Data Visualisation 2019 - Assignment 5

List of R colors:

http://www.stat.columbia.edu/~tzheng/files/Rcolor.pdf

Simple Linear Regression

• Given a set of data \[ (x_1,y_1), (x_2,y_2),\ldots,(x_n,y_n) \] a regression analysis is a means of determining if there is a correlation between the \(x\)-data and the \(y\)-data.

• A Scatter Plot is a means of tentatively determining whether or not the data is correlated, simply by observing the shape of the plot

Example 1 - Strong Linear Correlation

Plot the following data points on a scatter plot

\[ (1.2, 10.3), (1.5,10.45), (1.6, 10.79), (2.1, 10.8), (2.2,11.0), (2.4, 11.2), (2.5,11.5), (2.9, 12.2), (3.1,12.4), (3.8,13.0), (4.1,13.4) \]

For this data set we create two data vectors, one for the \(x\)-variables and one for the \(y\)-variables, and plot the result, with the \(x\)-data plotted along the horizontal axis, and the \(y\)-data plotted along the vertical axis.

X1<-c(1.2,1.5,1.6,2.1,2.1,2.4,2.5,2.9,3.1,3.8,4.1)
Y1<-c(10.3,10.45,10.79,10.8,11.0,11.2,11.5,12.2,12.4,13.0,13.4)
plot(X1,Y1,xlab="x",ylab="y",col="red", main="Linearly Correlated Data",pch=18)

Remarks:

The argument pch allows us to change the type of marker used to plot the data points.

1. There a 25 options available, just by selecting the values from 1 to 25.
2. Alternatively, it is possible to set each marker to a single letter or symbol. For instance to set each marker to £ we could use pch=“£”

Correlated Data

• It is clear from the scatter plot that as \(x\) increases the \(y\) values also tend to increase.

• It also seems clear from the plot that a straight line can be drawn through the data points, which we would interpret to mean there exists a linear relationship between the \(x\) and \(y\) variables, of the form \[ y=a +b x. \]

Example 2 - Non-correlated Data

In this example we will plot the following data set \[ (-1.4,2.4), (0.4,1.3), (1.4,6.5), (2.1,3.4), (4.5,12.3), (5.2, 2.1), (5.8,3.1), (6.5,5.4), (6.9, 6.7) \]

X2<-c(-1.4,0.4,1.4,2.1,4.5,5.2,5.8,6.5,6.9)
Y2<-c(2.4,1.3,6.5,3.4,12.3,2.1,3.1,13.4,6.7)
plot(X2,Y2,xlab="x",ylab="y",col="red", main="Non-correlated Data",pch=19)

• It is clear from the scatter plot that there is no immediately obvious trend relating the \(x\) and \(y\) variables in this plot, in which case we say the \(x\) and \(y\) variables are not correlated.

• While the scatter plot doesn’t rule out a correlation between the \(x\) and \(y\) variables, it does give an initial indication as to whether or not a relationship exists.

The Line of Best Fit

• Given the data set above, there are values of the estimated parameters \(\hat{\beta}_0\) an \(\hat{\beta}_1\) which will yield a line which best fits the points of the data plot.

• One way to find a line which passes through or close to these data point is to obtain the lie slope from the first and last data points as follows: \[ b=\frac{6.7-2.4}{6.9+1.4}=\frac{4.3}{8.3}=0.518. \]

• Taking the mean of the \(x\) and \(y\) data, we have

mean(X1)

[1] 2.481818

mean(Y1)

[1] 11.54909

• This now allows us to find an estimate for the parameter \(\beta_0\) as follows \[ \bar{y}=a+b\bar{x}\Rightarrow 11.54909=a+0.518\times 2.481818 \Rightarrow a=10.2635. \]

• Our first approximation at a linear model relating the \(x\) and \(y\) data is now given by \[y=10.2635+0.518x\]

• We now plot this line through the data points as follows: \[x=0\Rightarrow y=10.2635\qquad x=5\Rightarrow y=12.8535\]

X1<-c(1.2,1.5,1.6,2.1,2.1,2.4,2.5,2.9,3.1,3.8,4.1)
Y1<-c(10.3,10.45,10.79,10.8,11.0,11.2,11.5,12.2,12.4,13.0,13.4)
plot(X1,Y1,xlab="x",ylab="y",col="red",main="Linearly Correlated Data",pch=18)
segments(0,10.2635,x1=5,y1=12.8535,col ="blue",lwd=2)

• Clearly, this line doesn’t fit the data very well, and there are better approximations available to represent the relationship between to \(x\) and \(y\) data.

