Business Analytics Lab Project 2
Predictive Analytics
Joseph Johnson
Date Here
About
This project will explore the concepts of simple linear regression, multiple linear regression and finding the optimum solution with possible business limitations.
Setup
Remember to always set your working directory to the source file location. Go to ‘Session’, scroll down to ‘Set Working Directory’, and click ‘To Source File Location’. Read carefully the below and follow the instructions to complete the tasks and answer any questions.
Submit your html, pdf or word output to Sakai, If you want you can also publish your work into rpubs and share the link on Sakai. If you worked as a team you can have 1 submission with all the names of the team on it.
Task 1: Simple Linear Regression (Total 2.5 points)
First, read in the marketing data that was used in the previous lab. Make sure the file is read in correctly. (0.25 points)
#Read data correctly
traindata = read.csv(file="data/marketing.csv", header=TRUE)
head(traindata)## case_number sales radio paper tv pos
## 1 1 11125 65 89 250 1.3
## 2 2 16121 73 55 260 1.6
## 3 3 16440 74 58 270 1.7
## 4 4 16876 75 82 270 1.3
## 5 5 13965 69 75 255 1.5
## 6 6 14999 70 71 255 2.1Next, apply the cor() function to the data to understand the correlations between variables. This is a great way to compare the correlations between all variables.(0.25 points)
#Correlation matrix of all columns in the data
corr=cor(traindata)
corr## case_number sales radio paper tv
## case_number 1.0000000 0.2402344 0.23586825 -0.36838393 0.22282482
## sales 0.2402344 1.0000000 0.97713807 -0.28306828 0.95797025
## radio 0.2358682 0.9771381 1.00000000 -0.23835848 0.96609579
## paper -0.3683839 -0.2830683 -0.23835848 1.00000000 -0.24587896
## tv 0.2228248 0.9579703 0.96609579 -0.24587896 1.00000000
## pos 0.0539763 0.0126486 0.06040209 -0.09006241 -0.03602314
## pos
## case_number 0.05397630
## sales 0.01264860
## radio 0.06040209
## paper -0.09006241
## tv -0.03602314
## pos 1.00000000#Correlation matrix of all columns except the first column. This is convenient since case_number is only an indicator for the month and should be excluded from the calculations.
corr=cor(traindata[2:6])
corr## sales radio paper tv pos
## sales 1.0000000 0.97713807 -0.28306828 0.95797025 0.01264860
## radio 0.9771381 1.00000000 -0.23835848 0.96609579 0.06040209
## paper -0.2830683 -0.23835848 1.00000000 -0.24587896 -0.09006241
## tv 0.9579703 0.96609579 -0.24587896 1.00000000 -0.03602314
## pos 0.0126486 0.06040209 -0.09006241 -0.03602314 1.00000000Q 1A) Why the value of “1.0” along the diagonal? (0.25 points)
A 1B) in this the sale and radio have a high correlation which the other two don’t do as well
Now, try to find a simple linear regression model between sales and marketing expences for radio commercials.(0.25 points)
#Simple Linear Regression
pos = traindata$pos
paper = traindata$paper
tv = traindata$tv
sales = traindata$sales
sales = traindata$sales
radio = traindata$radio
reg <- lm(sales ~ radio)
reg##
## Call:
## lm(formula = sales ~ radio)
##
## Coefficients:
## (Intercept) radio
## -9741.9 347.7#Summary of ModelQ 1C) What is the values for the intercept and the slope for radio? Write down the equation for the linear regression model and your interpretation for the intercept and slope.(0.5 points)
A 1C) intercept is -9741.9 slope is 347.7. equation is y = 347.7(x) - 9741.9
Given this equation we can predict the value of sales for any given value of radio ads. Use your equation to calculate the predicted sales value for 75 (investing $75,000 in radio ads).(0.25 points)
sales_predicted = -9741.92 + 347.69 * (75)
sales_predicted## [1] 16334.83### Sales_predicted ~ Radio expencesA high R-Squared value indicates that the model is a good fit, but not perfect. For the case of Sales versus Radio we will overlay the trend line representing the regression equation over the original plot. This will show how far the calculated results are from the actual value. The difference between the actual sales (circles) and the fitted sales (solid line) is captured in the residual error calculations.
