EXERCISE 1

Find your own data set that contains at least 20 units and at least 4 variables (most of which are numeric, but it is good to have at least one categorical variable as well). Perform the following steps using the R program:

mydata <- force(airquality)

Explain the data set (the variables used in the analysis).

#Interpretation of the data. We can see that some values are missing, which we might remove later.
head(mydata)
##   Ozone Solar.R Wind Temp Month Day
## 1    41     190  7.4   67     5   1
## 2    36     118  8.0   72     5   2
## 3    12     149 12.6   74     5   3
## 4    18     313 11.5   62     5   4
## 5    NA      NA 14.3   56     5   5
## 6    28      NA 14.9   66     5   6

Explanation of variables:

  • Ozone: Mean ozone in parts per billion from 1300 to 1500 hours at Roosevelt Island
  • Solar.R: Solar radiation in Langleys in the frequency band 4000–7700 Angstroms from 0800 to 1200 hours at Central Park
  • Wind: Average wind speed in miles per hour at 0700 and 1000 hours at LaGuardia Airport
  • Temp: Maximum daily temperature in degrees Fahrenheit at LaGuardia Airport.
  • Month: Month of the year 1973
  • Day: Day of month

Perform some data manipulations (create new variable, delete some units due to missing data, rename variables, create new data.frame based on conditions, etc.).

#Replacing the name of variable, so it has better visual representation in dataframe.

colnames(mydata)[2]<-"SolarRadiation"
#Changing the the number of the month to factor 

mydata$MonthN <- factor(mydata$Month, 
                           levels = c(5, 6, 7, 8, 9), 
                           labels = c("May", "June", "July", "August", "September"))
library(pastecs)
round(stat.desc(mydata[,c(-5,-6,-7)]),2)
##                Ozone SolarRadiation    Wind     Temp
## nbr.val       116.00         146.00  153.00   153.00
## nbr.null        0.00           0.00    0.00     0.00
## nbr.na         37.00           7.00    0.00     0.00
## min             1.00           7.00    1.70    56.00
## max           168.00         334.00   20.70    97.00
## range         167.00         327.00   19.00    41.00
## sum          4887.00       27146.00 1523.50 11916.00
## median         31.50         205.00    9.70    79.00
## mean           42.13         185.93    9.96    77.88
## SE.mean         3.06           7.45    0.28     0.77
## CI.mean.0.95    6.07          14.73    0.56     1.51
## var          1088.20        8110.52   12.41    89.59
## std.dev        32.99          90.06    3.52     9.47
## coef.var        0.78           0.48    0.35     0.12
library(tidyr)
## 
## Attaching package: 'tidyr'
## The following object is masked from 'package:pastecs':
## 
##     extract
mydata1 <- drop_na(mydata) #Removing units with missing values

library(pastecs)
round(stat.desc(mydata1[,c(-5,-6,-7)]),2)
##                Ozone SolarRadiation    Wind    Temp
## nbr.val       111.00         111.00  111.00  111.00
## nbr.null        0.00           0.00    0.00    0.00
## nbr.na          0.00           0.00    0.00    0.00
## min             1.00           7.00    2.30   57.00
## max           168.00         334.00   20.70   97.00
## range         167.00         327.00   18.40   40.00
## sum          4673.00       20513.00 1103.30 8635.00
## median         31.00         207.00    9.70   79.00
## mean           42.10         184.80    9.94   77.79
## SE.mean         3.16           8.65    0.34    0.90
## CI.mean.0.95    6.26          17.15    0.67    1.79
## var          1107.29        8308.74   12.66   90.82
## std.dev        33.28          91.15    3.56    9.53
## coef.var        0.79           0.49    0.36    0.12

As we can see, statistics does not really change that much when we remove the missing values, therefore we will keep the missing values in order to get better representation of the months (for example, if we remove missing data, there will be observed only 9 days of June).