The Method of Least Squares

X1<-c(1.2,1.5,1.6,2.1,2.1,2.4,2.5,2.9,3.1,3.8,4.1)
Y1<-c(10.3,10.45,10.79,10.8,11.0,11.2,11.5,12.2,12.4,13.0,13.4)
plot(X1,Y1,xlab="x",ylab="y",col="red",main="Linearly Correlated Data",pch=18)
segments(X1,Y1,x1=X1,y1=10.2635+0.518*X1,col ="darkgreen",lwd=2)

segments(0,10.2635,x1=5,y1=12.8535,col ="blue",lwd=2)

• The Method of Least Squares is method to choose values of the parameters \(a\) and \(b\) which returns the simple linear regression model \[y=a+bx.\]
• Evaluating the values of \(y\) at each of the data values \(x_1,\ldots,x_{11}\) gives corresponding values \(\hat{y}_1=a+bx_1,\)…, \(\hat{y}_{11}=a+bx_{11}\)
• The difference between these predicted values and the actual data values is given by \(\hat{y}_1-y_1\),…,\(\hat{y}_{11}-y_{11}\).
• The method of least squares is the means of choosing values for \(a\) and \(b\) so that the sum of the squares of these differences is minimised, i.e. \[(\hat{y}_1-y_1)^2+\ldots+(\hat{y}_{11}-y_{11})^2\] is the smallest value possible.
• In terms of the graph above it means we have chosen the blue line so that the sum of the lenghts of each green line segment squared is the minimum possible value.
• The value for \(b\) is obtained by means of fairly straight forward calculus, which we omit here, but which gives the result \[b=\frac{n\sum_{i=1}^{n}x_iy_i-\sum_{i=1}^{n}x_i\sum_{i=1}^{n}y_i}{n\sum_{i=1}^{n}x_i^2-\left(\sum_{i=1}^{n}x_i\right)^2}.\]

• The value of \(a\) is obtained from \[\bar{y}=a+b\bar{x}\Rightarrow a=\bar{y}-b\bar{x}.\]

• These parameters are obtained automatically for us from the data set using the lm() function:

Model1<-lm(Y1 ~ X1)
Model1

Call:
lm(formula = Y1 ~ X1)

Coefficients:
(Intercept)           X1  
      8.784        1.114  

• Hence the values of the parameters are given by \[ a=8.784\qquad b=1.114 \]
• Lastly, we can plot the simple linear regression model \[y=8.784+1.114x\] onto our scatter plot as follows:

plot(X1,Y1,col="red",pch=18)
abline(Model1,col="darkgreen")

Example 3 - From Lectures

We were given the following data relating Weekly Natural Gas Consumption in a U.S. city along with the Hourly Average Temperature in that city for the same week.

Week	Hourly Avg. Temp. (\(^{\circ}\)F)	Weekly Natural Gas Consumption (MMcf)
1	28.0	12.4
2	28.0	11.7
3	32.5	12.4
4	39.0	10.8
5	45.9	9.4
6	57.8	9.5
7	58.1	8.0
8	62.5	7.5

Using this data answer the following:

1. Create a data file to represent this data.
2. Import this data into R.
3. Create two data vectors corresponding to temperature and consumption.
4. Create a scatter plot to represent this data
5. Use the lm() function to create a linear model for these data vectors.
6. Using this model estimate the parameters \(\hat{\beta}_0\) and \(\hat{\beta}_1\) for the simple linear model.
7. Plot the line of best fit along with the scatter plot for the data set.

Solution:

1. The data file for this data set is available on Moodle and is called FuelConsumption.csv

FuelData<-read.csv("FuelConsumption.csv")

FuelData

Temp<-FuelData$Temp
Temp

[1] 28.0 28.0 32.5 39.0 45.9 57.8 58.1 62.5

Fuel<-FuelData$Fuel
Fuel

[1] 12.4 11.7 12.4 10.8  9.4  9.5  8.0  7.5

plot(Temp,Fuel,col="red",pch=18)

FuelModel<-lm(Fuel ~ Temp)
FuelModel

Call:
lm(formula = Fuel ~ Temp)

Coefficients:
(Intercept)         Temp  
    15.8379      -0.1279  

6. The parameters of the Fuel Model are \[a=15.8379\quad b=-0.1279.\]

plot(Temp,Fuel,col="red",pch=18)
abline(FuelModel,col="darkgreen")

Exercises

Exercise 1 - Moodle - Problem Set 5 -Exercise 1

Using the data table given in Problem Set 5 – Exercise 1 answer the following:

1. Identify the dependent and independent variables.
2. Create a .csv file to represent this data.
3. Import this data file into R.
4. Create two data vectors from these data set.
5. Create a scatter plot using these data vectors.
6. Use the lm() function to create a linear model for these data vectors.
7. Using this model estimate the parameters \(\hat{\beta}_0\) and \(\hat{\beta}_1\) for line of best fit.
8. Plot the line of best fit along with the scatter plot for the data set.

Exercise 2 - Moodle - Problem Set 5 - Exercise 2

Using the data table given in Problem Set 5 – Exercise 2 answer the following:

1. Identify the dependent and independent variables.
2. Create a .csv file to represent this data.
3. Import this data file into R.
4. Create two data vectors from this data set.
5. Create a scatter plot using these data vectors.
6. Use the lm() function to create a linear model for these data vectors.
7. Using this model estimate the parameters \(\hat{\beta}_0\) and \(\hat{\beta}_1\) for line of best fit.
8. Plot the line of best fit along with the scatter plot for the data set.

Exercise 3

Using the data in the files IrelandInflation and IrelandEmissions available on Moodle answer the following

1. Import these two data files into R as two separate data structures.
2. Extract the data for CO2 Emissions and the data for InflationRate as two separate data structures.
3. Create a scatter plot for each of these data vectors.
4. Does there appear to be correlation between inflation and CO2 emissions, from the data?
5. Use the lm() function to create a linear model for these data vectors.
6. Use this model to estimate the parameters for the line of best fit.
7. Plot this line of best fit along with the scatter plot.