#Plot Radio and Sales
plot(radio,sales)
scatter.smooth(radio,sales)
#Add a trend line plot using the linear model we created above
abline(reg, col="blue",lwd=2) 
Now you can derive the linear regression model for Sales versus TV (0.25 points)
#Simple Linear Regression
reg <- lm(sales ~ tv)
reg##
## Call:
## lm(formula = sales ~ tv)
##
## Coefficients:
## (Intercept) tv
## -42229.2 221.1#Summary of Model
summ <- summary(reg)
summ##
## Call:
## lm(formula = sales ~ tv)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1921.87 -412.24 7.02 581.59 1081.61
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -42229.21 4164.12 -10.14 7.19e-09 ***
## tv 221.10 15.61 14.17 3.34e-11 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 771.3 on 18 degrees of freedom
## Multiple R-squared: 0.9177, Adjusted R-squared: 0.9131
## F-statistic: 200.7 on 1 and 18 DF, p-value: 3.336e-11Q 1D) Write down the equation for the linear regression model.(0.25 points)
A 1D) y = 221.10 - 42229.21
Task 2: Multiple Linear Regression (Total 1.5 points)
Create a multiple linear regression predicting sales using both independent variables radio and tv. (0.25 points)
#Multiple Linear Regression Model
mlr1 <- lm(sales ~ radio + tv)
mlr1##
## Call:
## lm(formula = sales ~ radio + tv)
##
## Coefficients:
## (Intercept) radio tv
## -17150.46 275.69 48.34#Summary of Multiple Linear Regression Model
summ = summary(mlr1)
summ##
## Call:
## lm(formula = sales ~ radio + tv)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1729.58 -205.97 56.95 335.15 759.26
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -17150.46 6965.59 -2.462 0.024791 *
## radio 275.69 68.73 4.011 0.000905 ***
## tv 48.34 44.58 1.084 0.293351
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 568.9 on 17 degrees of freedom
## Multiple R-squared: 0.9577, Adjusted R-squared: 0.9527
## F-statistic: 192.6 on 2 and 17 DF, p-value: 2.098e-12Q 2A) Write down the equation for the linear regression model and your interpretation for the intercept and the slopes. (0.5 points)
A 2A) y = 275.69(x) + 48.34(z) -17150.46
The predicted sales can again be calculated given the coefficients of the regression model.
Calculate the predicted sales for TV = 270 and Radio = 75 (0.25 points)
# sales_predicted = radio + tv
sales_predicted = coef(mlr1)[1] + coef(mlr1)[2]*(75) + coef(mlr1)[3]*(270)
sales_predicted## (Intercept)
## 16578.3Create a multiple linear regression model for each of the following, and display the summary statistics (0.25 points)
#Multiple regression model for sales predicted by radio, tv, and pos
mlr1 <- lm( sales ~ radio + tv + pos)
mlr1##
## Call:
## lm(formula = sales ~ radio + tv + pos)
##
## Coefficients:
## (Intercept) radio tv pos
## -15491.23 291.36 38.26 -107.62#Summary of Multiple Linear Regression Model
summ = summary(mlr1)
summ##
## Call:
## lm(formula = sales ~ radio + tv + pos)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1748.20 -187.42 -61.14 352.07 734.20
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -15491.23 7697.08 -2.013 0.06130 .
## radio 291.36 75.48 3.860 0.00139 **
## tv 38.26 48.90 0.782 0.44538
## pos -107.62 191.25 -0.563 0.58142
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 580.7 on 16 degrees of freedom
## Multiple R-squared: 0.9585, Adjusted R-squared: 0.9508
## F-statistic: 123.3 on 3 and 16 DF, p-value: 2.859e-11#Multiple regression model for sales predicted by radio, tv, pos, and paper
mlr2 <- lm( sales ~ radio + tv + pos + paper)
mlr2##
## Call:
## lm(formula = sales ~ radio + tv + pos + paper)
##
## Coefficients:
## (Intercept) radio tv pos paper
## -13801.015 294.224 33.369 -128.875 -9.159#Summary of Multiple Linear Regression Model
summ = summary(mlr2)
summ##
## Call:
## lm(formula = sales ~ radio + tv + pos + paper)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1558.13 -239.35 7.25 387.02 728.02
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -13801.015 7865.017 -1.755 0.09970 .