Present the descriptive statistics for the selected variables and explain at least 3 sample statistics (mean, median, etc.).

library(pastecs)
round(stat.desc(mydata[,c(-5,-6,-7)]),2)
##                Ozone SolarRadiation    Wind     Temp
## nbr.val       116.00         146.00  153.00   153.00
## nbr.null        0.00           0.00    0.00     0.00
## nbr.na         37.00           7.00    0.00     0.00
## min             1.00           7.00    1.70    56.00
## max           168.00         334.00   20.70    97.00
## range         167.00         327.00   19.00    41.00
## sum          4887.00       27146.00 1523.50 11916.00
## median         31.50         205.00    9.70    79.00
## mean           42.13         185.93    9.96    77.88
## SE.mean         3.06           7.45    0.28     0.77
## CI.mean.0.95    6.07          14.73    0.56     1.51
## var          1088.20        8110.52   12.41    89.59
## std.dev        32.99          90.06    3.52     9.47
## coef.var        0.78           0.48    0.35     0.12

From May to the end of September, mean value of ozone equals to 42,13 ppb, solar radiation 185,93 lang, average wind speed is 9,96 mph and average temperature is 77,88 °F.

Standard deviation tells us, that Solar radiation has biggest oscilating values around its mean (90,06), since the values of Solar radiation are quite bigger, while the wind has the smallest oscilation of values around the mean (3,52).

Range of values in the dataset is the largest for variable called Solar radiation (range = 327 lang), while the smallest range has the wind (range = 19 mph).

Graph the distribution of the variables using histograms, scatterplots, and/or boxplots. Explain the results.

library(ggplot2)
library(gridExtra)  # For arranging multiple plots

# Histogram for Ozone
p1 <- ggplot(mydata, aes(x = Ozone)) +
  geom_histogram(binwidth = 10, fill = "skyblue", color = "black") +labs(title = "Ozone frequency", x = "Ozone Levels", y = "Frequency") + scale_x_continuous( breaks = seq(0,170,30))+
theme_minimal()

# Histogram for Solar Radiation
p2 <- ggplot(mydata, aes(x = SolarRadiation)) +
  geom_histogram(binwidth = 10, fill = "lightgreen", color = "black") + labs(title = "Solar Radiation frequency", x = "Solar Radiation", y = "Frequency") + theme_minimal()+ scale_x_continuous( breaks = seq(0,300,50))

# Histogram for Wind speed
p3 <- ggplot(mydata, aes(x = Wind)) + geom_histogram(binwidth = 1, fill = "orange", color = "black") + labs(title = " Wind speed frequency", x = "Wind Speed", y = "Frequency") + theme_minimal()

# Histogram for Temperature
p4 <- ggplot(mydata, aes(x = Temp)) + geom_histogram(binwidth = 2, fill = "lightcoral", color = "black") + labs(title = "Temperature frequency", x = "Temperature", y = "Frequency") + theme_minimal()

# Arrange all histograms in a 2x2 grid
grid.arrange(p1, p2, p3, p4, ncol = 2)
## Warning: Removed 37 rows containing non-finite outside the scale range
## (`stat_bin()`).
## Warning: Removed 7 rows containing non-finite outside the scale range
## (`stat_bin()`).

From the set of histograms we can spot, that throughout those 5 months, the most frequent values for different variables are:

  • most frequent value for ozone is 23 ppb,
  • most frequent value fro Solar radiation is 191 lang,
  • most frequent value for Wind Speed is 11.5 mph and
  • most frequent value for Temperature is 81 °F.
library(ggplot2)
library(tidyr)


# Create scatter plots for each variable
p1 <- ggplot(mydata, aes(x = Day, y = Ozone, color = MonthN)) +
  geom_point(size = 1) +
  labs(title = "Ozone Levels (Month)", x = "Day of the month", y = "Ozone Levels [ppb]") +
  theme_minimal()

p2 <- ggplot(mydata, aes(x = Day, y = SolarRadiation, color = MonthN)) +
  geom_point(size = 1) +
  labs(title = "Solar Radiation (Month)", x = "Day of the month", y = "Solar Radiation [lang]") +
  theme_minimal()

p3 <- ggplot(mydata, aes(x = Day, y = Wind, color = MonthN)) +
  geom_point(size = 1) +
  labs(title = "Wind speed (Month)", x = "Day of the month", y = "Wind speed [mph]") +
  theme_minimal()

p4 <- ggplot(mydata, aes(x = Day, y = Temp, color = MonthN)) +
  geom_point(size = 1) +
  labs(title = "Temperature (Month)", x = "Day of the month", y = "Temperature [°F]") +
  theme_minimal()

# Arrange all scatter plots in a 2x2 grid
grid.arrange(p1, p2, p3, p4, ncol = 2)
## Warning: Removed 37 rows containing missing values or values outside the scale range
## (`geom_point()`).
## Warning: Removed 7 rows containing missing values or values outside the scale range
## (`geom_point()`).

