Homework 3
Tammy Brockman
2022-10-11
Week 3 Homework Part 1
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). Using your results, you should be able to write an equation to predict Overall_Satisfaction_with_Electronic_Trades as: B1(Satisfaction_with_Speed_of_Execution) + B2(Satisfaction_with_Trade_Price) + B3, where B1, B2, and B3 are real numbers Fill-in the blanks for the following statements:
library(readxl)
broker <- read_xlsx("C:/Users/justt/Desktop/School/621/Assignment/Homework 3/Brokerage Satisfaction.xlsx")
colnames(broker)## [1] " Brokerage"
## [2] "Satisfaction_with_Trade_Price"
## [3] "Satisfaction_with_Speed_of_Execution"
## [4] "Overall_Satisfaction_with_Electronic_Trades"# Brokerage is not relevant as it functions as more of a label and will not be used in these models.
broker <- broker[,-1]
colnames(broker)## [1] "Satisfaction_with_Trade_Price"
## [2] "Satisfaction_with_Speed_of_Execution"
## [3] "Overall_Satisfaction_with_Electronic_Trades"#B1 - Satisfaction_with_Speed_of_Execution
#B2 - Satisfaction_with_Trade_Price
#B3 - Overall_Satisfaction_with_Electronic_Trades
colnames(broker) <- c("B2", "B1", "B3")
model1 <- lm(B3 ~ B1 + B2, data = broker)
summary(model1)##
## Call:
## lm(formula = B3 ~ B1 + B2, data = broker)
##
## 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
## B1 0.4897 0.2016 2.429 0.033469 *
## B2 0.7746 0.1521 5.093 0.000348 ***
## ---
## 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: 0.0008157Both coefficients, B1 and B2, are statistically significant to Overall_Satisfaction_with_Electronic_Trades.
confint(model1, level = 0.80)## 10 % 90 %
## (Intercept) -1.7879306 0.4612749
## B1 0.2148115 0.7645252
## B2 0.5672241 0.9819956- There is an 80% probability that the number B1 will fall between _______ and ________.
There is an 80% probability that the number B1 will fall between 0.2148115 and 0.7645252.
- There is an 80% probability that the number B2 will fall between _______ and ________.
There is an 80% probability that the number B2 will fall between 0.5672241 and 0.9819956.
- Fill in the blanks:
- Suppose that we want to use the regression model created in the preceding question to predict the Overall_Satisfaction_with_Electronic_Trades when the Satisfaction_with_Speed_of_Execution is 3, and the Satisfaction_with_Trade_Price is 4. There is a 90% chance that this prediction will fall between _______ and _______.
B3_pred = data.frame(B1 = c(3), B2 = c(4))
B3_pred## B1 B2
## 1 3 4predict(model1, B3_pred, type = "response")## 1
## 3.904117predict(model1, B3_pred, interval = "prediction", level = 0.90, type = "response")## fit lwr upr
## 1 3.904117 3.174452 4.633781There is a 90% chance that this prediction will fall between 3.174452 and 4.633781.
- When the Satisfaction_with_Speed_of_Execution is 3 and the Satisfaction_with_Trade_Price is 4, there is a 90% chance that the mean response (i.e., mean value of the target variable) will fall between _________ and ___________.
predict(model1, B3_pred, interval = "confidence", level = 0.90, type = "response")## fit lwr upr
## 1 3.904117 3.514362 4.293871When the Satisfaction_with_Speed_of_Execution is 3 and the Satisfaction_with_Trade_Price is 4, there is a 90% chance that the mean response will fall between 3.514362 and 4.293871.
- Suppose that we want to use the regression model created in the preceding question to predict the Overall_Satisfaction_with_Electronic_Trades when the Satisfaction_with_Speed_of_Execution is 2, and the Satisfaction_with_Trade_Price is 3. There is an 85% chance that this prediction will fall between ______ and _______.
B3_pred = data.frame(B1 = c(2), B2 = c(3))
B3_pred## B1 B2
## 1 2 3predict(model1, B3_pred, type = "response")## 1
## 2.639838predict(model1, B3_pred, interval = "prediction", level = 0.85, type = "response")## fit lwr upr
## 1 2.639838 1.965909 3.313768There is an 85% chance that this prediction will fall between 1.965909 and 3.313768.
- When the Satisfaction_with_Speed_of_Execution is 2 and the Satisfaction_with_Trade_Price is 3, there is an 85% chance that the mean response will fall between _____________ and ____________.
predict(model1, B3_pred, interval = "confidence", level = 0.85, type = "response")## fit lwr upr
## 1 2.639838 2.225554 3.054123There is an 85% chance that the mean response will fall between 2.225554 and 3.054123.
- Use unit normal scaling to calculate standardized regression coefficients for the model that you created in #1.