## radio 294.224 75.442 3.900 0.00142 **
## tv 33.369 49.080 0.680 0.50693
## pos -128.875 192.156 -0.671 0.51262
## paper -9.159 8.991 -1.019 0.32449
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 580 on 15 degrees of freedom
## Multiple R-squared: 0.9612, Adjusted R-squared: 0.9509
## F-statistic: 92.96 on 4 and 15 DF, p-value: 2.13e-10Q 2B) Which of the three multiple linear regression models is best in predicting sales. Explain why. (0.25 points)
A 2B) the first one as it has the highest adjusted R squared
###Task 3: Linear Programming & Optimization (total 4 points) For this task, we need to install an optimization package in R.
if(!require("lpSolveAPI",quietly = TRUE))
install.packages("lpSolveAPI",dependencies = TRUE, repos = "https://cloud.r-project.org")
#install special package required for the solver
#install.packages("lpSolveAPI", repos = "https://cran.us.r-project.org")
#install.packages("lpSolveAPI", repos = "https://cloud.r-project.org")# load package library
library("lpSolveAPI") Solving Marketing Model
First create the linear programming model object in R. This is the starting point. The object will eventually contain all definitions and results. (0.25 points)
# Start creating the linear programming model
lpmark <- make.lp(0,2)Next, define the type of optimization, set the objective function, and add the constraints to our model object.(0.5 points)
# Define type of optimization,and dump the screen output into a variable `dump`
dump = lp.control(lpmark, sense="max")
# Set the objective function
set.objfn(lpmark, c(275.691, 48.341))
# add constraints
add.constraint(lpmark, c(1, 1), "<=", 350000)
add.constraint(lpmark, c(1, 0), ">", 15000)
add.constraint(lpmark, c(0, 1), ">", 75000)
add.constraint(lpmark, c(2, -1), "=", 0)
add.constraint(lpmark, c(1, 0), ">=", 0)
add.constraint(lpmark, c(0, 1), ">=", 0)Finally solve the model. (0.5 points)
# View the problem formulation in tabular/matrix form
lpmark## Model name:
## C1 C2
## Maximize 275.691 48.341
## R1 1 1 <= 350000
## R2 1 0 >= 15000
## R3 0 1 >= 75000
## R4 2 -1 = 0
## R5 1 0 >= 0
## R6 0 1 >= 0
## Kind Std Std
## Type Real Real
## Upper Inf Inf
## Lower 0 0# Solve
solve(lpmark)## [1] 0# Display the objective function optimum value
get.objective(lpmark)## [1] 43443517# Display the decision variables optimum values
get.variables(lpmark)## [1] 116666.7 233333.3# Display the constraints
get.constraints(lpmark)## [1] 350000.0 116666.7 233333.3 0.0 116666.7 233333.3Q 3A) Write down and clearly mark the optimum values for sales, radio, and tv ads. Show how the optimum values satisfy all constraints (0.5 points)
A 3A) the optimal amount for sales would be 350000 the optimal amount for radio would be 116666.72 for tv it would be 23333.3
Display the sensitivity of the coefficiants(0.25 points)
# display sensitivity to coefficients of objective function.
get.sensitivity.obj(lpmark)## $objfrom
## [1] -96.6820 -137.8455
##
## $objtill
## [1] 1e+30 1e+30Q 3B) Explain in coincise manner what the sensitivity results represent in reference to the marketing model.(1 points)
A 3B) this means that the the low end of the radio would be -96.682 and the low end of tv would be -137.8455. anything beneath those would cause the optimal function to change
Display the sensitivity to right hand side constraints (0.25 points)
# display sensitivity to right hand side constraints.
# There will be a total of m+n values where m is the number of contraints and n is the number of decision variables
get.sensitivity.rhs(lpmark)## $duals
## [1] 124.12433 0.00000 0.00000 75.78333 0.00000 0.00000 0.00000
## [8] 0.00000
##
## $dualsfrom
## [1] 1.125e+05 -1.000e+30 -1.000e+30 -3.050e+05 -1.000e+30 -1.000e+30 -1.000e+30
## [8] -1.000e+30
##
## $dualstill
## [1] 1.00e+30 1.00e+30 1.00e+30 4.75e+05 1.00e+30 1.00e+30 1.00e+30 1.00e+30