Inja Dovšak, 30. 9. 2022, R take-home exam

Task #1

The World Happiness Report uses global survey data to report how people evaluate their own lives in the countries worldwide.The survey measure of SWB (Subjective Well-being) is from the Gallup World Poll (GWP).

library(readr)
library(pastecs)
library(psych)
library(dplyr)

## 
## Attaching package: 'dplyr'

## 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

library(car)

## Loading required package: carData

## 
## Attaching package: 'car'

## The following object is masked from 'package:dplyr':
## 
##     recode

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

library(ggplot2)

## 
## Attaching package: 'ggplot2'

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

library(car)
library(reshape2)
library(readxl)

podatki <- read.table("~/Library/CloudStorage/OneDrive-Personal/IMB/Bootcamp R/R data exam/WorldHappinessReport2022.csv",
                      header = TRUE,
                      sep = ";",
                      dec = ",")

head(podatki)

##   ID Happiness.Rank     Country         Region Happiness.Score
## 1  0              1     Finland Western Europe           7.821
## 2  1              2     Denmark Western Europe           7.636
## 3  2              3     Iceland Western Europe           7.557
## 4  3              4 Switzerland Western Europe           7.512
## 5  4              5 Netherlands Western Europe           7.415
## 6  5              6 Luxembourg*              -           7.404
##   Economy..GDP.per.Capita. Family..Social.Support. Health..Life.Expectancy.
## 1                    1.892                   1.258                    0.775
## 2                    1.953                   1.243                    0.777
## 3                    1.936                   1.320                    0.803
## 4                    2.026                   1.226                    0.822
## 5                    1.945                   1.206                    0.787
## 6                    2.209                   1.155                    0.790
##   Freedom Trust..Government.Corruption. Generosity Year
## 1   0.736                         0.534      0.109 2022
## 2   0.719                         0.532      0.188 2022
## 3   0.718                         0.191      0.270 2022
## 4   0.677                         0.461      0.147 2022
## 5   0.651                         0.419      0.271 2022
## 6   0.700                         0.388      0.120 2022

colnames(podatki) <- c("ID", "HappinessRank", "Country", "Region", "HappinessScore", "GDP", "FamilySocialSupport" , "LifeExpectancy", "Freedom", "Trust", "Generosity", "Year")

podatki1 <- podatki[, c(1, 3, 4, 2, 5, 6, 7, 8, 9, 10, 11, 12)]
head(podatki1)

##   ID     Country         Region HappinessRank HappinessScore   GDP
## 1  0     Finland Western Europe             1          7.821 1.892
## 2  1     Denmark Western Europe             2          7.636 1.953
## 3  2     Iceland Western Europe             3          7.557 1.936
## 4  3 Switzerland Western Europe             4          7.512 2.026
## 5  4 Netherlands Western Europe             5          7.415 1.945
## 6  5 Luxembourg*              -             6          7.404 2.209
##   FamilySocialSupport LifeExpectancy Freedom Trust Generosity Year
## 1               1.258          0.775   0.736 0.534      0.109 2022
## 2               1.243          0.777   0.719 0.532      0.188 2022
## 3               1.320          0.803   0.718 0.191      0.270 2022
## 4               1.226          0.822   0.677 0.461      0.147 2022
## 5               1.206          0.787   0.651 0.419      0.271 2022
## 6               1.155          0.790   0.700 0.388      0.120 2022

Description:

This is the data from World Happiness Report 2022.