From the graphs above we can see:

  • that May and September had lower ozone levels as June,July and August (density of data is there the biggest),
  • solar radiation was on the average highest in May, June and July,
  • wind speed was through mentioned months oscillating equaly,
  • temperature was the lowest in may and the highest in July and August.

EXERCISE 2

You have a data set for 100 MBA students of the current generation. In the previous year, the average grade in this program was 74.

#Importing the dataset Business School.xlsx

library(readxl)
mydata <- read_xlsx("./Business School.xlsx")

mydata <- as.data.frame(mydata) 

#head(mydata)
colnames(mydata) <- gsub(" ", "", colnames(mydata)) #removes the space between variables for further use
head(mydata)
##   StudentID  UndergradDegree UndergradGrade MBAGrade WorkExperience
## 1         1         Business           68.4     90.2             No
## 2         2 Computer Science           70.2     68.7            Yes
## 3         3          Finance           76.4     83.3             No
## 4         4         Business           82.6     88.7             No
## 5         5          Finance           76.9     75.4             No
## 6         6 Computer Science           83.3     82.1             No
##   Employability(Before) Employability(After) Status AnnualSalary
## 1                   252                  276 Placed       111000
## 2                   101                  119 Placed       107000
## 3                   401                  462 Placed       109000
## 4                   287                  342 Placed       148000
## 5                   275                  347 Placed       255500
## 6                   254                  313 Placed       103500

Graph the distribution of undergrad degrees using the ggplot function. Which degree is the most common?

library(ggplot2)
ggplot(mydata, aes(x = UndergradDegree)) + 
  geom_bar(colour = "black") +
  ylab("Frequency")+
  xlab("Undergrad Degree")+
  theme(plot.title = element_text(hjust = 0.5))+
  ggtitle("Frequency distribution of Undergrad Degree")

The most common degree is Business degree.

Show the descriptive statistics of the Annual Salary and its distribution with the histogram (use the ggplot function). Describe the distribution.

library(pastecs)
round(stat.desc(mydata$AnnualSalary))
##      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            0
library(ggplot2)
ggplot(mydata, aes(x = AnnualSalary)) + 
  geom_histogram(colour = "black") +
  scale_x_continuous(labels = scales::label_number())+
  ylab("Frequency")+
  xlab("Annual salary")+
  theme(plot.title = element_text(hjust = 0.5))+
  scale_x_continuous( breaks = seq(0,400000,50000))+
  ggtitle("Frequency distribution of annual salary")
## Scale for x is already present.
## Adding another scale for x, which will replace the existing scale.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.

Distribution is asymmetrical to the right side (positive coefficient of assimetricity).The most frequent annual salary is 100k € per year.

Test the following hypothesis: 𝐻0:𝜇MBA Grade=74. Explain the result and interpret the effect size.

t.test(mydata$UndergradGrade,
       mu = 74,
       alternative = "two.sided")
## 
##  One Sample t-test
## 
## data:  mydata$UndergradGrade
## t = 3.8851, df = 99, p-value = 0.0001849
## alternative hypothesis: true mean is not equal to 74
## 95 percent confidence interval:
##  75.41841 78.37959
## sample estimates:
## mean of x 
##    76.899
library(effectsize)

effectsize::cohens_d(mydata$UndergradGrade,mu = 74)
## Cohen's d |       95% CI
## ------------------------
## 0.39      | [0.18, 0.59]
## 
## - Deviation from a difference of 74.
interpret_cohens_d(0.39,rules = "sawilowsky2009")
## [1] "small"
## (Rules: sawilowsky2009)

We can conclude that we must reject null hypothesis, since p < 0.001 , which is smaller as 0.05.

Average grade is higher - the average grade increase is “small” sized.

EXERCISE 3

You analyze the price per m2 for a sample of apartments in the Ljubljana region (Apartments.xlsx in Task 3 folder). Follow the questions in R Markdown included in the folder.