# Need for Standardized regression coefficients and unit normal scaling
head(broker)## # A tibble: 6 × 3
## B2 B1 B3
## <dbl> <dbl> <dbl>
## 1 3.2 3.1 3.2
## 2 3.3 3.1 3.2
## 3 3.1 3.3 4
## 4 2.8 3.5 3.7
## 5 2.9 3.2 3
## 6 2.4 3.2 2.7summary(model1)##
## Call:
## lm(formula = B3 ~ B1 + B2, data = broker)
##
## 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
## B1 0.4897 0.2016 2.429 0.033469 *
## B2 0.7746 0.1521 5.093 0.000348 ***
## ---
## 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: 0.0008157# transform the data using unit normal scaling
broker_unit_normal = as.data.frame(apply(broker, 2, function(x){(x - mean(x))/sd(x)}))
# redo regression
#B1 - Satisfaction_with_Speed_of_Execution
#B2 - Satisfaction_with_Trade_Price
#B3 - Overall_Satisfaction_with_Electronic_Trades
model1_unit_normal <- lm(B3 ~ B1 + B2, data = broker_unit_normal)
#obtain standardized regression coefficients
model1_unit_normal##
## Call:
## lm(formula = B3 ~ B1 + B2, data = broker_unit_normal)
##
## Coefficients:
## (Intercept) B1 B2
## 4.115e-16 3.870e-01 8.115e-01Based on these coefficients, which covariate is more influential in predicting overall satisfaction? Is the Satisfaction_with_Speed_of_Execution more influential than the Satisfaction_with_Trade_Price? Or is the Satisfaction_with_Trade_Price more influential than the Satisfaction_with_Speed_of_Execution?
Based on these normalized coefficients, we can see that the Satisfaction_with_Trade_Price, at 8.115e-01, is more influential in predicting overall satisfaction than Satisfaction_with_Speed_of_Execution, at 3.870e-01.
Week 3 Homework Part 2
Use the data_RocketProp csv 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.
rocket <- read.csv("C:/Users/justt/Desktop/School/621/Assignment/Homework 3/data_RocketProp.csv")
colnames(rocket)## [1] "y" "x"- Create a linear regression to predict y based on x
model2 <- lm(y ~ x, data = rocket)
summary(model2)##
## Call:
## lm(formula = y ~ x, data = rocket)
##
## Residuals:
## Min 1Q Median 3Q Max
## -215.98 -50.68 28.74 66.61 106.76
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 2627.822 44.184 59.48 < 2e-16 ***
## x -37.154 2.889 -12.86 1.64e-10 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 96.11 on 18 degrees of freedom
## Multiple R-squared: 0.9018, Adjusted R-squared: 0.8964
## F-statistic: 165.4 on 1 and 18 DF, p-value: 1.643e-10- Create the design matrix for your regression model.
X = model.matrix(model2)
X## (Intercept) x
## 1 1 15.50
## 2 1 23.75
## 3 1 8.00
## 4 1 17.00
## 5 1 5.50
## 6 1 19.00
## 7 1 24.00
## 8 1 2.50
## 9 1 7.50
## 10 1 11.00
## 11 1 13.00
## 12 1 3.75
## 13 1 25.00
## 14 1 9.75
## 15 1 22.00
## 16 1 18.00
## 17 1 6.00
## 18 1 12.50
## 19 1 2.00
## 20 1 21.50
## attr(,"assign")
## [1] 0 1- Calculate the leverage of all datapoints in the data_RocketProp csv file.
hatvalues(model2)## 1 2 3 4 5 6 7
## 0.05412893 0.14750959 0.07598722 0.06195725 0.10586587 0.07872092 0.15225968
## 8 9 10 11 12 13 14
## 0.15663134 0.08105925 0.05504393 0.05011875 0.13350221 0.17238964 0.06179345
## 15 16 17 18 19 20
## 0.11742196 0.06943538 0.09898644 0.05067227 0.16667373 0.10984216The leverage of all the data points in this data set are in the table above.
- What is the maximum leverage calculated in part a?
max(hatvalues(model2))## [1] 0.1723896The max leverage is 0.17238964, and is rounded up to 0.1723896 in the above code.
- Suppose that we want to use the regression created in #1 to predict the value of y when x is 25.5. Would this prediction be considered extrapolation?
x_new = c(1, 25.5)
t(x_new)%*%solve(t(X)%*%X)%*%x_new## [,1]
## [1,] 0.1831324Yes. The predicted value of y, when x = 25.5, is 0.1831324. Since the max leverage is 0.17238964, and this new predicted value is larger than the max leverage, then this prediction is considered extrapolation.
- Suppose that we want to use the regression created in #1 to predict the value of y when x is 15. Would this prediction be considered extrapolation?
x_new = c(1, 15)
t(x_new)%*%solve(t(X)%*%X)%*%x_new## [,1]
## [1,] 0.05242319No. The predicted value of y, when x = 15, is 0.05242319. Since the max leverage is 0.17238964, and this new predicted value is smaller than the max leverage, then this prediction is not considered extrapolation.
- Calculate Cook’s Distance for all datapoints in the data_RocketProp csv file.
cooks.distance(model2)## 1 2 3 4 5 6
## 0.0373281981 0.0497291858 0.0010260760 0.0161482719 0.3343768993 0.2290842436
## 7 8 9 10 11 12
## 0.0270491200 0.0191323748 0.0003959877 0.0047094549 0.0012482345 0.0761514881
## 13 14 15 16 17 18
## 0.0889892211 0.0192517639 0.0166302585 0.0387158541 0.0005955991 0.0041888627
## 19 20
## 0.1317143774 0.0425721512The Cook’s Distance for all datapoints in this data set are in the table above.
- What is the maximum Cook’s Distance calculated in part a?
max(cooks.distance(model2))## [1] 0.3343769The max Cook’s Distance is 0.3343768993, and is rounded up to 0.3343769 in the above code.
- Based on your answer to part b, are there any outliers in the dataset that we should be concerned about?
No, all points are below the 1.0 value for Cook’s Distance, so there are no outliers that we need to be concerned about.
plot(model2)