ID - Represents an identification number of input. Happiness.Rank - The ranking number of a country according to Happiness score

Country - Variable states the country for data

Region - Variable states the region of a country for data

HappinessRank - the ranking number or a country based on HappineessScore

HappinessScore - score of subjective well being (1-10)

GDP - Economy..GDP.per.Capita.: The statistics of GDP per capita in purchasing power parity

FamilySocialSupport. - Variable shows the answers to questions measuring having someone to count on in times of trouble. It is the national average of the binary responses (either 0 or 1)

LifeExpectancy. - Healthy life expectancies at birth that are based on the data extracted from the World Health Organization’s (WHO) Global Health Observatory data repository

Freedom - National average of responses to the GWP question “Are you satisfied or dissatisfied with your freedom to choose what you do with your life?” (0 -1)

Trust - Trust government corruption: national average of the survey responses to two questions in the GWP: “Is corruption widespread throughout the government or not” and “Is corruption widespread within businesses or not?” The overall perception is the average of the binary 0-or-1 responses.

Generosity - Residual of regressing national average of response to the GWP question “Have you donated money to a charity in the past month?” on GDP per capita.

Year - Year of the report and data results

summary(podatki1 [,c(5, 6, 7, 8, 9, 10, 11)])

##  HappinessScore       GDP        FamilySocialSupport LifeExpectancy  
##  Min.   :2.404   Min.   :0.000   Min.   :0.0000      Min.   :0.0000  
##  1st Qu.:4.889   1st Qu.:1.095   1st Qu.:0.7320      1st Qu.:0.4632  
##  Median :5.569   Median :1.446   Median :0.9575      Median :0.6215  
##  Mean   :5.554   Mean   :1.410   Mean   :0.9059      Mean   :0.5862  
##  3rd Qu.:6.305   3rd Qu.:1.785   3rd Qu.:1.1142      3rd Qu.:0.7198  
##  Max.   :7.821   Max.   :2.209   Max.   :1.3200      Max.   :0.9420  
##     Freedom           Trust           Generosity    
##  Min.   :0.0000   Min.   :0.00000   Min.   :0.0000  
##  1st Qu.:0.4405   1st Qu.:0.06825   1st Qu.:0.0890  
##  Median :0.5435   Median :0.11950   Median :0.1325  
##  Mean   :0.5172   Mean   :0.15478   Mean   :0.1474  
##  3rd Qu.:0.6260   3rd Qu.:0.19850   3rd Qu.:0.1978  
##  Max.   :0.7400   Max.   :0.58700   Max.   :0.4680

mean(podatki1$HappinessScore)

## [1] 5.553575

mean(podatki1$Freedom)

## [1] 0.517226

The average Happiness score across countries is 5,55 and average response across all countries to questions regarding Freedom is 0,52.

median(podatki1$HappinessScore)

## [1] 5.5685

median(podatki1$Freedom)

## [1] 0.5435

Half of the countries have a HappinessScore lower than 5,57 and half have a HappinessScore higher than 5,57. Half of the countries had a response to the measure of Freedom lower than 0,54 and half higher than 0,54.

hist(podatki1$HappinessScore, 
     main = "HappinessScore frequency", 
     xlab = "Happiness Score", 
     ylab = "Frequency")

We can see that the histogram of HappinessScore frequency is slightly skewed to the left which tells us that HappinessScore distribution is higher on the right side of the graph.

scatterplot(y =podatki1$Freedom,
            x =podatki1$HappinessScore,
            ylab= "Satisfaction of freedom",
            xlab= "Happiness Score", 
            smooth=FALSE)

Happiness score is increasing with the feeling of satisfaction with freedom to choose what to do in life.

ggplot(podatki1, aes(x=Region, y=HappinessScore)) + 
  geom_boxplot() +
  xlab("Region")

We can see that HappinessScore median is the highest in the region of New Zeland, Europe and North America. In Australia there is one country with extremely high Happiness Score. East and Northern Africa is the region with the highest distribution od Happiness Score rates across countries.

