HWK 2
Randy Reichenbach
2023-10-26
Week 2
Use the Brokerage Satisfaction excel file to answer the following questions in R. Create an R Markdown file to answer the questions, and then “knit” your file to create an HTML document. Your HTML document should contain both textual explanations of your answers, as well as all R code needed to support your work. Submit both your HTML document and original R Markdown file (i.e., your Rmd file).
- Read the excel file titled Brokerage Satisfaction into R. You will use regression to predict the Overall_Satisfaction_with_Electronic_Trades. Remove any column(s) from the dataset that will likely not be useful in performing this task, and use all remaining columns of the dataset to perform a single regression to predict the overall satisfaction. Because the dataset is small, use all rows of the dataset as your training set (i.e., use all data rows to build the regression model).
brokerage <- read.csv("C:/Users/raze1/OneDrive/Desktop/UIndy/MSDA 621-50/Homework/HWK 2/BrokerageSatisfaction.csv")
brokerage## X..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 2.0summary(brokerage)## X..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. :4.000sample_brokerage <- sample(nrow(brokerage), nrow(brokerage)*1)
brokerage_train <- brokerage[sample_brokerage,]
brokerage_test <- brokerage[-sample_brokerage]
brokerage_train## X..Brokerage Satisfaction_with_Trade_Price
## 5 E*Trade Financial 2.9
## 1 Scottrade, Inc. 3.2
## 7 Vanguard Brokerage Services 2.7
## 12 Zecco.com 2.5
## 11 Interactive Brokers 3.7
## 10 Wells Fargo Investments 2.3
## 6 (Not listed) 2.4
## 14 Banc of America Investment Services 1.0
## 8 USAA Brokerage Services 2.4
## 9 Thinkorswim 2.6
## 13 Firstrade Securities 3.0
## 3 Fidelity Brokerage Services 3.1
## 4 TD Ameritrade 2.8
## 2 Charles Schwab 3.3
## Satisfaction_with_Speed_of_Execution
## 5 3.2
## 1 3.1
## 7 3.8
## 12 2.5
## 11 3.9
## 10 2.7
## 6 3.2
## 14 4.0
## 8 3.7
## 9 2.6
## 13 3.0
## 3 3.3
## 4 3.5
## 2 3.1
## Overall_Satisfaction_with_Electronic_Trades
## 5 3.0
## 1 3.2
## 7 2.7
## 12 2.5
## 11 4.0
## 10 2.3
## 6 2.7
## 14 2.0
## 8 3.4
## 9 2.7
## 13 3.0
## 3 4.0
## 4 3.7
## 2 3.2colnames(brokerage_train)## [1] "X..Brokerage"
## [2] "Satisfaction_with_Trade_Price"
## [3] "Satisfaction_with_Speed_of_Execution"
## [4] "Overall_Satisfaction_with_Electronic_Trades"brokerage_train$X..Brokerage <- NULL
brokerage_train## Satisfaction_with_Trade_Price Satisfaction_with_Speed_of_Execution
## 5 2.9 3.2
## 1 3.2 3.1
## 7 2.7 3.8
## 12 2.5 2.5
## 11 3.7 3.9
## 10 2.3 2.7
## 6 2.4 3.2
## 14 1.0 4.0
## 8 2.4 3.7
## 9 2.6 2.6
## 13 3.0 3.0
## 3 3.1 3.3
## 4 2.8 3.5
## 2 3.3 3.1
## Overall_Satisfaction_with_Electronic_Trades
## 5 3.0
## 1 3.2
## 7 2.7
## 12 2.5
## 11 4.0
## 10 2.3
## 6 2.7
## 14 2.0
## 8 3.4
## 9 2.7
## 13 3.0
## 3 4.0
## 4 3.7
## 2 3.2brokerage_model <- lm(Overall_Satisfaction_with_Electronic_Trades ~ ., data = brokerage_train)
summary(brokerage_model)##
## Call:
## lm(formula = Overall_Satisfaction_with_Electronic_Trades ~ .,
## data = brokerage_train)
##
## 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: 0.0008157print(brokerage_model)##
## Call:
## lm(formula = Overall_Satisfaction_with_Electronic_Trades ~ .,
## data = brokerage_train)
##
## Coefficients:
## (Intercept) Satisfaction_with_Trade_Price
## -0.6633 0.7746
## Satisfaction_with_Speed_of_Execution
## 0.4897- Based on your resulting p-values, is there likely a relationship between the overall satisfaction and the Satisfaction_with_Speed_of_Execution?
