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.

anscombe

##    x1 x2 x3 x4    y1   y2    y3    y4
## 1  10 10 10  8  8.04 9.14  7.46  6.58
## 2   8  8  8  8  6.95 8.14  6.77  5.76
## 3  13 13 13  8  7.58 8.74 12.74  7.71
## 4   9  9  9  8  8.81 8.77  7.11  8.84
## 5  11 11 11  8  8.33 9.26  7.81  8.47
## 6  14 14 14  8  9.96 8.10  8.84  7.04
## 7   6  6  6  8  7.24 6.13  6.08  5.25
## 8   4  4  4 19  4.26 3.10  5.39 12.50
## 9  12 12 12  8 10.84 9.13  8.15  5.56
## 10  7  7  7  8  4.82 7.26  6.42  7.91
## 11  5  5  5  8  5.68 4.74  5.73  6.89

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!)

mean(data$x1)

## [1] 9

var(data$x1)

## [1] 11

mean(data$x2)

## [1] 9

var(data$x2)

## [1] 11

mean(data$x3)

## [1] 9

var(data$x3)

## [1] 11

mean(data$x4)

## [1] 9

var(data$x4)

## [1] 11

mean(data$y1)

## [1] 7.500909

var(data$y1)

## [1] 4.127269

mean(data$y2)

## [1] 7.500909

var(data$y2)

## [1] 4.127629

mean(data$y3)

## [1] 7.5

var(data$y3)

## [1] 4.12262

mean(data$y4)

## [1] 7.500909

var(data$y4)

## [1] 4.123249

library('fBasics')

## Loading required package: timeDate

## Loading required package: timeSeries

##

## Rmetrics Package fBasics

## Analysing Markets and calculating Basic Statistics

## Copyright (C) 2005-2014 Rmetrics Association Zurich

## Educational Software for Financial Engineering and Computational Science

## Rmetrics is free software and comes with ABSOLUTELY NO WARRANTY.

## https://www.rmetrics.org --- Mail to: info@rmetrics.org

cor.test(data$x1,data$y1)

## 
##  Pearson's product-moment correlation
## 
## data:  data$x1 and data$y1
## t = 4.2415, df = 9, p-value = 0.00217
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
##  0.4243912 0.9506933
## sample estimates:
##       cor 
## 0.8164205

cor.test(data$x2,data$y2)

## 
##  Pearson's product-moment correlation
## 
## data:  data$x2 and data$y2
## t = 4.2386, df = 9, p-value = 0.002179
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
##  0.4239389 0.9506402
## sample estimates:
##       cor 
## 0.8162365

cor.test(data$x3,data$y3)

## 
##  Pearson's product-moment correlation
## 
## data:  data$x3 and data$y3
## t = 4.2394, df = 9, p-value = 0.002176
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
##  0.4240623 0.9506547
## sample estimates:
##       cor 
## 0.8162867

cor.test(data$x4,data$y4)

## 
##  Pearson's product-moment correlation
## 
## data:  data$x4 and data$y4
## t = 4.243, df = 9, p-value = 0.002165
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
##  0.4246394 0.9507224
## sample estimates:
##       cor 
## 0.8165214

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

plot(data$x1,data$y1)

plot(data$x2,data$y2)

plot(data$x3,data$y3)

plot(data$x4,data$y4)

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,pch=19)
plot(data$x2,data$y2,pch=19)
plot(data$x3,data$y3,pch=19)
plot(data$x4,data$y4,pch=19)

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

fit1<-lm(data$y1~data$x1)
fit2<-lm(data$y2~data$x2)
fit3<-lm(data$y3~data$x3)
fit4<-lm(data$y4~data$x4)

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(data$x1,data$y1,pch=19)
abline(fit1)
plot(data$x2,data$y2,pch=19)
abline(fit2)
plot(data$x3,data$y3,pch=19)
abline(fit3)
plot(data$x4,data$y4,pch=19)
abline(fit4)

Now compare the model fits for each model object.

anova(fit1)

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(fit2)

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(fit3)

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(fit4)

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.

When we explore the relationship between independent variable and dependent variable, it does give us a lot information from data itself, i.e. coefficient, etc. Data visualization makes this relationship more direct and visual. In question 6, it is much easier to tell which data pair has linear relationship and which data pair has strongest linear relationship. Relating this to business world, business moves very fast and leadership group wants to see more visual report than plain data, which makes the effect of data visualization more significant.

ANLY 512 - Problem Set 2

Anscombe’s quartet

Shan Huang

2017-05-26

Objectives

Questions