Task #2

A researcher examined the effects of online schooling on body weight in ninth graders (Body mass.csv in Task 2 folder). Based on data from the 2018/2019 school year, the average weight was 59.5 kg. At the beginning of the 2021/2022 school year, researchers randomly selected 50 ninth graders and measured their weight. 1 Show the descriptive statistics of body mass and its distribution with the histogram. 2 Test the following hypothesis with program R: 𝐻0:𝜇 = 59.5. 3 Explain the results and interpret the effect size.

mydatat2 <- read.table("~/Library/CloudStorage/OneDrive-Personal/IMB/Bootcamp R/R data exam/Task 2/Body mass.csv",
                      header = TRUE,
                      sep = ";",
                      dec = ",")

head(mydatat2)

##   ID Mass
## 1  1 62.1
## 2  2 64.5
## 3  3 56.5
## 4  4 53.4
## 5  5 61.3
## 6  6 62.2

Description:

ID: Identification number of a ninth grader randomly selected by researchers in the beginning of the 2021/2022 school year to measure the weight of 50 ninth graders.

Mass: Weight of a ninth grader in measured in Kg at the beginning of 2021/2022 school year

1) Show the descriptive statistics of body mass and its distribution with the histogram.

round(stat.desc(mydatat2$Mass), 2)

##      nbr.val     nbr.null       nbr.na          min          max        range 
##        50.00         0.00         0.00        49.70        83.20        33.50 
##          sum       median         mean      SE.mean CI.mean.0.95          var 
##      3143.80        62.80        62.88         0.85         1.71        36.14 
##      std.dev     coef.var 
##         6.01         0.10

describe(mydatat2[ , -1])

##    vars  n  mean   sd median trimmed  mad  min  max range skew kurtosis   se
## X1    1 50 62.88 6.01   62.8   62.56 3.34 49.7 83.2  33.5 0.85     2.11 0.85

This data set contains a sample of 50 observations. We can see that in this sample the student with the lowest weight weighs 49,70 Kg and the student who weighs the most, weighs 83,20 Kg. Half (50 %) of the students weigh 62.80 Kg or less and half of students in the sample weigh more than 62.80 Kg. The estimated average weight of a 9th grader at the beginning of 2021/2022 school year is 62.88 kg. Strandard deviation in this sample is 6.01 Kg which tells us that values vary around the mean value for 6.01 Kg up or down.

hist(mydatat2$Mass,
     main = "Distribution of weight",
     xlab = "Weight",
     ylab = "Frequency",
     breaks = seq(from = 0, to = 100, by = 5))

Distribution of this sample as seen on the historam is asymetric to right (positive skewness).

2) Test the following hypothesis with program R: 𝐻0: 𝜇 = 59.5.

t.test(mydatat2$Mass,
       mu = 59.5)

## 
##  One Sample t-test
## 
## data:  mydatat2$Mass
## t = 3.9711, df = 49, p-value = 0.000234
## alternative hypothesis: true mean is not equal to 59.5
## 95 percent confidence interval:
##  61.16758 64.58442
## sample estimates:
## mean of x 
##    62.876

t <- 3.9711
r <- sqrt((t^2) / (t^2 + 49))
r

## [1] 0.4934295

3) Explain the results and interpret the effect size.

With one sample t-test we tested the hypothesis and got the p-value, which tells us the probability that the null hypothesis (H0) is TRUE.

In our case the p-value is 0.02 %. We can say that H0, saying that average height is 59.5 Kg, is extremely unlikely and we can reject the H0 at p < 0.001.

We can say with 95% certainty that the true arithmetic mean is between 61.17 and 64.58.

The calculated effect size is 0.49 which means that it is medium (0.3 to 0.5).

Task 3

Import the dataset Apartments.xlsx

mydatat3 <- read_xlsx("~/Library/CloudStorage/OneDrive-Personal/IMB/Bootcamp R/R data exam/Task 3/Apartments.xlsx")

head(mydatat3)

## # A tibble: 6 × 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

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

mydatat3$ParkingFactor <- factor(mydatat3$Parking,
                           levels = c(0, 1),
                           labels = c("No", "Yes"))

mydatat3$BalconyFactor <- factor(mydatat3$Balcony,
                           levels = c(0, 1),
                           labels = c("No", "Yes"))

head(mydatat3)