Yes, but it is a weak relationship.
- What p-value did you use to answer part a?
0.033469
- Is there likely a relationship between the overall satisfaction and the Satisfaction_with_Trade_Price?
Yes, it is a strong relationship
- What p-value did you use to answer part c?
000348
- How much of the variation in the sample values of Overall_Satisfaction_with_Electronic_Trades does the estimated regression model explain?
The Adjusted R-squared is 0.6757, so approximately 67.57% is explained by the model.
- Calculate the residuals of the data points in the Brokerage Satisfaction excel file.
brokerage_model$residuals## 5 1 7 12 11 10
## -0.149979422 -0.133395541 -0.588858465 0.002632365 -0.112435149 -0.140379336
## 6 14 8 9 13 3
## -0.062674498 -0.069955392 0.392491325 0.076204545 -0.129506736 0.646131773
## 4 2
## 0.480581057 -0.210856526BSRes = summary(brokerage_model)$sigma^2
BSRes## [1] 0.1179775standardized_res = brokerage_model$residuals/summary(brokerage_model)$sigma
standardized_res## 5 1 7 12 11 10
## -0.436648567 -0.388366425 -1.714396556 0.007663841 -0.327342551 -0.408698974
## 6 14 8 9 13 3
## -0.182469896 -0.203667419 1.142695261 0.221861138 -0.377044597 1.881141485
## 4 2
## 1.399158810 -0.613885550mean(standardized_res)## [1] -2.0383e-17var(standardized_res)## [1] 0.8461538- Check the normality of the residuals by creating a QQ-plot. Based on the plot, are the residuals normally distributed?
plot(brokerage_model)
qqnorm(brokerage_model$residuals, main = "Brokerage_Model")
qqline(brokerage_model$residuals)
The residuals are not normally distributed; per the reading,
the residuals would be considered heavily tailed.
- Examine the leverage of the datapoints by doing the following:
- Determine the average leverage of the datapoints in the Brokerage Satisfaction excel file.
hatvalues(brokerage_model)## 5 1 7 12 11 10 6
## 0.07909281 0.12226809 0.17268548 0.28805237 0.45080022 0.22639250 0.09225511
## 14 8 9 13 3 4 2
## 0.74621276 0.14817354 0.22725402 0.10586879 0.10348052 0.09498002 0.14248379mean(hatvalues(brokerage_model))## [1] 0.2142857nrow(brokerage_test)## [1] 143/14## [1] 0.2142857- If a data point’s leverage is more than twice the average, it is generally considered to be a high leverage point. Using the average that you found in part a, determine which data points are considered to have high leverage.
hatvalues(brokerage_model)## 5 1 7 12 11 10 6
## 0.07909281 0.12226809 0.17268548 0.28805237 0.45080022 0.22639250 0.09225511
## 14 8 9 13 3 4 2
## 0.74621276 0.14817354 0.22725402 0.10586879 0.10348052 0.09498002 0.14248379Data points 11 (0.45080022) and 14 (0.74621276) have high leverage.
- If a data point with unusually high leverage changes our regression or other predictive model greatly, we may possibly consider re-building the model without that abnormal data point. To determine the extent to which a data point impacts our model, we need to calculate the influence of the data points on our model. One measure for this influence is Cook’s Distance. If Cook’s Distance is greater than 1, the point is influential enough that we may want to consider either removing it from the model, or at the very least, treating it differently than the other data. Based on Cook’s Distance, are there outliers that may possibly need to be removed and/or treated differently in the data set? Create a plot using R to support your answer.
cooks.distance(brokerage_model)## 5 1 7 12 11 10
## 0.0059271795 0.0079790547 0.2471814307 0.0000111262 0.0533835032 0.0210623352
## 6 14 8 9 13 3
## 0.0012425789 0.1601935254 0.0888808107 0.0062442356 0.0062752299 0.1518660850
## 4 2
## 0.0756708677 0.0243407712plot(brokerage_model)
“Cook’s Distance is a measure used to determine outliers
that are so unusual, we need to give them special attention. This is the
case when Cook’s Distance is bigger than 1.” None of that data points,
when Cook’s Distance is calculated, is greater than 1 and thus do not
need to be treated differently or removed.