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

Change categorical variables into factors.

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

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

summary(mydata[c(6,7)])
##  ParkingN BalconyN
##  No :42   No :48  
##  Yes:43   Yes:37

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

t.test(mydata$Price,
       mu = 1900,
       alternative = "two.sided")
## 
##  One Sample t-test
## 
## data:  mydata$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
library(effectsize)

effectsize::cohens_d(mydata$Price,mu = 1900)
## Cohen's d |       95% CI
## ------------------------
## 0.31      | [0.10, 0.53]
## 
## - Deviation from a difference of 1900.
interpret_cohens_d(0.31,rules = "sawilowsky2009")
## [1] "small"
## (Rules: sawilowsky2009)

We can conclude that we must reject null hypothesis, since p = 0.0047, which is smaller as 0.05.

Price increased - the price increase is “small” sized.

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.

library(car)
## Loading required package: carData
scatterplot(x=mydata$Age,
            y=mydata$Price,
            xlab="Age of the apartment",
            ylab = "Price of the apartment",
            smooth = FALSE,
            boxplot = FALSE)

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
cor(mydata$Price, mydata$Age)
## [1] -0.230255

  • Price_estimated = 2185.455 - 8.975 * Age

  • If age goes up by one year, the price of the apartment decreases by 8.975€/m^2 on average (p = 0.034).

  • The linear relationship between age and price of the apartment is negative and weak (|r| = 0,23).

  • 5.3% of variability in price is explained by linear effect of the age of the apartment.

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

library(car)
scatterplotMatrix(mydata[ , c(1,2,3)], #Scatterplot matrix
                  smooth = FALSE)

Based on the dispersion of the values between different variables we can assume we don’t have problems with multicolinearity.

library(Hmisc)
## 
## Attaching package: 'Hmisc'
## The following objects are masked from 'package:base':
## 
##     format.pval, units
rcorr(as.matrix(mydata[ , c(1,2,3)])) #Correlation matrix
##            Age Distance Price
## Age       1.00     0.04 -0.23
## Distance  0.04     1.00 -0.63
## Price    -0.23    -0.63  1.00
## 
## n= 85 
## 
## 
## P
##          Age    Distance Price 
## Age             0.6966   0.0340
## Distance 0.6966          0.0000
## Price    0.0340 0.0000

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

fit2 <- lm(Price ~ Age + Distance, #Dependent and explanatory variables
           data = mydata) #Name of the data frame
fit4 <- lm(Price ~ Age + Distance, #for later
           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

Check the multicolinearity with VIF statistics. Explain the findings.

vif(fit2) #Multicolinearity
##      Age Distance 
## 1.001845 1.001845

Since VIF values for both explainatory variables are 1, it means that there is no strong multicolinearity between those 2 variables (they don’t strongly influence each other).

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

mydata$StdResid <- round(rstandard(fit2), 3) #Standardized residuals
mydata$CooksD <- round(cooks.distance(fit2), 3) #Cooks distances

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

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

head(mydata[order(-mydata$CooksD),c(-4,-5,-6,-7)], 6) #Six units with highest value of Cooks distance
##    Age Distance Price StdResid CooksD
## 38   5       45  2180    2.577  0.320
## 55  43       37  1740    1.445  0.104
## 33   2       11  2790    2.051  0.069
## 53   7        2  1760   -2.152  0.066
## 22  37        3  2540    1.576  0.061
## 39  40        2  2400    1.091  0.038
mydata$ID <- 1:nrow(mydata) #added ID variable
library(dplyr)
## 
## Attaching package: 'dplyr'
## The following objects are masked from 'package:Hmisc':
## 
##     src, summarize
## The following object is masked from 'package:car':
## 
##     recode
## The following object is masked from 'package:gridExtra':
## 
##     combine
## The following objects are masked from 'package:pastecs':
## 
##     first, last
## The following objects are masked from 'package:stats':
## 
##     filter, lag
## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union
mydata <- mydata %>%
  filter (!ID %in% c(33,38,53,55)) #Removing units, which have big impact 

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

#repeating the calculations of Residuals and Cooks Distances
mydata$StdResid <- round(rstandard(fit2), 3) 
mydata$CooksD <- round(cooks.distance(fit2), 3) 

#repeating the graphical interpretation
hist(mydata$StdResid, 
     xlab = "Standardized residuals", 
     ylab = "Frequency", 
     main = "Histogram of standardized residuals")

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

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

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)

library(olsrr)
## 
## Attaching package: 'olsrr'
## The following object is masked from 'package:datasets':
## 
##     rivers
ols_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          =    1.64762 
##  Prob > Chi2   =    0.1992832

P value from Breusch-Pagan test indicates, that we cannot reject null hypothesis (p = 0,199 (which is bigger as 0.05)), therefore we can assume homoskedasticity.