## # A tibble: 6 × 7
##     Age Distance Price Parking Balcony ParkingFactor BalconyFactor
##   <dbl>    <dbl> <dbl>   <dbl>   <dbl> <fct>         <fct>        
## 1     7       28  1640       0       1 No            Yes          
## 2    18        1  2800       1       0 Yes           No           
## 3     7       28  1660       0       0 No            No           
## 4    28       29  1850       0       1 No            Yes          
## 5    18       18  1640       1       1 Yes           Yes          
## 6    28       12  1770       0       1 No            Yes

mydatat3$ID <- seq.int(nrow(mydatat3))

mydatat3 <- mydatat3[ , c(1,2,3,4,5,6,7,8)]

print(mydatat3)

## # A tibble: 85 × 8
##      Age Distance Price Parking Balcony ParkingFactor BalconyFactor    ID
##    <dbl>    <dbl> <dbl>   <dbl>   <dbl> <fct>         <fct>         <int>
##  1     7       28  1640       0       1 No            Yes               1
##  2    18        1  2800       1       0 Yes           No                2
##  3     7       28  1660       0       0 No            No                3
##  4    28       29  1850       0       1 No            Yes               4
##  5    18       18  1640       1       1 Yes           Yes               5
##  6    28       12  1770       0       1 No            Yes               6
##  7    14       20  1850       0       1 No            Yes               7
##  8    18        6  1970       1       1 Yes           Yes               8
##  9    22        7  2270       1       0 Yes           No                9
## 10    25        2  2570       1       0 Yes           No               10
## # … with 75 more rows

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

t.test(mydatat3$Price ,
       mu = 1900 )

## 
##  One Sample t-test
## 
## data:  mydatat3$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

We reject H0 at p-value p-value = 0.047 or 0.47 % taking 5 % as a cut point. It is unlikely that average price of an apartment is 1900 eur/m2.Therefore we accept the alternative hypothesis that says: “The average price is not equal to 1900 eur.”.

With can say with 95 % confidence, that the average price is between 1937.4 eur and 2100.4 eur, which is higher than sugested by the null hypothesis.

Sample estimated mean is 2018.9 eur.

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 = mydatat3) 

summary(fit1)

## 
## Call:
## lm(formula = Price ~ Age, data = mydatat3)
## 
## 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

Explanation:

Estimate of regression coefficient is -8.975, which means that with every additional year of age of apartment, average price decreases by 8.975 eur per m2, assuming that everything else remains unchanged (p < 0.05).

Coefficient of determination is 0.053, which means that 5.3 % of variability of Price is explained by the effect of Age.

sqrt(0.05302)

## [1] 0.2302607

Coefficient of correlation is -0.23, which tells us that the higher the age of apartment, the lower the price per m2.

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

library(car)
scatterplotMatrix(mydatat3 [ , c(3, 1, 2)],
                  smooth = FALSE)

According to the scatterplot matrix there is no sign of potential problem with multicollinearty as values are distanced and scattered around the line.

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

fit2 <- lm(Price ~ Age + Distance,
          data = mydatat3) 

summary(fit2)

## 
## Call:
## lm(formula = Price ~ Age + Distance, data = mydatat3)
## 
## 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

Chech the multicolinearity with VIF statistics. Explain the findings.

vif(fit2)

##      Age Distance 
## 1.001845 1.001845

mean(vif(fit2))

## [1] 1.001845

All values of VIF < 5 , therefore there is no multicolinearity or stron relationship between variables.

