FALSE,
            regLine = FALSE,
            smooth = FALSE)

The graph does not indicate any heteroskedasticity

11. Are standardized residuals ditributed normally? Show the graph and formally test it. Explain the findings. (0.5 p)

hist(mydata$StdResid,
     xlab = "Standardized residuals",
     ylab = "Frequency",
     main = "Histogram of standardized residuals")

shapiro.test(mydata$StdResid)

## 
##  Shapiro-Wilk normality test
## 
## data:  mydata$StdResid
## W = 0.94963, p-value = 0.002636

H0: error is normally distributed H1: error is not normally distributed

we reject H0 at p = 0.003, meaning that error is not normally distributed

12. Estimate the fit2 again without potentially excluded units and show the summary of the model. Explain all coefficients. (1 p)

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

summary(fit2)

## 
## Call:
## lm(formula = Price ~ Age + Distance, data = mydata)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -627.27 -212.96  -46.23  205.05  578.98 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 2490.112     76.189  32.684  < 2e-16 ***
## Age           -7.850      3.244  -2.420   0.0178 *  
## Distance     -23.945      2.826  -8.473 9.53e-13 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 273.5 on 80 degrees of freedom
## Multiple R-squared:  0.4968, Adjusted R-squared:  0.4842 
## F-statistic: 39.49 on 2 and 80 DF,  p-value: 1.173e-12

Interpretation

If apartment age is increased by 1 year, the apartment price on average decreases by 7.850 euro, holding other variable constant (p = 0.018)
If the distance between apartment and city center increases by 1km, the apartment price on average decreases by 23.945 euro, holding other variable constant (p < 0.001)
The correlation coefficient is 0.7, meaning that the linear relationship between price, age and distance is semi strong
49.68% of the variability in apartment prices is explained by the linear effect of age and distance

13. Estimate the linear regression function Price = f(Age, Distance, Parking and Balcony). Be careful to correctly include categorical variables. Save the object named fit3. (0.5 p)

fit3 <- lm(Price ~ Age + Distance + ParkingF + BalconyF,
           data = mydata)

summary(fit3)

## 
## Call:
## lm(formula = Price ~ Age + Distance + ParkingF + BalconyF, data = mydata)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -499.06 -194.33  -32.04  219.03  544.31 
## 
## Coefficients:
##                 Estimate Std. Error t value Pr(>|t|)    
## (Intercept)     2358.900     93.664  25.185  < 2e-16 ***
## Age               -7.197      3.148  -2.286  0.02499 *  
## Distance         -21.241      2.911  -7.296 2.14e-10 ***
## ParkingFParking  168.921     62.166   2.717  0.00811 ** 
## BalconyFBalcony   -6.985     58.745  -0.119  0.90566    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 264.5 on 78 degrees of freedom
## Multiple R-squared:  0.5408, Adjusted R-squared:  0.5173 
## F-statistic: 22.97 on 4 and 78 DF,  p-value: 1.449e-12

14. With function anova check if model fit3 fits data better than model fit2. (0.5 p)

15. 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? (1 p)

summary(fit3)

## 
## Call:
## lm(formula = Price ~ Age + Distance + ParkingF + BalconyF, data = mydata)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -499.06 -194.33  -32.04  219.03  544.31 
## 
## Coefficients:
##                 Estimate Std. Error t value Pr(>|t|)    
## (Intercept)     2358.900     93.664  25.185  < 2e-16 ***
## Age               -7.197      3.148  -2.286  0.02499 *  
## Distance         -21.241      2.911  -7.296 2.14e-10 ***
## ParkingFParking  168.921     62.166   2.717  0.00811 ** 
## BalconyFBalcony   -6.985     58.745  -0.119  0.90566    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 264.5 on 78 degrees of freedom
## Multiple R-squared:  0.5408, Adjusted R-squared:  0.5173 
## F-statistic: 22.97 on 4 and 78 DF,  p-value: 1.449e-12

Interpretation

If apartment age is increased by 1 year, the apartment price on average decreases by 7.197 euro, holding other variable constant (p = 0.025)
If the distance between apartment and city center increases by 1km, the apartment price on average decreases by 21.241 euro, holding other variable constant (p < 0.001)

Categorical Variables Interpretation

Given the values of apartment age and distance from city center, apartments with parking space on average have a higher price by 168.921 euro compared with apartments with no parking (p = 0.009)
We can not say the price of apartments with balcony and without balcony are different
The correlation coefficient is 0.735, meaning that the linear relationship between price, age and distance is strong
54.08% of the variability in apartment prices is explained by the linear effect of age and distance

HW 3

Stella Maria Lourdes Kusumawardhani

1. Import the dataset Apartments.xlsx

2.What could be a possible research question given the data you analyze? (1 p)

3. Change categorical variables into factors. (0.5 p)

4. Test the hypothesis H0: Mu_Price = 1900 eur. What can you conclude? (1 p)

H0: On average the price of apartments is equal to 1900 euro

H1: On average the price of apartments is not equal to 1900 euro

Normality Assumption Test

Wilcoxon Signed Rank Test

Effect Size

Conclusion

5. 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. (1 p)

Interpretation

6. Show the scateerplot matrix between Price, Age and Distance. Based on the matrix determine if there is potential problem with multicolinearity. (0.5 p)

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

Interpretation

8. Check the multicolinearity with VIF statistics. Explain the findings. (0.5 p)

9. Calculate standardized residuals and Cooks Distances for model fit2. Remove any potentially problematic units (outliers or units with high influence). (1 p)

10. Check for potential heteroskedasticity with scatterplot between standarized residuals and standrdized fitted values. Explain the findings. (0.5 p)

11. Are standardized residuals ditributed normally? Show the graph and formally test it. Explain the findings. (0.5 p)

12. Estimate the fit2 again without potentially excluded units and show the summary of the model. Explain all coefficients. (1 p)

Interpretation

13. Estimate the linear regression function Price = f(Age, Distance, Parking and Balcony). Be careful to correctly include categorical variables. Save the object named fit3. (0.5 p)

14. With function anova check if model fit3 fits data better than model fit2. (0.5 p)

15. 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? (1 p)

Interpretation

Categorical Variables Interpretation

16. Save fitted values and calculate the residual for apartment ID2. (0.5 p)