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?

A correlation, for example, of sales with sales will obviously be 1. It is a correlation with itself so it will always be 1.

1B) Which pairs has the strongest correlations?

The pair with the strongest correlation is sales and radio because it is the closest correlation to 1.

corr = cor( mydata[ c(2,3) ] )
corr
##           sales     radio
## sales 1.0000000 0.9771381
## radio 0.9771381 1.0000000

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/g5/t1y3ltbj1_zgv8c32k_rp9rh0000gn/T//Rtmpfr1sX5/downloaded_packages
## 
## The downloaded binary packages are in
##  /var/folders/g5/t1y3ltbj1_zgv8c32k_rp9rh0000gn/T//Rtmpfr1sX5/downloaded_packages
## Warning: package 'corrplot' was built under R version 3.4.2
## 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)

When knitting the file, the correlation matrix looks wierd, so I included a screen shot of what they look like in RStudio itself. I was told by fellow students that this may have happened because I use a Mac.

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
scatter.smooth(tv,sales)

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

The intercept value is -42229.21 and the slope is 221.10. So the equation is Sales Predicted = -42229.21 + 221.10 x XTV

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)

I was having the same problem again. When knitting the file, the blue line would not appear. But the blue line appears within rstudio. I heard from fellow students that this is because I am on a Mac computer. But I have inserted the screen shot of how this graph looks in Rstudio

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)
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 = Sales predicted by radio, tv, pos, and paper
#Summary of Multiple Linear Regression Model

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 Equation for predicted sales is = -17150.46 + 275.69a(xradio) + 48.34b(xtv) R-Squared is 0.9577, Adjusted R-Squared is 0.9527.

mlr2 Equation for predicted sales is = -15491.23 + 291.36a(xradio) + 38.26b(xtv) + -107.62c(xpos) R-Squared is 0.9585, Adjusted R-Squared is 0.9508.

mlr3 Equation for predicted sales is = -13801.015 + 294.224a(xradio) + 33.369b(xtv) + -128.875c(xpos) + -9.159d(xpaper) R-Squared is 0.9612, Adjusted R-Squared is 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.

When looking at the R-Squared, the mlr3 model seems the best because it is closest to 1. The more variables we add the model, the higher r-squared becomes. So we want the model with the least amount of variables with the highest adjusted squared. That is because it will be the best predictor. So that mlr1 is the best predictor.

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.
#mlr1
Sales_Predicteda = coef(mlr1)[1] + coef(mlr1)[2]*69 + coef(mlr1)[3]*255
Sales_Predicteda
## (Intercept) 
##    14199.05
sales_predictedb = coef(mlr2)[1] + coef(mlr2)[2]*69 + coef(mlr2)[3]*255 + coef(mlr2)[4]*1.5
sales_predictedb
## (Intercept) 
##    14208.44
sales_predictedc = coef(mlr3)[1] + coef(mlr3)[2]*69 + coef(mlr3)[3]*255 + coef(mlr3)[4]*1.5 + coef(mlr3)[5]*75
sales_predictedc
## (Intercept) 
##    14129.32

These values show the the best predictor is mlr3. So we will use the error sqaured eqaution using our previous choice of mlr1.

#mlr1
(14199.05-13965)^2
## [1] 54779.4
#mlr2
(14208.44-13965)^2
## [1] 59263.03
#mlr3
(14129.32-13965)^2
## [1] 27001.06

The model with the smallest error is the model mlr3. It is the closest to the real sales number. So this shows us that the multiple regression model cant be 100% accurate.

Task 3: Watson Analytics Predict

Watson Analytics contains a module for Predict, besides the two others for Explore and Assemble. Although Watson is a useful tool it should never be treated as a black box. In other words there are many circumstances where Watson can fail to make the proper diagnosis and suggest the correct remedies. This is why it is critical, as more of the cognitive analytics and advanced machine learning tools invade our environment, that we also allow room for human intervention and learn how to best interact.

There are many details in the Predict module. We will keep it relatively simple for this introductory phase. To complete this task, follow the directions below. Make sure to capture any screenshots and to answer any questions where requested

  1. Logon to your Watson Analytics account at http://watsonanalytics.com
  2. Upload the file marketing.csv unless already in your folder
  3. Select the Predict module to analyze the data

You will be prompted to select the data set and to enter a name for your workbook. By default the target to predict is correctly set to Sales.

3A) Capture a screen shot of the top predictors of sales. Consider the one field predictive model only. Note the predictive power strengths.

Note that Watson follows a different approach to identify and quantify the predictive strength of each variable. Such details are outside the scope of this exercise. Watson provides a lot of on-line help, and the interested reader is welcome to explore the many capabilities of Watson on their own.

3B) How do Watson results reconcile with your findings based on the R regression analysis in task 2? Explain.

Watson shows that radio drives sales. This matches my findings in saying that Radio has the highest correlation in task #1. The only difference is the percentages. Here it’s 93% but in Rstudio it was 97%. The correlation for TV and sales are also different from RStudio. Watson showsTV to be 83% while Rstudio shows 95%.