Calculate standardized residuals and Cooks Distances for model fit2. Remove any potentially problematic case (outlier or unit with big influence).

mydatat3$StdResid <- round(rstandard(fit2), 3) 

head(mydatat3[order(-mydatat3$StdResid),])

## # A tibble: 6 × 9
##     Age Distance Price Parking Balcony ParkingFactor BalconyFactor    ID StdRe…¹
##   <dbl>    <dbl> <dbl>   <dbl>   <dbl> <fct>         <fct>         <int>   <dbl>
## 1     5       45  2180       1       1 Yes           Yes              38    2.58
## 2     2       11  2790       1       0 Yes           No               33    2.05
## 3    18        1  2800       1       0 Yes           No                2    1.78
## 4    18        1  2800       1       1 Yes           Yes              61    1.78
## 5     8        2  2820       1       0 Yes           No               58    1.66
## 6    10        1  2810       0       0 No            No               57    1.60
## # … with abbreviated variable name ¹StdResid

hist(mydatat3$StdResid, 
     xlab = "Standardized residuals", 
     ylab = "Frequency", 
     main = "Histogram of standardized residuals")

All values of standardized residuals are between -3 and 3 and therefore we will not remove any unit.

mydatat3$CooksD <- round(cooks.distance(fit2), 3) 

head(mydatat3[order(-mydatat3$CooksD),])

## # A tibble: 6 × 10
##     Age Distance Price Parking Balcony ParkingFac…¹ Balco…²    ID StdRe…³ CooksD
##   <dbl>    <dbl> <dbl>   <dbl>   <dbl> <fct>        <fct>   <int>   <dbl>  <dbl>
## 1     5       45  2180       1       1 Yes          Yes        38    2.58  0.32 
## 2    43       37  1740       0       0 No           No         55    1.44  0.104
## 3     2       11  2790       1       0 Yes          No         33    2.05  0.069
## 4     7        2  1760       0       1 No           Yes        53   -2.15  0.066
## 5    37        3  2540       1       1 Yes          Yes        22    1.58  0.061
## 6    40        2  2400       0       1 No           Yes        39    1.09  0.038
## # … with abbreviated variable names ¹ParkingFactor, ²BalconyFactor, ³StdResid

hist(mydatat3$CooksD, 
     xlab = "Cooks distance", 
     ylab = "Frequency", 
     main = "Histogram of Cooks distances")

We see there is a unit with significantly higher Cooks distance, according to the histogram above. We see that the Cooks distance of 0.32 that is significantly higher belongs to the apartment with ID = 38. We will remove this unit.

mydatat3 <- mydatat3[c(-38), ] 
head(mydatat3[order(-mydatat3$CooksD),])

## # A tibble: 6 × 10
##     Age Distance Price Parking Balcony ParkingFac…¹ Balco…²    ID StdRe…³ CooksD
##   <dbl>    <dbl> <dbl>   <dbl>   <dbl> <fct>        <fct>   <int>   <dbl>  <dbl>
## 1    43       37  1740       0       0 No           No         55    1.44  0.104
## 2     2       11  2790       1       0 Yes          No         33    2.05  0.069
## 3     7        2  1760       0       1 No           Yes        53   -2.15  0.066
## 4    37        3  2540       1       1 Yes          Yes        22    1.58  0.061
## 5    40        2  2400       0       1 No           Yes        39    1.09  0.038
## 6     8        2  2820       1       0 Yes          No         58    1.66  0.037
## # … with abbreviated variable names ¹ParkingFactor, ²BalconyFactor, ³StdResid

fit2 <- lm(Price ~ Age + Distance,
          data = mydatat3)

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

mydatat3$StdfittedValues <- scale(fit2$fitted.values)

library(car)
scatterplot(y = mydatat3$StdResid, x= mydatat3$StdfittedValues,
            xlab = "Standarized fitted values",
            ylab = "Standarized residuals",
            boxplots = FALSE, 
            regLine = FALSE, 
            smooth = FALSE)

According to the scatterplot we can not see that there is heteroskedasticity. We run further testing with the Breusch Pagan Test

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          =    2.927455 
##  Prob > Chi2   =    0.08708469

We can see that p >0.05 and therefore we can not reject H0 that states that the variance is constant (homoskedasticity). Therefore we conclude that there is no heteroskedasticity.

