Objectives

The objectives of this problem set is to orient you to a number of activities in R and to conduct a thoughtful exercise in appreciating the importance of data visualization. For each question enter your code or text response in the code chunk that completes/answers the activity or question requested. To submit this homework you will create the document in Rstudio, using the knitr package (button included in Rstudio) and then submit the document to your Rpubs account. Once uploaded you will submit the link to that document on Canvas. Please make sure that this link is hyper linked and that I can see the visualization and the code required to create it. Each question is worth 5 points.

Questions

  1. Anscombe’s 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.
library(datasets)
data<-anscombe
attach(data)
  1. 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 dplyr package!)
library(dplyr)
sapply(data, var)
##        x1        x2        x3        x4        y1        y2        y3        y4 
## 11.000000 11.000000 11.000000 11.000000  4.127269  4.127629  4.122620  4.123249
sapply(data, mean)
##       x1       x2       x3       x4       y1       y2       y3       y4 
## 9.000000 9.000000 9.000000 9.000000 7.500909 7.500909 7.500000 7.500909
cor(data[,1:4],data[,5:8])
##            y1         y2         y3         y4
## x1  0.8164205  0.8162365  0.8162867 -0.3140467
## x2  0.8164205  0.8162365  0.8162867 -0.3140467
## x3  0.8164205  0.8162365  0.8162867 -0.3140467
## x4 -0.5290927 -0.7184365 -0.3446610  0.8165214
  1. Using ggplot, create scatter plots for each \(x, y\) pair of data (maybe use ‘facet_grid’ or ‘facet_wrap’).
library(ggplot2)
par(mfrow=c(2,2))
plot(x1,y1,main="x1 against y1", xlab = "x1", ylab = "y1")
plot(x1,y1,main="x1 against y1", xlab = "x1", ylab = "y1")
plot(x2,y2,main="x2 against y2", xlab = "x2", ylab = "y2")
plot(x3,y3,main="x3 against y3", xlab = "x3", ylab = "y3")

plot(x4,y4,main="x4 against y4", xlab = "x4", ylab = "y4")

  1. Now change the symbols on the scatter plots to solid blue circles.
par(mfrow=c(2,2))
plot(x1,y1,main="x1 against y1", xlab = "x1", ylab = "y1", col = "blue")
plot(x2,y2,main="x2 against y2", xlab = "x2", ylab = "y2", col = "blue")
plot(x3,y3,main="x3 against y3", xlab = "x3", ylab = "y3", col = "blue")
plot(x4,y4,main="x4 against y4", xlab = "x4", ylab = "y4", col = "blue")

  1. Now fit a linear model to each data set using the lm() function.
model1<-lm(y1~x1)
model2<-lm(y2~x2)
model3<-lm(y3~x3)
model4<-lm(y4~x4)
  1. 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(x1,y1,main="x1 against y1", xlab = "x1", ylab = "y1", col = "blue", abline(model1,col="red"))
plot(x2,y2,main="x2 against y2", xlab = "x2", ylab = "y2", col = "blue", abline(model2,col="red"))
plot(x3,y3,main="x3 against y3", xlab = "x3", ylab = "y3", col = "blue", abline(model3,col="red"))
plot(x4,y4,main="x4 against y4", xlab = "x4", ylab = "y4", col = "blue", abline(model4,col="red"))

  1. Now compare the model fits for each model object.
summary(model1)

Call: lm(formula = y1 ~ x1)

Residuals: Min 1Q Median 3Q Max -1.92127 -0.45577 -0.04136 0.70941 1.83882

Coefficients: Estimate Std. Error t value Pr(>|t|)
(Intercept) 3.0001 1.1247 2.667 0.02573 * x1 0.5001 0.1179 4.241 0.00217 ** — Signif. codes: 0 ‘’ 0.001 ’’ 0.01 ’’ 0.05 ‘.’ 0.1 ’ ’ 1

Residual standard error: 1.237 on 9 degrees of freedom Multiple R-squared: 0.6665, Adjusted R-squared: 0.6295 F-statistic: 17.99 on 1 and 9 DF, p-value: 0.00217

summary(model2)

Call: lm(formula = y2 ~ x2)

Residuals: Min 1Q Median 3Q Max -1.9009 -0.7609 0.1291 0.9491 1.2691

Coefficients: Estimate Std. Error t value Pr(>|t|)
(Intercept) 3.001 1.125 2.667 0.02576 * x2 0.500 0.118 4.239 0.00218 ** — Signif. codes: 0 ‘’ 0.001 ’’ 0.01 ’’ 0.05 ‘.’ 0.1 ’ ’ 1

Residual standard error: 1.237 on 9 degrees of freedom Multiple R-squared: 0.6662, Adjusted R-squared: 0.6292 F-statistic: 17.97 on 1 and 9 DF, p-value: 0.002179

summary(model3)

Call: lm(formula = y3 ~ x3)

Residuals: Min 1Q Median 3Q Max -1.1586 -0.6146 -0.2303 0.1540 3.2411

Coefficients: Estimate Std. Error t value Pr(>|t|)
(Intercept) 3.0025 1.1245 2.670 0.02562 * x3 0.4997 0.1179 4.239 0.00218 ** — Signif. codes: 0 ‘’ 0.001 ’’ 0.01 ’’ 0.05 ‘.’ 0.1 ’ ’ 1

Residual standard error: 1.236 on 9 degrees of freedom Multiple R-squared: 0.6663, Adjusted R-squared: 0.6292 F-statistic: 17.97 on 1 and 9 DF, p-value: 0.002176

summary(model4)

Call: lm(formula = y4 ~ x4)

Residuals: Min 1Q Median 3Q Max -1.751 -0.831 0.000 0.809 1.839

Coefficients: Estimate Std. Error t value Pr(>|t|)
(Intercept) 3.0017 1.1239 2.671 0.02559 * x4 0.4999 0.1178 4.243 0.00216 ** — Signif. codes: 0 ‘’ 0.001 ’’ 0.01 ’’ 0.05 ‘.’ 0.1 ’ ’ 1

Residual standard error: 1.236 on 9 degrees of freedom Multiple R-squared: 0.6667, Adjusted R-squared: 0.6297 F-statistic: 18 on 1 and 9 DF, p-value: 0.002165

  1. In text, summarize the lesson of Anscombe’s Quartet and what it says about the value of data visualization.
#Data visualization is a very important aspect of statistical analysis. Often, knowing only a correlation coefficient is not enough to understand the relationship between the variables fully. In this lesson, we found that the correlation coefficients between each pair of xi and yi (i = 1,2,3,4) are equal. However, the visualization showed us how much the situation is varied. In the first plot, there was a significantly dispersed data, but the points were quite balanced above and below the line, and therefore the correlation was strong. In the second example, we clearly see a curvilinear association, and the linear approximation is not an appropriate option for this relationship. Without visualization, we could not know about that. The third plot shows how one outlier may ruin the whole picture of the relationship. Without that point, we could have found a functional relationship (with 100% R-squared). Finally, the last plot illustrates another influence of an outlier - when it creates an impression about an existent association. As we can see, the other points form a vertical line parallel to y-axis. This means that the variables are not correlated. However, an outlier affects the linear trend and makes a significant correlation and linear model.
#Summing up, we should always support our statistical inference procedures with appropriate plots. Otherwise, we may conclude a misleading information about the relationships in the data.