I have two variables which are measured on a Likert scale. I ran the Spearman correlation test to test the relationship between them and they have a negative correlation coefficient (-0.729 with a p-value < 1.2e-12). I am pretty new to plotting with R and my question is, is there any way I can plot the relationship between the two variables? I am having difficulties with what kind of plot can depict the relationship since they are discrete.


1 Answer 1


One idea is to do a jittered scatterplot.

I simulated 100 Likert scores x from 1 to 7, and roughly positively associated Likert scores y. They are summarized and tabulated below.

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   1.00    3.00    5.00    4.48    6.00    7.00 
 1  2  3  4  5  6  7 
 5  8 13 22 24 16 12 

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   1.00    3.00    4.00    3.94    5.00    7.00 
 1  2  3  4  5  6  7 
 9 11 19 23 19 14  5 

cor(x, y, meth="s")
[1] 0.9044931

A standard scatterplot shows locations of the 100 $(x,y)$ pairs, but there is a lot of over-plotting and we can't get an idea how many individuals are represented by each point.

plot(x, y)

enter image description here

By jittering (adding a small amount of random noise to) each score, I can see how many individuals correspond to each location.

xj = x + runif(100, -.2,.2)
yj = y + runif(100, -.2,.2)
plot(xj, yj, pch=20)
 abline(v=1:7, col="green2")
 abline(h=1:7, col="green2")

enter image description here

Here I have jittered by $\pm 0.2$ or less. One can choose the amount of jittering---depending on the sample size and the strength of the association---to seek the best graphical effect. If you have a very large sample size, then you may want to use small dots for individual subjects, by using parameter pch-"." in plot,


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.