About

In a given year, if it rains more, we may see that there might be an increase in crop production. This is because more water may lead to more plants. This is a direct relationship; the number of fruits may be able to be predicted by amount of waterfall in a certain year. This example represents simple linear regression, which is an extremely useful concept that allows us to predict values of a certain variable based off another variable.

This lab will explore the concepts of simple linear regression, multiple linear regression, and watson analytics.

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 work to RPubs as detailed in previous notes.

Note

For your assignment you may be using different data sets than what is included here. Always read carefully the instructions on Sakai. Tasks/questions to be completed/answered are highlighted in larger bolded fonts and numbered according to their particular placement in the task section.


Task 1: Simple Linear Regression

First, read in the marketing data that was used in the previous lab. Make sure the file is read in correctly.

#Read data correctly
mydata = read.csv(file="data/marketing.csv")
head(mydata)
##   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 2.1

Next, 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.

#Correlation matrix of all columns in the data
corr = cor(mydata)
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 columns 2,4, and 6
corr = cor( mydata[ c(2,4,6) ] )
corr
##            sales       paper         pos
## sales  1.0000000 -0.28306828  0.01264860
## paper -0.2830683  1.00000000 -0.09006241
## pos    0.0126486 -0.09006241  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( mydata[ 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  1.00000000
1A) Why the value of “1.0” along the diagonal?

The value 1 represents the correlations between the variables. According to the diagonal, when the the variable meet the same variable will have the correlation 1. For instance, when sales meet sales there is a correlation 1. When TV meets TV, there is also a correlation 1. This means that the same variables are equal to each other, therefore, they have perfect correlations.These are called reflecive property.

1B) Which pairs has the strongest correlations?

Excepting reflexive correlations, the highest correlations are 0.97713807 which are sales and radio. Radio and TV also have high correlations which are 0.9609578. Tv and sales also have strong correlations which are tv and sales with 0.957973.

Next we will create a visual diagram of the correlation matrix called a corrgram where the correlations strength are represented by colors intensity. To do this we need first to install two packages in R-Studio as executed by the command lines below

## 
## The downloaded binary packages are in
##  /var/folders/y0/tv09q3b13x17z8xmy93f7k0m0000gn/T//Rtmpy4eZGU/downloaded_packages
## 
## The downloaded binary packages are in
##  /var/folders/y0/tv09q3b13x17z8xmy93f7k0m0000gn/T//Rtmpy4eZGU/downloaded_packages
## corrplot 0.84 loaded
# Generates a corrgram of last computed correlation matrix
corrgram(corr)

# Generates a corrplot, similar a corrgram, but with a different visual display
corrplot(corr)

From the matrix, its clear that Sales, Radio and TV have the strongest correlations. Let’s create now few scatterplots to visualize the data and trending lines. First we need to extract the columns from the data file.

#Extract all variables
pos  = mydata$pos
paper = mydata$paper
tv = mydata$tv
sales = mydata$sales
radio = mydata$radio
#Plot of Radio and Sales using plot command from Worksheet 4
scatter.smooth(radio,sales)

From this plot, it seems the points are scattered in an almost linear way. So, we will try to fit a simple linear regression model to the graph.

The lm() linear modeling function is a very useful one. The function is set up as lm(y ~ x) where the x variable, the independent variable, predicts values of the y variable, or the dependent variable.

In the simple linear regression model below, we are using radio ads to predict sales. We print out a summary to view the quantitative facts about the linear model.

#Simple Linear Regression - the R function lm()
reg <- lm(sales ~ radio)

#Summary of Model
summary(reg)
## 
## Call:
## lm(formula = sales ~ radio)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -1732.85  -198.88    62.64   415.26   637.70 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -9741.92    1362.94  -7.148 1.17e-06 ***
## radio         347.69      17.83  19.499 1.49e-13 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 571.6 on 18 degrees of freedom
## Multiple R-squared:  0.9548, Adjusted R-squared:  0.9523 
## F-statistic: 380.2 on 1 and 18 DF,  p-value: 1.492e-13

As indicated by the summary report the intercept value is -9741.92 and the slope for radio is 347.69. We can therefore write the following equation for the linear regression model predicting Sales based on Radio Sales_predicted = -9741.92 + 347.69 * Xradio. Given this equation we can predict the value of sales for any given value of radio like for example 75 (investing $75,000 in radio ads)

### Linear model  ( Y = b +- mx ) 
### Sales_predicted  ~ Radio_X1

sales_predicted = -9741.92 + 347.69 * (75)
sales_predicted
## [1] 16334.83
### Another way to write the equation is to refer to the coefficients of the equation instead of typing the actual values out.
### This is demonstrated below.

sales_predicted = coef(reg)[1] + coef(reg)[2] * 75
coef(reg)[1] # intercept
## (Intercept) 
##   -9741.921
coef(reg)[2] # slope
##    radio 
## 347.6888
sales_predicted
## (Intercept) 
##    16334.74
1C) Repeat the above calculations to derive the linear regression model for Sales versus TV ?
reg <- lm(sales ~ tv)
summary(reg)
## 
## 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-11
plot(tv,sales)
abline(reg, col = "blue", lwd=2)

1D) Write down the equation for the linear regression model. Note the values for the intercept and the slope

The equation is Ysales =-42229.21+221.1X, where the -42229.21 is the intercept and 221.1 is the intercept.

