In our last post, we calculated Pearson and Spearman correlation coefficients in R and got a surprising result.
So let’s investigate the data a little more with a scatter plot.
We use the same version of the data set of tourists. We have data on tourists from different nations, their gender, number of children, and how much they spent on their trip.
Again we copy and paste the following array into R.
M <- structure(list(COUNTRY = structure(c(3L, 3L, 3L, 3L, 1L, 3L, 2L, 3L, 1L, 3L, 3L, 1L, 2L, 2L, 3L, 3L, 3L, 2L, 3L, 1L, 1L, 3L,
1L, 2L), .Label = c("AUS", "JAPAN", "USA"), class = "factor"),GENDER = structure(c(2L, 1L, 2L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 2L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 1L, 2L), .Label = c("F", "M"), class = "factor"), CHILDREN = c(2L, 1L, 3L, 2L, 2L, 3L, 1L, 0L, 1L, 0L, 1L, 2L, 2L, 1L, 1L, 1L, 0L, 2L, 1L, 2L, 4L, 2L, 5L, 1L), SPEND = c(8500L, 23000L, 4000L, 9800L, 2200L, 4800L, 12300L, 8000L, 7100L, 10000L, 7800L, 7100L, 7900L, 7000L, 14200L, 11000L, 7900L, 2300L, 7000L, 8800L, 7500L, 15300L, 8000L, 7900L)), .Names = c("COUNTRY", "GENDER", "CHILDREN", "SPEND"), class = "data.frame", row.names = c(NA, -24L))
M
attach(M)
plot(CHILDREN, SPEND)
(more…)
Let’s use R to explore bivariate relationships among variables.
Part 7 of this series showed how to do a nice bivariate plot, but it’s also useful to have a correlation statistic.
We use a new version of the data set we used in Part 20 of tourists from different nations, their gender, and number of children. Here, we have a new variable – the amount of money they spend while on vacation.
First, if the data object (A) for the previous version of the tourists data set is present in your R workspace, it is a good idea to remove it because it has some of the same variable names as the data set that you are about to read in. We remove A as follows:
rm(A)
Removing the object A ensures no confusion between different data objects that contain variables with similar names.
Now copy and paste the following array into R.
M <- structure(list(COUNTRY = structure(c(3L, 3L, 3L, 3L, 1L, 3L, 2L, 3L, 1L, 3L, 3L, 1L, 2L, 2L, 3L, 3L, 3L, 2L, 3L, 1L, 1L, 3L,
1L, 2L), .Label = c("AUS", "JAPAN", "USA"), class = "factor"),GENDER = structure(c(2L, 1L, 2L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 2L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 1L, 2L), .Label = c("F", "M"), class = "factor"), CHILDREN = c(2L, 1L, 3L, 2L, 2L, 3L, 1L, 0L, 1L, 0L, 1L, 2L, 2L, 1L, 1L, 1L, 0L, 2L, 1L, 2L, 4L, 2L, 5L, 1L), SPEND = c(8500L, 23000L, 4000L, 9800L, 2200L, 4800L, 12300L, 8000L, 7100L, 10000L, 7800L, 7100L, 7900L, 7000L, 14200L, 11000L, 7900L, 2300L, 7000L, 8800L, 7500L, 15300L, 8000L, 7900L)), .Names = c("COUNTRY", "GENDER", "CHILDREN", "SPEND"), class = "data.frame", row.names = c(NA, -24L))
M
attach(M)
Do tourists with greater numbers of children spend more? Let’s calculate the correlation between CHILDREN and SPEND, using the cor() function.
R <- cor(CHILDREN, SPEND)
[1] -0.2612796
We have a weak correlation, but it’s negative! Tourists with a greater number of children tend to spend less rather than more!
(Even so, we’ll plot this in our next post to explore this unexpected finding).
We can round to any number of decimal places using the round() command.
round(R, 2)
[1] -0.26
The percentage of shared variance (100*r2) is:
100 * (R**2)
[1] 6.826704
To test whether your correlation coefficient differs from 0, use the cor.test() command.
cor.test(CHILDREN, SPEND)
Pearson's product-moment correlation
data: CHILDREN and SPEND
t = -1.2696, df = 22, p-value = 0.2175
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
-0.6012997 0.1588609
sample estimates:
cor
-0.2612796
The cor.test() command returns the correlation coefficient, but also gives the p-value for the correlation. In this case, we see that the correlation is not significantly different from 0 (p is approximately 0.22).
Of course we have only a few values of the variable CHILDREN, and this fact will influence the correlation. Just how many values of CHILDREN do we have? Can we use the levels() command directly? (Recall that the term “level” has a few meanings in statistics, once of which is the values of a categorical variable, aka “factor“).
levels(CHILDREN)
NULL
R does not recognize CHILDREN as a factor. In order to use the levels() command, we must turn CHILDREN into a factor temporarily, using as.factor().
levels(as.factor(CHILDREN))
[1] "0" "1" "2" "3" "4" "5"
So we have six levels of CHILDREN. CHILDREN is a discrete variable without many values, so a Spearman correlation can be a better option. Let’s see how to implement a Spearman correlation:
cor(CHILDREN, SPEND, method ="spearman")
[1] -0.3116905
We have obtained a similar but slightly different correlation coefficient estimate because the Spearman correlation is indeed calculated differently than the Pearson.
Why not plot the data? We will do so in our next post.
About the Author: David Lillis has taught R to many researchers and statisticians. His company, Sigma Statistics and Research Limited, provides both on-line instruction and face-to-face workshops on R, and coding services in R. David holds a doctorate in applied statistics.
See our full R Tutorial Series and other blog posts regarding R programming.
