Homework 2
Bartlomiej Swiderski
2022-09-15
Exercise 1
- a)
library(readxl)
brokerage <- read_excel("/Users/macuser/Downloads/Brokerage Satisfaction.xlsx")
as.data.frame(brokerage)## Brokerage Satisfaction_with_Trade_Price
## 1 Scottrade, Inc. 3.2
## 2 Charles Schwab 3.3
## 3 Fidelity Brokerage Services 3.1
## 4 TD Ameritrade 2.8
## 5 E*Trade Financial 2.9
## 6 (Not listed) 2.4
## 7 Vanguard Brokerage Services 2.7
## 8 USAA Brokerage Services 2.4
## 9 Thinkorswim 2.6
## 10 Wells Fargo Investments 2.3
## 11 Interactive Brokers 3.7
## 12 Zecco.com 2.5
## 13 Firstrade Securities 3.0
## 14 Banc of America Investment Services 1.0
## Satisfaction_with_Speed_of_Execution
## 1 3.1
## 2 3.1
## 3 3.3
## 4 3.5
## 5 3.2
## 6 3.2
## 7 3.8
## 8 3.7
## 9 2.6
## 10 2.7
## 11 3.9
## 12 2.5
## 13 3.0
## 14 4.0
## Overall_Satisfaction_with_Electronic_Trades
## 1 3.2
## 2 3.2
## 3 4.0
## 4 3.7
## 5 3.0
## 6 2.7
## 7 2.7
## 8 3.4
## 9 2.7
## 10 2.3
## 11 4.0
## 12 2.5
## 13 3.0
## 14 2.0colnames(brokerage)## [1] " Brokerage"
## [2] "Satisfaction_with_Trade_Price"
## [3] "Satisfaction_with_Speed_of_Execution"
## [4] "Overall_Satisfaction_with_Electronic_Trades"summary(brokerage)## Brokerage Satisfaction_with_Trade_Price
## Length:14 Min. :1.000
## Class :character 1st Qu.:2.425
## Mode :character Median :2.750
## Mean :2.707
## 3rd Qu.:3.075
## Max. :3.700
## Satisfaction_with_Speed_of_Execution
## Min. :2.500
## 1st Qu.:3.025
## Median :3.200
## Mean :3.257
## 3rd Qu.:3.650
## Max. :4.000
## Overall_Satisfaction_with_Electronic_Trades
## Min. :2.000
## 1st Qu.:2.700
## Median :3.000
## Mean :3.029
## 3rd Qu.:3.350
## Max. :4.000brokerage_lm1 <- lm(Overall_Satisfaction_with_Electronic_Trades~Satisfaction_with_Trade_Price+Satisfaction_with_Speed_of_Execution, data=brokerage)
summary(brokerage_lm1)##
## Call:
## lm(formula = Overall_Satisfaction_with_Electronic_Trades ~ Satisfaction_with_Trade_Price +
## Satisfaction_with_Speed_of_Execution, data = brokerage)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.58886 -0.13863 -0.09120 0.05781 0.64613
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -0.6633 0.8248 -0.804 0.438318
## Satisfaction_with_Trade_Price 0.7746 0.1521 5.093 0.000348 ***
## Satisfaction_with_Speed_of_Execution 0.4897 0.2016 2.429 0.033469 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.3435 on 11 degrees of freedom
## Multiple R-squared: 0.7256, Adjusted R-squared: 0.6757
## F-statistic: 14.54 on 2 and 11 DF, p-value: 0.0008157Based on the results from regression model, I can state that there is
a relation between Overall Satisfaction and
Satisfaction_with_Speed_of_Execution.
b) The p-value equals to 0.033469.
c) Based on the results from regression model, I can state that there is a relation between Overall Satisfaction and
Satisfaction_with_Trade_Price.d) The p-value equals to 0.000348
e)
pi <- predict(object = brokerage_lm1, newdata = brokerage)
#SST
SST <- sum((brokerage$Overall_Satisfaction_with_Electronic_Trades - mean(brokerage$Overall_Satisfaction_with_Electronic_Trades))^2)
SST## [1] 4.728571#SSR
SSR <- sum((pi - mean(brokerage$Overall_Satisfaction_with_Electronic_Trades))^2)
SSR## [1] 3.430819R_Squared <- SSR/SST
R_Squared## [1] 0.7255508The regression model explains 72.5% of the variation in the samples.
f)
g)
qqnorm(brokerage_lm1$residuals)
qqline(brokerage_lm1$residuals)
Based on the Q-Q plot, the residuals are not normally distributed although most of the residuals seems to aline the dotted line which represents the normal distribution. However, data points number 3,4, and 7 are far from being normally distributed.
