I am trying to find correlation between job satisfaction and intention to stay. In order to do so, I am trying to find correlation between Likert scale (agree, disagree, strongly disagree) and nominal question (Yes and No). Should I use Kendall or Pearson to find the correlation between the two?

  • $\begingroup$ Isn't this a point-biserial assuming your likert can be interpreted as continuous? $\endgroup$ – ReliableResearch Jul 1 '13 at 21:36
  • $\begingroup$ Have you done any study on the kinds of variables that you have and how they can be treated? $\endgroup$ – John Jul 1 '13 at 21:55

Neither Pearson nor Kendall correlation really seems suited here.

It sounds like you want a regression where the dependent variable is intention to stay and the independent variable is satisfaction. This could be handled with a logistic regression, although the independent variable should probably be treated as categorical.

However, in my experience, it is is rare for such studies to be based on a single Likert type question. Do you have only one?

  • $\begingroup$ Neither Pearson nor Kendall correlation really seems suited here Why so, Peter? Would you support your claim? $\endgroup$ – ttnphns Jul 1 '13 at 21:48
  • $\begingroup$ Because 1) Neither variable is numeric; point biserial would work if one was numeric and one was binary. 2) Regression seems to be what is needed, as there is a clear DV. $\endgroup$ – Peter Flom - Reinstate Monica Jul 1 '13 at 22:30
  • $\begingroup$ The Likert-type rating scale could be assumed to be ordinal or inteval. The dichotomous question is formally nominal, but since it is of 2 categories it is statistically equivalent to interval or ordinal (proof 1: if both variables are dichotomous, Phi = Pearson r = Kendall tau-b = Spearman rho = Eta; proof 2: dichotomous IV behaves identically as factor and as [when coded binary] covariate). So, your statement "neither Pearson nor Kendall suits here" is wrong. $\endgroup$ – ttnphns Jul 1 '13 at 23:18
  • $\begingroup$ Well, thank you for your answer. Actually, my independent variable is job satisfaction and my dependent variable is intention to stay. $\endgroup$ – Elle Jul 1 '13 at 23:32
  • 2
    $\begingroup$ @ttnphns, the job satisfaction variable is a single Likert item w/ only 3 levels. I fail to see how treating that as interval data is the best way to go. Logistic regression seems reasonable. $\endgroup$ – gung - Reinstate Monica Jul 2 '13 at 2:07

Your scale has only 3 levels so it's very unlikely to be conducive to any kind of regular least squares analysis. You cannot treat it as continuous. And, you can't really treat it as equal intervals either without some additional theory about the particular measure and evidence.

The easiest thing here looks to be a chi-square. That would tell you if there is a relationship between your measures of any shape. Unfortunately your measures aren't entirely independent so a regular chi-square test doesn't work and neither does a McNemar test which requires 2x2 tables. I believe that there are generalizations of the McNemar test that you could use but I'm not familiar with them. I don't see a nonparametric bootstrapping solution because there's nothing really to randomize within subjects.

If you're willing to place some restrictions on what you think happens to the relationship you have more options.

The suggestion of a logistic regression with the rating as a categorical predictor and intention to stay as a response is good. You'd need to turn the intention to stay into a numeric variable with just 0's and 1's.

Alternatively you might try an ordinal regression with your rating as response and intention to stay as a predictor. You should be expecting the relationship to be one of moving higher or lower in the ratings and not some other kind of change (like a more concentrated distribution in the middle). If your data appear to conform to these limitations you should be OK using either analysis.

So, that's how you should approach analysis. If you just want some kind of correlation type measure to compare across experiments, conditions, or other similar kinds of correlations, then you need one based on chi-square. One of the ones below will work.

# (you pass these functions the columns in a long format data.frame 
# that you want to assess, not the counts.)
# Pearson contingency coefficient function.
C <- function(v1, v2) {
    x <- chisq.test(v1, v2, correct = FALSE)$statistic
    sqrt( x/(length(v1) + x) )

# Cramer's V function
V <- function(v1, v2){
    tab <- table(v1, v2)
    X <- chisq.test(tab)$statistic
    n <- length(v1)
    q <- min(nrow(tab), ncol(tab))
    sqrt( X / (n*q) )
  • $\begingroup$ So, what i should use are not Kendall or Pearson. But, i could is logistic regression right? $\endgroup$ – Elle Jul 1 '13 at 23:37
  • $\begingroup$ Logistic regression is OK if you remember that it's supposed to be about a linear effect. It can't be Pearson because you have no continuous variables. You might be able to argue for Kendall by assigning numerical order to the variables but while correlation in concept can be analyzed none of the typical correlation coefficients work in this case. $\endgroup$ – John Jul 2 '13 at 0:32

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