Notebook Instructions

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.

Load Packages in R/RStudio

We are going to use tidyverse a collection of R packages designed for data science.

## Loading required package: tidyverse
## -- Attaching packages ----------------------------------------------------------------------------------- tidyverse 1.2.1 --
## v ggplot2 2.2.1     v purrr   0.2.4
## v tibble  1.4.2     v dplyr   0.7.4
## v tidyr   0.7.2     v stringr 1.2.0
## v readr   1.1.1     v forcats 0.2.0
## -- Conflicts -------------------------------------------------------------------------------------- tidyverse_conflicts() --
## x dplyr::filter() masks stats::filter()
## x dplyr::lag()    masks stats::lag()
## Loading required package: plotly
## 
## Attaching package: 'plotly'
## The following object is masked from 'package:ggplot2':
## 
##     last_plot
## The following object is masked from 'package:stats':
## 
##     filter
## The following object is masked from 'package:graphics':
## 
##     layout

Task 1: Correlation Analysis

1A) Read the csv file into R Studio and display the dataset.

  • Name your dataset ‘mydata’ so it easy to work with.

  • Commands: read_csv() rename() head()

Extract the assigned features (columns) to perform some analytics.

mydata = read.csv(file="data/Advertising.csv")

1B) Create a correlation table for your to compare the correlations between all variables. Remove any variables where correlation between variables is irrelevant or inaccurate

mydata <- rename(mydata, "case_number" = "X")
head(mydata)
##   case_number    TV radio newspaper sales
## 1           1 230.1  37.8      69.2  22.1
## 2           2  44.5  39.3      45.1  10.4
## 3           3  17.2  45.9      69.3   9.3
## 4           4 151.5  41.3      58.5  18.5
## 5           5 180.8  10.8      58.4  12.9
## 6           6   8.7  48.9      75.0   7.2
  • Commands: cor() mydata[ -c(“COLUMN_NAME OR COLUMN_NUMBER”) ]
#corr = cor( MYDATA )
#corr
corr = cor(mydata[ -c(1) ])
corr
##                   TV      radio  newspaper     sales
## TV        1.00000000 0.05480866 0.05664787 0.7822244
## radio     0.05480866 1.00000000 0.35410375 0.5762226
## newspaper 0.05664787 0.35410375 1.00000000 0.2282990
## sales     0.78222442 0.57622257 0.22829903 1.0000000

1C) Why is the value “1.0” down the diagonal? Which pairs seem to have the strongest correlations, list the pairs.

This just means that each variable is perfectly correlated with itself, which makes sense. Aside from all the 1.0, the strongest correlation is between TV and sales.

1D-a) Identifying the dependent variable (y) and one independent variable (x_i) using the correlation table to identify a variable with a coefficient greater than 0.20 and lower than 0.60. Use those two variables to create a scatterplot to visualize the data. Note any patterns or relation between the two variables

  • Commands: qplot( x = VARIABLE, y = VARIABLE, data = mydata)
qplot( x = radio, y = sales, data = mydata)

As radios increase, the sales seems to increase as well.

1D-b) Create a 3D scatterplot between the two of the strongest correlated variables to the dependent variable. Note any patterns and the coordinates of three points with the heights values (x,y,z)

#p <- plot_ly(mydata, x = ~VARIBLE_1, y = ~VARIBLE_2, z = ~VARIBLE_3, marker = list(size = 5)) %>%
#  add_markers() %>% 
#p
p <- plot_ly(mydata, x = ~TV, y = ~radio, z = ~sales, marker = list(size = 5))
add_markers(p)

TV and sales clearly have a strong positive correlation between one another. However, TV and radio do not seem to affect one another at all.

Task 2: Regression Analysis

2A) Create a linear regression model by identifying the dependent variable (y) and for independent variable (x_i) use the correlation table to identify a variable with a coefficient greater than 0.20 and lower than 0.60. (same variables as 1D-a)

  • Commands: lm( y ~ x )
#Simple Linear Regression Model

#reg <- lm( DEPENDENT_VARIABLE ~ INDEPENDENT_VARIABLE )
reg <- lm( sales ~ radio, data = mydata )
reg
## 
## Call:
## lm(formula = sales ~ radio, data = mydata)
## 
## Coefficients:
## (Intercept)        radio  
##      9.3116       0.2025

2B) Use the regression model to create a report. Note the R-Squared and Adjusted R-Squared values, determine if this is a good or bad fit for your data?

  • Commands: Use the summary() function to create a report for the linear model
#Summary of Simple Linear Regression Model

#summary(MODEL)
summary(reg)
## 
## Call:
## lm(formula = sales ~ radio, data = mydata)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -15.7305  -2.1324   0.7707   2.7775   8.1810 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  9.31164    0.56290  16.542   <2e-16 ***
## radio        0.20250    0.02041   9.921   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 4.275 on 198 degrees of freedom
## Multiple R-squared:  0.332,  Adjusted R-squared:  0.3287 
## F-statistic: 98.42 on 1 and 198 DF,  p-value: < 2.2e-16

The multiple R-squared and adjusted R-squared are both around .33 which means that most of the data is not on the regression line.

2C) Create a plot for the dependent (y) and independent (x) variables Note any patterns or relation between the two variables describe the trend line.