- Use the regression model that you created in #1 to predict the Overall_Satisfaction_with_Electronic_Trades in each of the following scenarios:
- Satisfaction_with_Speed_of_Execution = 2, Satisfaction_with_Trade_Price = 4
predict(brokerage_model, data.frame(Satisfaction_with_Trade_Price = 4, Satisfaction_with_Speed_of_Execution=2))## 1
## 3.414448- Satisfaction_with_Speed_of_Execution = 3, Satisfaction_with_Trade_Price = 5
predict(brokerage_model, data.frame(Satisfaction_with_Trade_Price = 5, Satisfaction_with_Speed_of_Execution=3))## 1
## 4.678726- Satisfaction_with_Speed_of_Execution = 3, Satisfaction_with_Trade_Price = 4
predict(brokerage_model, data.frame(Satisfaction_with_Trade_Price = 4, Satisfaction_with_Speed_of_Execution=3))## 1
## 3.904117- Satisfaction_with_Speed_of_Execution = 2, Satisfaction_with_Trade_Price = 3
predict(brokerage_model, data.frame(Satisfaction_with_Trade_Price = 3, Satisfaction_with_Speed_of_Execution=2))## 1
## 2.639838- Extrapolation occurs when a prediction is made outside of the region of data used to train the model. This is “risky” because we do not have any prior data to see what tends to happen in the particular situation that we are trying to predict. To determine if a prediction will involve extrapolation, we need to first determine the largest leverage of all datapoints. What is the maximum leverage of the datapoints in the Brokerage Satisfaction excel file?
X = model.matrix(brokerage_model)
X## (Intercept) Satisfaction_with_Trade_Price
## 5 1 2.9
## 1 1 3.2
## 7 1 2.7
## 12 1 2.5
## 11 1 3.7
## 10 1 2.3
## 6 1 2.4
## 14 1 1.0
## 8 1 2.4
## 9 1 2.6
## 13 1 3.0
## 3 1 3.1
## 4 1 2.8
## 2 1 3.3
## Satisfaction_with_Speed_of_Execution
## 5 3.2
## 1 3.1
## 7 3.8
## 12 2.5
## 11 3.9
## 10 2.7
## 6 3.2
## 14 4.0
## 8 3.7
## 9 2.6
## 13 3.0
## 3 3.3
## 4 3.5
## 2 3.1
## attr(,"assign")
## [1] 0 1 2max(hatvalues(brokerage_model))## [1] 0.7462128- If a new datapoint has a leverage higher than the maximum (calculated in #4), then the prediction will involve extrapolation.
- Was the prediction made in part a of #3 considered to be extrapolation? (Note that predictions made with extrapolation are not extremely trustworthy.)
x_newa = c(1, 4, 2)
t(x_newa)%*%solve(t(X)%*%X)%*%x_newa## [,1]
## [1,] 0.8323367Yes, 0.8323367 > 0.7462128
- Was the prediction made in part b of #3 considered to be extrapolation?
x_newb = c(1, 5, 3)
t(x_newb)%*%solve(t(X)%*%X)%*%x_newb## [,1]
## [1,] 1.08481Yes, 1.08481 > 0.7462128
- Was the prediction made in part c of #3 considered to be extrapolation?
x_newc = c(1, 4, 3)
t(x_newc)%*%solve(t(X)%*%X)%*%x_newc## [,1]
## [1,] 0.3992325No, 0.3992325 < 0.7462128
- Was the prediction made in part d of #3 considered to be extrapolation?
x_newd = c(1, 3, 2)
t(x_newd)%*%solve(t(X)%*%X)%*%x_newd## [,1]
## [1,] 0.6074396No, 0.6074396 < 0.7462128