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

hist(mydatat3$StdResid,
     xlab = "Standardized residuals",
     ylab = "Frequency",
     main = "Histogram of standardized residuals")

shapiro.test(mydatat3$StdResid)

## 
##  Shapiro-Wilk normality test
## 
## data:  mydatat3$StdResid
## W = 0.94879, p-value = 0.002187

We can see from the histogram that the distribution of standardized residuals is not normal.

Shapiro-Wilk normallity test tested the H0: Variable is normally distributed H1: Variable is NOT normally distributed

With p-value: 0.0018 (p < 0.05) we can reject the null hypothesis and accept the alternative -> variable is NOT normally distributed.

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

fit2 <- lm(Price ~ Age + Distance,
          data = mydatat3)

summary(fit2)

## 
## Call:
## lm(formula = Price ~ Age + Distance, data = mydatat3)
## 
## 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

sqrt(0.4831)

## [1] 0.695054

Explanation:

Estimate of regression coefficient for Age is -6.46, which means that with every additional year of age of apartment, average price decreases by -6.46 eur per m2, assuming that everything else remains unchanged (p < 0.05).

Estimate of regression coefficient for the explanatory variable Distance is -22.562, which means that with every 1 Km increase in Distance, the Price on average decreases by 22.562 eur per m2 (p < 0.001), assuming that all other variables remain unchanged.

Coefficient of determination is 0.4831, which tells us that 48.31 % of variability of Price is explained by linear effect of Age and Distance.

Multiple coefficient of correlation is -0.695, which means that linear relationship between Price and the two explanatory variables (Age and Distance) is medium strong.

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 + ParkingFactor + BalconyFactor,
          data = mydatat3)

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 + ParkingFactor + BalconyFactor
##   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 ' ' 1

H0: fit2 fits data better than fit3 H1: fit3 fits data better than fit2

p-value is 0.018, therefore we can reject H0 and accept H1 and so conclude that fit3 fits the data better than 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 + ParkingFactor + BalconyFactor, 
##     data = mydatat3)
## 
## 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 ***
## ParkingFactorYes  167.531     62.864   2.665  0.00933 ** 
## BalconyFactorYes  -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

We can see that given the Age, Distance and Balcony, the apartments with Parking, are on average priced 180.399 eur higher per m2 than those without Parking. (p < 0.01).

Estimate of regression coefficient for the explanatory variable BalconyFactor is -26.178, but p-value is 0.66 (p > 0,05), therefore we can not say that Balcony ha affect on the Price.

ANOVA H0: R2 = 0 H1: R2 > 0

P-value: 3.018e-12 (p < 0.001). We reject H0 which means there is linear correlation between dependent (Price) and at least one explanatory variable. Model is good (R squared > 0)

Save fitted values and claculate the residual for apartment ID2.

mydatat3$FittedValues <- fitted.values(fit3) 

mydatat3$Residuals <- residuals(fit3)

head(fit3$residuals)

##          1          2          3          4          5          6 
##  -65.95206  427.80292  -61.15894  286.56890 -372.24396 -138.17733

Dovsak_R-exam_1-3

Inja Dovšak

2022-09-30

Inja Dovšak, 30. 9. 2022, R take-home exam

Task #1

The World Happiness Report uses global survey data to report how people evaluate their own lives in the countries worldwide.The survey measure of SWB (Subjective Well-being) is from the Gallup World Poll (GWP).

Task #2

1) Show the descriptive statistics of body mass and its distribution with the histogram.

2) Test the following hypothesis with program R: 𝐻0: 𝜇 = 59.5.

3) Explain the results and interpret the effect size.

Task 3

Import the dataset Apartments.xlsx

Change categorical variables into factors

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

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.

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

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

Chech the multicolinearity with VIF statistics. Explain the findings.

Calculate standardized residuals and Cooks Distances for model fit2. Remove any potentially problematic case (outlier or unit with big influence).

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

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

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

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

With function anova check if model 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?

Save fitted values and claculate the residual for apartment ID2.