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I would like to compare a continuous variable across 5 health status groups, but one of the groups has only 1 observation. Would an ANOVA/Kruskal-Wallis be valid? What can I do about the group with only 1 observation?

The boxplots for the continuous variable across the groups look as follows: enter image description here

The group sizes are:

A: 12
B: 8
C: 9
D: 7
E: 1

Background info: I work in biostatistics and my datasets are fairly small, and some outcomes are rare in biology. Collecting samples is also dependent on patients and is an expensive procedure, so getting a larger sample is not a feasible solution.

Edit: What I currently do is to omit the group with only one observation in my analysis, however I am not sure about the validity of that.

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  • $\begingroup$ What happens if you run your ANOVA/Kruskal-Wallis with all five groups? $\endgroup$ – amoeba Apr 26 '16 at 21:01
  • $\begingroup$ @amoeba If I run it with all 5 groups, I still get a p-value output (ANOVA 0.49 and Kruskal-Wallis 0.26) but I am not sure of the validity of the output. $\endgroup$ – Sheep Apr 27 '16 at 16:58
  • $\begingroup$ It's valid all right. You have less power as compared to the situation when groups had more balanced size (but the same total size), but this is not something you can change now. Kicking out this one group with 1 observation will probably give you even less power. $\endgroup$ – amoeba May 1 '16 at 23:33
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You can perform inference in one-way-ANOVA type designs where there's a group with only one observation if you make the equal-variance assumption. If you don't assume equal variance (or some other informative variance structure), then you won't have information about variance in the singleton group.

Not all packages will deal with it in their implementation of ANOVA, it depends on how they're set up, but that doesn't mean it can't be done. [I gave an example of performing a t-test with a singleton in another answer on site.]

Here's an example in R with a singleton group both included and omitted for both one-way ANOVA and Kruskal-Wallis:

x=rnorm(100)
g=as.factor(rep(1:5,c(40,30,20,9,1)))
anova(lm(x~g))
Analysis of Variance Table

Response: x
          Df Sum Sq Mean Sq F value Pr(>F)
g          4  6.554 1.63839  1.6588  0.166
Residuals 95 93.830 0.98769               

anova(lm(x[-100]~g[-100]))
Analysis of Variance Table

Response: x[-100]
          Df Sum Sq Mean Sq F value Pr(>F)
g[-100]    3  3.498 1.16608  1.1806 0.3213
Residuals 95 93.830 0.98769               

kruskal.test(x~g)

        Kruskal-Wallis rank sum test

data:  x by g
Kruskal-Wallis chi-squared = 5.9232, df = 4, p-value = 0.205

kruskal.test(x[-100]~g[-100])

        Kruskal-Wallis rank sum test

data:  x[-100] by g[-100]
Kruskal-Wallis chi-squared = 3.1894, df = 3, p-value = 0.3633
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