ANLY 512 - Problem Set 2

Questions

Anscombes quartet is a set of 4 \(x,y\) data sets that were published by Francis Anscombe in a 1973 paper Graphs in statistical analysis. For this first question load the anscombe data that is part of the library(datasets) in R. And assign that data to a new object called data.

data <- anscombe

Summarise the data by calculating the mean, variance, for each column and the correlation between each pair (eg. x1 and y1, x2 and y2, etc) (Hint: use the fBasics() package!)

summary(data)

##        x1             x2             x3             x4    
##  Min.   : 4.0   Min.   : 4.0   Min.   : 4.0   Min.   : 8  
##  1st Qu.: 6.5   1st Qu.: 6.5   1st Qu.: 6.5   1st Qu.: 8  
##  Median : 9.0   Median : 9.0   Median : 9.0   Median : 8  
##  Mean   : 9.0   Mean   : 9.0   Mean   : 9.0   Mean   : 9  
##  3rd Qu.:11.5   3rd Qu.:11.5   3rd Qu.:11.5   3rd Qu.: 8  
##  Max.   :14.0   Max.   :14.0   Max.   :14.0   Max.   :19  
##        y1               y2              y3              y4        
##  Min.   : 4.260   Min.   :3.100   Min.   : 5.39   Min.   : 5.250  
##  1st Qu.: 6.315   1st Qu.:6.695   1st Qu.: 6.25   1st Qu.: 6.170  
##  Median : 7.580   Median :8.140   Median : 7.11   Median : 7.040  
##  Mean   : 7.501   Mean   :7.501   Mean   : 7.50   Mean   : 7.501  
##  3rd Qu.: 8.570   3rd Qu.:8.950   3rd Qu.: 7.98   3rd Qu.: 8.190  
##  Max.   :10.840   Max.   :9.260   Max.   :12.74   Max.   :12.500

var(data$x1)

## [1] 11

var(data$x2)

## [1] 11

var(data$x3)

## [1] 11

var(data$x4)

## [1] 11

var(data$y1)

## [1] 4.127269

var(data$y2)

## [1] 4.127629

var(data$y3)

## [1] 4.12262

var(data$y4)

## [1] 4.123249

cor(data)

##            x1         x2         x3         x4         y1         y2
## x1  1.0000000  1.0000000  1.0000000 -0.5000000  0.8164205  0.8162365
## x2  1.0000000  1.0000000  1.0000000 -0.5000000  0.8164205  0.8162365
## x3  1.0000000  1.0000000  1.0000000 -0.5000000  0.8164205  0.8162365
## x4 -0.5000000 -0.5000000 -0.5000000  1.0000000 -0.5290927 -0.7184365
## y1  0.8164205  0.8164205  0.8164205 -0.5290927  1.0000000  0.7500054
## y2  0.8162365  0.8162365  0.8162365 -0.7184365  0.7500054  1.0000000
## y3  0.8162867  0.8162867  0.8162867 -0.3446610  0.4687167  0.5879193
## y4 -0.3140467 -0.3140467 -0.3140467  0.8165214 -0.4891162 -0.4780949
##            y3         y4
## x1  0.8162867 -0.3140467
## x2  0.8162867 -0.3140467
## x3  0.8162867 -0.3140467
## x4 -0.3446610  0.8165214
## y1  0.4687167 -0.4891162
## y2  0.5879193 -0.4780949
## y3  1.0000000 -0.1554718
## y4 -0.1554718  1.0000000

Create scatter plots for each \(x, y\) pair of data.

plot(data$x1,data$y1,main="Scatter Plots for each X and Y",xlab="x1",ylab="y1",pch=0)

plot(data$x2,data$y2,main="Scatter Plots for each X and Y",xlab="x2",ylab="y2",pch=0)

plot(data$x3,data$y3,main="Scatter Plots for each X and Y",xlab="x3",ylab="y3",pch=0)

plot(data$x4,data$y4,main="Scatter Plots for each X and Y",xlab="x4",ylab="y4",pch=0)

Now change the symbols on the scatter plots to solid circles and plot them together as a 4 panel graphic

par(mfrow=c(2,2))
plot(data$x1,data$y1, main="Scatterplot between x1,y1",pch=19) 
plot(data$x2,data$y2, main="Scatterplot between x2,y2",pch=19) 
plot(data$x3,data$y3, main="Scatterplot between x3,y3",pch=19) 
plot(data$x4,data$y4, main="Scatterplot between x4,y4",pch=19)

Now fit a linear model to each data set using the lm() function.

model1<-lm(data$y1~data$x1)
model1

## 
## Call:
## lm(formula = data$y1 ~ data$x1)
## 
## Coefficients:
## (Intercept)      data$x1  
##      3.0001       0.5001

model2<-lm(data$y2~data$x2)
model2

## 
## Call:
## lm(formula = data$y2 ~ data$x2)
## 
## Coefficients:
## (Intercept)      data$x2  
##       3.001        0.500

model3<-lm(data$y3~data$x3)
model3

## 
## Call:
## lm(formula = data$y3 ~ data$x3)
## 
## Coefficients:
## (Intercept)      data$x3  
##      3.0025       0.4997

model4<-lm(data$y4~data$x4)
model4

## 
## Call:
## lm(formula = data$y4 ~ data$x4)
## 
## Coefficients:
## (Intercept)      data$x4  
##      3.0017       0.4999

Now combine the last two tasks. Create a four panel scatter plot matrix that has both the data points and the regression lines. (hint: the model objects will carry over chunks!)

par(mfrow=c(2,2))
plot(model1)

plot(model2)

plot(model3)

plot(model4)

## Warning: not plotting observations with leverage one:
##   8

## Warning: not plotting observations with leverage one:
##   8

Now compare the model fits for each model object.

anova(model1)

Analysis of Variance Table

Response: data\(y1 Df Sum Sq Mean Sq F value Pr(>F) data\)x1 1 27.510 27.5100 17.99 0.00217 ** Residuals 9 13.763 1.5292
— Signif. codes: 0 ‘’ 0.001 ’’ 0.01 ’’ 0.05 ‘.’ 0.1 ‘’ 1

anova(model2)

Analysis of Variance Table

Response: data\(y2 Df Sum Sq Mean Sq F value Pr(>F) data\)x2 1 27.500 27.5000 17.966 0.002179 ** Residuals 9 13.776 1.5307
— Signif. codes: 0 ‘’ 0.001 ’’ 0.01 ’’ 0.05 ‘.’ 0.1 ‘’ 1

anova(model3)

Analysis of Variance Table

Response: data\(y3 Df Sum Sq Mean Sq F value Pr(>F) data\)x3 1 27.470 27.4700 17.972 0.002176 ** Residuals 9 13.756 1.5285
— Signif. codes: 0 ‘’ 0.001 ’’ 0.01 ’’ 0.05 ‘.’ 0.1 ‘’ 1

anova(model4)

Analysis of Variance Table

Response: data\(y4 Df Sum Sq Mean Sq F value Pr(>F) data\)x4 1 27.490 27.4900 18.003 0.002165 ** Residuals 9 13.742 1.5269
— Signif. codes: 0 ‘’ 0.001 ’’ 0.01 ’’ 0.05 ‘.’ 0.1 ‘’ 1

In text, summarize the lesson of Anscombe’s Quartet and what it says about the value of data visualization. It stresses the importance of visualization in understanding data

ANLY 512 - Problem Set 2

Anscombe’s quartet

Your Name

2018-06-17

Objectives

Questions