Exercise 2
- a)
mean(hatvalues(brokerage_lm1))## [1] 0.2142857The average leverage of datapoints in Brokerage Satisfaction excel file equals to 0.2142857
- b)
hatvalues(brokerage_lm1)## 1 2 3 4 5 6 7
## 0.12226809 0.14248379 0.10348052 0.09498002 0.07909281 0.09225511 0.17268548
## 8 9 10 11 12 13 14
## 0.14817354 0.22725402 0.22639250 0.45080022 0.28805237 0.10586879 0.74621276hatvalues(brokerage_lm1) > 2*mean(hatvalues(brokerage_lm1))## 1 2 3 4 5 6 7 8 9 10 11 12 13
## FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE TRUE FALSE FALSE
## 14
## TRUEData points number 11 and 14 are considered high leverage points.
- c)
cooks.distance(brokerage_lm1)## 1 2 3 4 5 6
## 0.0079790547 0.0243407712 0.1518660850 0.0756708677 0.0059271795 0.0012425789
## 7 8 9 10 11 12
## 0.2471814307 0.0888808107 0.0062442356 0.0210623352 0.0533835032 0.0000111262
## 13 14
## 0.0062752299 0.1601935254cooks.distance(brokerage_lm1)>1## 1 2 3 4 5 6 7 8 9 10 11 12 13
## FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
## 14
## FALSEplot(brokerage_lm1)
As we can see, none of the data points has a cook’s distance greater than 1. Moreover, the fourth graph that shows cook’s distance for each data point, proves that none of the data points exceeded allowed range. This means none of the data points should be considered an outlier.
Exercise 3
- a)
x_newa <- c(1,4,2)
coefficients <- as.vector(brokerage_lm1$coefficients)
coefficients## [1] -0.6633279 0.7746098 0.4896684t(x_newa)%*%coefficients## [,1]
## [1,] 3.414448The Overall_Satisfaction_with_Electronic_Trades equals
to 3.414448
- b)
x_newb <- c(1,5,3)
t(x_newb)%*%coefficients## [,1]
## [1,] 4.678726The Overall_Satisfaction_with_Electronic_Trades equals
to 4.678726
- c)
x_newc <- c(1,4,3)
t(x_newc)%*%coefficients## [,1]
## [1,] 3.904117The Overall_Satisfaction_with_Electronic_Trades equals
to 3.904117
- d)
x_newd <- c(1,3,2)
t(x_newd)%*%coefficients## [,1]
## [1,] 2.639838The Overall_Satisfaction_with_Electronic_Trades equals
to 2.639838
Exercise 4
max(hatvalues(brokerage_lm1))## [1] 0.7462128The maximum leverage of our data points is equal to the highest leverage of a single data point in the train dataset. Hence, the maximum leverage equals to 0.7462128.
Exercise 5
- a)
X=model.matrix(brokerage_lm1)
t(x_newa)%*%solve(t(X)%*%X)%*%x_newa## [,1]
## [1,] 0.8323367t(x_newa)%*%solve(t(X)%*%X)%*%x_newa > max(hatvalues(brokerage_lm1))## [,1]
## [1,] TRUEThe new data point is a point of extrapolation since its leverage is higher than maximum leverage of the dataset.
- b)
t(x_newb)%*%solve(t(X)%*%X)%*%x_newb## [,1]
## [1,] 1.08481t(x_newb)%*%solve(t(X)%*%X)%*%x_newb > max(hatvalues(brokerage_lm1))## [,1]
## [1,] TRUEThe new data point is a point of extrapolation since its leverage is higher than maximum leverage of the dataset.
- c)
t(x_newc)%*%solve(t(X)%*%X)%*%x_newc## [,1]
## [1,] 0.3992325t(x_newc)%*%solve(t(X)%*%X)%*%x_newc > max(hatvalues(brokerage_lm1))## [,1]
## [1,] FALSEThe new data point is not a point of extrapolation since its leverage is lower than maximum leverage of the dataset.
- d)
t(x_newd)%*%solve(t(X)%*%X)%*%x_newd## [,1]
## [1,] 0.6074396t(x_newd)%*%solve(t(X)%*%X)%*%x_newd > max(hatvalues(brokerage_lm1))## [,1]
## [1,] FALSEThe new data point is not a point of extrapolation since its leverage is lower than maximum leverage of the dataset.