A 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 predictions are from the actual value. The difference between the actual sales (circles) and the predicted sales (solid line) is captured in the residual error calculations as reported by the summary function.

#Plot Radio and Sales 
plot(radio,sales)

#Add a trend line plot using the linear model we created above
abline(reg, col="blue",lwd=2) 


Task 2: Multiple Linear Regression

Many times there are more than one factor or variable that affect the prediction of an outcome. While increased rainfall is a good predictor of increased crop supply, decreased herbivores can also result in an increase of crops. This idea is a loose metaphor for multiple linear regression.

In R, multiple linear regression takes the form of lm(y ~ x0 + x1 + x2 + ... ), where y is the value that is being predicted, or the dependent variable and the x variables are the predictors or the independent variables

Lets create a multiple linear regression predicting sales using both independent variables radio and tv.

#Multiple Linear Regression Model
mlr1 <-lm(sales ~ radio + tv)

#Summary of Multiple Linear Regression Model
summary(mlr1)
## 
## 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-12

Note the values of 0.9577 for R-squared and 0.9527 for the Adj R-squared. The predicted sales can again be calculated given the coefficients of the regression model. The example below shows the predicted sales for TV = 270 and Radio = 75.

# sales_predicted = radio + tv
sales_predicted = coef(mlr1)[1] + coef(mlr1)[2]*(75) + coef(mlr1)[3]*(270)
sales_predicted
## (Intercept) 
##     16578.3
2A) Create a multiple linear regression model for each of the following, and display the summary statistics.
#mlr2 = Sales predicted by radio, tv, and pos
#Summary of Multiple Linear Regression Model
mlr2<-lm(sales ~ radio + tv +pos)
#mlr3 = Sales predicted by radio, tv, pos, and paper
#Summary of Multiple Linear Regression Model
summary(mlr2)
## 
## 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
mlr3<-lm(sales ~ radio + tv +pos +paper)
summary(mlr3)
## 
## 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-10
2B) Write down the regression equation and the values for R-Squared and Adj R-Squared for each of the three cases mlr1, mlr2, and mlr3

mlr1: Ysales=-17150.46+275.69Xradio+48.34Xtv R-square:0.9577 Adjusted R-square:0.9527 mlr2: Ysales=-15491.23+291.26Xradio+36.26Xtv-107.62Xpos R-square:0.9585 Ajusted R-square:0.9508 mlr3: Ysales= -13801.015+294.224Xradio+33.369Xtv-128.875Xpos-9.159Xpaper R-aquare:0.9612 Adjusted R-square:0.9509

2C) Based solely on the values for R-Squared and Adj R-Squared, which of the three multiple linear regression models mlr1, mlr2, mlr3 is best in predicting sales. Explain why.

The best predicted model is model 1 because it has the highest adjusted r-square value. This means that adjusted r-square will only increase if the added values provide a better predicition

2D) Calculate the predicted sales for each of the three models given that Radio = 69 , TV = 255 , POS = 1.5, and Paper = 75. Compare your predicted sales to the actual value as obtained from the data file. Which model is best at predicting the actual sales value? Quantify your answer by calculating the error squared. Compare your results to 2C) and share any observations.
sales_predicted=-17150.46 + 275.69 * (69) + 48.34 * (255)
sales_predicted
## [1] 14198.85
sales_predicted=-15491.23 + 291.36 * (69) + 38.26 * (255) - 107.62 * (1.5)
sales_predicted
## [1] 14207.48
sales_predicted=-13801.015 + 294.224 * (69) + 33.369 * (255) - 128.875 * (1.5) - 9.159 * (75)
sales_predicted
## [1] 14129.3
(13965-14198.85)^2
## [1] 54685.82
(13965-14207.48)^2
## [1] 58796.55
(13965-14129.3)^2
## [1] 26994.49

The best fitting is model 3 which provides the clostest value to the actual value. This is also supported by the error squared since error squared shows the difference between the actual value and predict value. In this case, by using the error squared model 3 is the best fit. However, this does not mean it is the best predictive value because at previous quesion 2D, we discussed that the highest adjusted r-square value is the best predictive value which is model 3.