# Correlation between dichotomous and continuous variable

I am trying to find the correlation between a dichotomous and a continuous variable.

From my ground work on this I found that I have to use independent t-test and the precondition for it is that the distribution of the variable has to be normal.

I performed Kolmogorov-Smirnov test for testing the normality and found that the continuous variable is non-normal and is skewed (for about 4,000 data points).

I did the Kolmogorov-Smirnov test for the entire range of variables. Should I split them into groups and do the test? I.e., say if I have risk level (0 = Not risky, 1 = Risky) and cholesterol levels, then should I:

• Divide them into two groups, like

Risk level =0 (Cholestrol level) -> Apply KS
Risk level =1 (Cholestrol level) -> Apply KS

• Take them together and apply the test? (I performed it on the whole dataset only.)

After that, what test should I do if it is still non-normal?

EDIT: The above scenario was just a description I tried to provide for my problem. I have a dataset which contains more than 1000 variables and about 4000 samples. They are either continuous or categorical in nature. My task is to predict a dichotomous variable based on these variables (maybe come up with a logistic regression model). So I thought the initial investigation would involve finding the correlation between dichotomous and a continuous variable.

I was trying to see how the distribution of the variables are and hence tried to go to t-test. Here I found the normality as an issue. The Kolmogorov-Smirnov test gave a significance value of 0.00 in most of these variables.

Should I assume normality here? The skewness and kurtosis of these variables also show that the data is skewed (>0) in almost all cases.

As per the note given below I will investigate the point-biserial correlation further. But about the distribution of variables I am still unsure.

• Correlation (of any sort) between a continuos and a binary (group) variable, is not much more (and maybe less ...) than just a comparison of means (some sort of mean ...) between the groups, so usually it should be better to just do that! Jun 10, 2014 at 11:54

I am a little confused; your title says "correlation" but your post refers to t-tests. A t-test is a test of central location - more specifically, is the mean of one set of data different from the mean of another set? Correlation, on the other hand, shows the relationship between two variables. There are a variety of correlation measures, it seems that point-biserial correlation is appropriate in your case.

You are correct that a t-test assumes normality; however, the tests of normality are likely to give significant results even for trivial non-normalities with an N of 4000. T-tests are fairly robust to modest deviations from normality if the variances of the two sets of data are roughly equal and the sample sizes roughly equal. But a nonparametric test is more robust to outliers and most of them have power almost as high as the t-test, even if the distributions are normal.

However, in your example, you use "cholesterol" as being risky or not-risky. This is almost certainly a bad idea. Dichotomizing a continuous variable invokes magical thinking. It says that, at some point, cholesterol goes from "not risky" to "risky". Suppose you used 200 as your cutoff - then you are saying that someone with cholesterol of 201 is just like someone with 400, and someone with 199 is just like someone with 100. This does not make sense.

• I agree, and I think most of us agree, that dichotomizing wastes information and that it can be a crude or coarse or clumsy method. I just think the "magical thinking" argument overreaches a bit. To choose to gloss over a difference is not the same as to believe there is no difference. I expect there will be times ahead when I'll find it convenient and worth the tradeoff to make categories out of some continuous variable, either for analytic or reporting purposes. Just my 2 cents. Mar 25, 2012 at 20:39
• Making categories out of continuous variables is worse than magical. Diabolical may be a better word. If you want to maximize model complexity, increase bias, and increase variance all at the same time, dichotomization is for you. [It maximizes complexity because the lost information due to categorization requires more variables to be added to the model to achieve the same $R^2$.] May 5, 2013 at 16:35

Let's simplify things. With N = 4,000 for cholesterol level, you should have no problem with your results being biased by outliers. Therefore you can use correlation itself, as implied by your initial sentence. It will make little difference whether you assess correlation via the Pearson, Spearman, or Point-Biserial method.

If instead you really need to phrase results in terms of typical cholesterol difference between High-Risk and Low-Risk groups, the Mann-Whitney U test is fine to use, but you may as well use the more informative t-test. With this N (and again, with astronomical outliers something you can no doubt rule out), you needn't worry that the lack of normality will compromise your results.

• Thanks for your reply. But If i have to know about the outliers makes a big distortion is it correct to use kurtosis and skewness to detect it? In case if this is true above what values of kurtosis and skewness should i assume that the distribution is not normal. Thanks for your reply Mar 25, 2012 at 16:35
• I'm assuming based on limited content knowledge that with cholesterol, you won't have any values that are many orders of magnitude higher than the others. That's why I think you can use a parametric method such as correlation or a t-test. It's not that I think the distribution is normal. You don't need it to be normal. By the way, in light of Peter's answer: I believed (and hope) that you had some source of the High/Low Risk status that was independent of cholesterol score. I agree that it's probably not helpful to dichotomize. Mar 25, 2012 at 20:30
• Can I suggest you add a section to your original question, marked "EDIT:....", that spells out what questions remain for you that have not been addressed by the answers and comments you've received so far. Mar 25, 2012 at 20:46
• Thanks for your suggestion.I have updated the same .Sorry for ambiguous question in first place.Thanks Mar 25, 2012 at 21:49