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

shapiro.test(mydata$StdResid)
## 
##  Shapiro-Wilk normality test
## 
## data:  mydata$StdResid
## W = 0.94297, p-value = 0.001286
  • Shapiro-Wilk normality test gives us p = 0,0013 and therefore we must reject null hypothesis - standard residuals distribution is not normal, but we do not need to worry, since our sample consists of 85 units, therefore we can assume t- distribution (Central Limit Theorem).

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 = mydata)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -408.72 -217.93  -38.86  220.60  506.18 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 2492.715     75.544  32.997  < 2e-16 ***
## Age           -7.496      3.169  -2.365   0.0205 *  
## Distance     -24.574      2.700  -9.100  6.9e-14 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 259.4 on 78 degrees of freedom
## Multiple R-squared:  0.5324, Adjusted R-squared:  0.5204 
## F-statistic:  44.4 on 2 and 78 DF,  p-value: 1.335e-13

  • If the age of the apartment increases by 1 year, the price of the apartment decreases by 7.934€ per m^2, assuming that the other explainatory variables remain unchanged.

  • If the distance of the apartment from the city center increases by 1 km, the price of the apartment decreases by 20.667€ per m^2, assuming that the other explainatory variables remain unchanged.

  • 53,24% of the variabillity of the price per m^2 is explained by the linear effect of age and distance of the apartment.

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 + ParkingN + BalconyN, 
           data = mydata)

With function anova check if model fit3 fits data better than model fit2.

anova(fit2, fit3) #Comparison of two regression models that differ in the number of expl. variables
## Analysis of Variance Table
## 
## Model 1: Price ~ Age + Distance
## Model 2: Price ~ Age + Distance + ParkingN + BalconyN
##   Res.Df     RSS Df Sum of Sq      F  Pr(>F)  
## 1     78 5248925                              
## 2     76 4915944  2    332981 2.5739 0.08287 .
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

By the p value we can see, that fit2 and fit3 models are equal ( p = 0.083)

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 + ParkingN + BalconyN, data = mydata)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -403.16 -204.08  -41.24  239.52  528.62 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 2372.693     93.344  25.419  < 2e-16 ***
## Age           -6.906      3.118  -2.215   0.0298 *  
## Distance     -22.254      2.839  -7.837 2.25e-11 ***
## ParkingNYes  137.479     60.854   2.259   0.0267 *  
## BalconyNYes   17.234     57.099   0.302   0.7636    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 254.3 on 76 degrees of freedom
## Multiple R-squared:  0.5621, Adjusted R-squared:  0.539 
## F-statistic: 24.38 on 4 and 76 DF,  p-value: 5.29e-13

  • Given the age, distance and availability of balcony of the apartment,the group of the apartments with the parking, are on average more expensive for 137,479€, then the group of the apartments without the parking.

  • Balcony - due to high p (0.76), the variable is not significant.

  • Hypothesis:

    • H0: dRo^2 = 0

    • H1: dRo^2 > 0

      The F- statistics shows that we can reject null hypothesis and this model explains at least some of the variability.

Save fitted values and claculate the residual for apartment ID2.

mydata$Fitted <- fitted.values(fit3) 
mydata$Residuals <- residuals(fit3)
head(mydata[ ,colnames(mydata)%in% c("Price", "Fitted", "Residuals")],3)
##   Price   Fitted Residuals
## 1  1640 1718.475 -78.47475
## 2  2800 2363.611 436.38945
## 3  1660 1701.241 -41.24101

Based on the estimation of regression function, we would expect the price for the apartment ID2 to be 2363.511 € per m^2, but the actual price of ID2 is 2800€ per m^2. Therefore the residual is the difference between those 2 values (436,3894).