  • The trend line will show how far the predictions are from the actual value
  • The distance from the actual versus the predicted is the residual
#p <- qplot( x = INDEPENDENT_VARIABLE, y = DEPENDENT_VARIABLE, data = mydata) + geom_point()

p <- qplot( x = radio, y = sales, data = mydata) + geom_point()

#Add a trend line plot using the a linear model
#p + geom_smooth(method = "lm", formula = y ~ x)
p + geom_smooth(method = "lm", formula = y ~ x)

The more and more radios there are the further the points are from the line.

2D-a) Create a Multiple linear regression model and summary report using the two strongest correlated variables and the dependent variable. Note the R-Squared and Adjusted R-Squared values, determine if this is a good or bad fit for your data? Compared this model to the previous model, which model is better?

  • Sometimes, one variable is very good at predicting another variable. But most times, there are more than one factors that affect the prediction of another variable.
  • 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.

  • Multiple linear regression lm(y ~ x_0 + x_1 + x_2 + … x_i )

  • Where y is the predicted/dependent variable and the x variables are the predictors/independent variable

  • commands: lm( y ~ x_1 + x_2 ) summary( reg_model )

#Multiple Linear Regression Model
#mlr1 <- lm( DEPENDENT_VARIABLE ~ INDEPENDENT_VARIABLE1 + INDEPENDENT_VARIABLE2 )
mlr1 <- lm( sales ~ radio + TV, data = mydata)
mlr1
## 
## Call:
## lm(formula = sales ~ radio + TV, data = mydata)
## 
## Coefficients:
## (Intercept)        radio           TV  
##     2.92110      0.18799      0.04575
#Summary of Multiple Linear Regression Model
summary(mlr1)
## 
## Call:
## lm(formula = sales ~ radio + TV, data = mydata)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -8.7977 -0.8752  0.2422  1.1708  2.8328 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  2.92110    0.29449   9.919   <2e-16 ***
## radio        0.18799    0.00804  23.382   <2e-16 ***
## TV           0.04575    0.00139  32.909   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1.681 on 197 degrees of freedom
## Multiple R-squared:  0.8972, Adjusted R-squared:  0.8962 
## F-statistic: 859.6 on 2 and 197 DF,  p-value: < 2.2e-16

The r-squared values in this model are nearly .90. This means by adding TV to radio the model becomes much stronger since the r-squared data went from .33 to .90.

2D-b) Create a Multiple Linear Regression Model using all relevant independent variables and the dependent variable. Note the R-Squared and Adjusted R-Squared values, determine if this is a good or bad fit for your data?

mlr2 <- lm( sales ~ radio + TV + newspaper, data = mydata)
mlr2
## 
## Call:
## lm(formula = sales ~ radio + TV + newspaper, data = mydata)
## 
## Coefficients:
## (Intercept)        radio           TV    newspaper  
##    2.938889     0.188530     0.045765    -0.001037
summary(mlr2)
## 
## Call:
## lm(formula = sales ~ radio + TV + newspaper, data = mydata)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -8.8277 -0.8908  0.2418  1.1893  2.8292 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  2.938889   0.311908   9.422   <2e-16 ***
## radio        0.188530   0.008611  21.893   <2e-16 ***
## TV           0.045765   0.001395  32.809   <2e-16 ***
## newspaper   -0.001037   0.005871  -0.177     0.86    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1.686 on 196 degrees of freedom
## Multiple R-squared:  0.8972, Adjusted R-squared:  0.8956 
## F-statistic: 570.3 on 3 and 196 DF,  p-value: < 2.2e-16

This model is actually a tad weaker than the previous model bbecause adjusted R-squared actually decreases by .0006. This means including newspapers is not as good of a fit for the data.

Based purely on the values for R-Squared and Adjusted R-Squared, which linear regression model is best in predicting the dependent variable? Explain why

The best at predicting the dependent variable is the radio and TV model. It has the higher overall R-squared calculations.

2E) Use the three different models to predicted the dependent variable for the given values of the independent variables.

  • Variable: Radio = 69
  • Variable: TV = 255
  • Variable: newspaper = 75

MODEL 1

Radio = 69
Model1 = 9.31164 + 0.20250 * (Radio)
Model1
## [1] 23.28414

MODEL 2

Radio = 69
TV = 255
Model2 = 2.92110 + 0.18799 * (Radio) + 0.04575 * (TV)
Model2
## [1] 27.55866

MODEL 3

Radio = 69
TV = 255
newspaper = 75
Model3 = 2.938889 + 0.188530 * (Radio) + 0.045765 *(TV) + -0.001037 * (newspaper)
Model3
## [1] 27.53976

Model 2 predicted the highest sales, while model 3, although it was close to Model 2, it predicted the second highest value. Model 1 was around 4 dollars lower than the other two models.

Task 3: Watson Analysis

To complete the last task, follow the directions found below. Make sure to screenshot and attach any pictures of the results obtained or any questions asked.

3A) Use the Predictive module to analyze the given data. Note any interesting patterns add an screenshot of what you found.

knitr::include_graphics("image2.png")

knitr::include_graphics("image3.png")

3B) Note the predictive power strength of reported variables. Consider the one field predictive model only, describe your findings and add and screenshot

knitr::include_graphics("image1.png")

Radio and TV are the best predictors of sales which confirms the data from earlier.

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

As I just mentioned, Watson found that using radio and TV is the best way to predict sales. Newspapers are not included in this calculation nor does newspaper alone have any noteworthy predictive strength in sales.