# Is it possible to calculate a confidence interval related to a significant p value from the kruskal wallis test?

I have run the nonparametric Kruskal Wallis test with pairwise comparisons in SPSS to understand if my 3 groups of survey participants were different in their responses to a range of 5 point ordinal scale questions (data is not normally distributed).

Some are responses significant at p < 0.05, some are not, as expected and the results fit with our what we expected to find. A particular journal I would like to submit to requires I report 95% confidence intervals with any p values reported, is this possible? It doesn't make sense to me as my understanding of the KW test was it uses ranks of the median. Thanks in advance.

To follow up and provide further information: I'm now running the Wilcoxon tests in R, this is anexample of an output where p<0.0167 but the confidence interval includes 0 as the CI itself is so small:

wilcox.test(Data$$Max[Data$$Cluster==1], Data$$Max[Data$$Cluster==3],conf.int=T,conf.lev=.983)

    Wilcoxon rank sum test with continuity correction


data: Data$$Max[Data$$Cluster == 1] and Data$$Max[Data$$Cluster == 3]
W = 19368, p-value = 0.01348
alternative hypothesis: true location shift is not equal to 0
98.3 percent confidence interval:
-2.390968e-05 4.505737e-05
sample estimates:
difference in location
6.994035e-06

The data are very skewed! Here is the table of my groups (cluster 1-3) x answers to the question "Max" 1 = Strongly oppose through to 5 = Strongly support

table(Data$$Cluster, Data$$Max)

  1   2   3   4   5


1 1 6 12 38 210
2 0 6 11 34 103
3 0 4 10 29 87

• Are you sure they require it in this case? For coefficients or other parameter estimates, it makes sense, but this is not a parameter estimate. Perhaps reporting where the 95th percentile of the $\chi^2_{n-1}$ distribution is as well as the KW test statistic itself would do the job (the $\chi^2_{n-1}$ distribution is the asymptotic distribution of the KW test statistic under the null hypothesis with $n$ groups.) – jbowman Sep 27 '18 at 2:32
• You can do what's effective a joint confidence region of location differences, but with three groups you'd be looking at two differences (e.g. G2 vs G1 and G3 vs G1). An example of a joint region is here stats.stackexchange.com/questions/76059/… ... alternatively you could set up some (marginal or joint) contrasts of interest and plot the acceptance regions for those (which would correspond to confidence intervals). ... ctd – Glen_b Sep 27 '18 at 2:41
• ctd... If something like that is not what you seek, you'll have to explain clearly what specific quantity you need a confidence interval for; is it possible to point to a specific statement by the journal about these requirements? Note that none of the intervals you would generate will be either a confidence interval for a median, or a difference of medians. – Glen_b Sep 27 '18 at 5:40

My interpretation of this is as follows. If I am wrong, please give the kind of clarification suggested by @whuber, perhaps along with some sample data to illustrate what you are doing.

If the Kruskal-Wallis test rejects the null hypothesis that the three medians are all equal, then you will use two-sample Wilcoxon tests to do multiple comparisons A vs B, B vs C and A vs C. In order to control the overall error rate for the three comparisons you might use the Bonferroni significance level $$.05/3 = .0167$$ for the comparisons.

Recently, some psychology and sociology journals have blamed irreproducibility of certain results on abuse of P-values, and ask for confidence intervals (CIs) in addition to or instead of P-values. (I'm not saying they are correct to deprecate P-values or that asking for CIs always makes sense, just stating what I have observed and heard.)

You might give $$(100 - 1.67)\% = 98.3\%$$ CIs for the differences in medians. Presumably, these could be CIs produced by Wilcoxon test procedures. A difficulty may be that a 5-point ordinal scale might produce some ties, but perhaps the approximate CIs given in spite of that would be useful.

I doubt that the journal is asking for a CI for the overall Kruskal-Wallis test, but if so, perhaps use @jbowman's suggestion.

In the tentative exploration (in R) below I use fake simulated data for groups A, B, and C. There are $$n = 50$$ responses in each group, summarized as follows:

summary(A)
Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
1.190   2.212   2.935   2.817   3.280   4.700
summary(B)
Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
1.350   2.695   3.450   3.331   4.107   4.690
summary(C)
Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
1.620   3.328   3.800   3.672   4.228   4.690


Concatenating the data to the vector X and making a group variable gp, we have the following notched boxplot. Notches in the sides of the boxes are approximate nonparametric CIs for individual group medians, calibrated so that two non-overlapping CIs indicate a significant difference. Roughly, it seems that A and B may differ significantly, that B and C clearly do not, and that A and C are obviously significantly different.

 kruskal.test(X ~ gp)

Kruskal-Wallis rank sum test

data:  X by gp
Kruskal-Wallis chi-squared = 17.887, df = 2, p-value = 0.0001306


So there is no doubt that the groups vary. Now we do three 2-sample Wilcoxon tests. Remember that we are looking for P-values below .0167 in order to declare significant differences.

wilcox.test(A, B, conf.int=T, conf.lev=.983)

Wilcoxon rank sum test with continuity correction

data:  A and B
W = 825, p-value = 0.003428
alternative hypothesis: true location shift is not equal to 0
98.3 percent confidence interval:
-1.0200705 -0.1199649
sample estimates:
difference in location
-0.5900226


.

wilcox.test(B, C, conf.int=T, conf.lev=.983)

Wilcoxon rank sum test with continuity correction

data:  B and C
W = 984, p-value = 0.06719
alternative hypothesis: true location shift is not equal to 0
98.3 percent confidence interval:
-0.72004953  0.09001254
sample estimates:
difference in location
-0.3000609


.

wilcox.test(A, C, conf.int=T, conf.lev=.983)

Wilcoxon rank sum test with continuity correction

data:  A and C
W = 537.5, p-value = 9.175e-07
alternative hypothesis: true location shift is not equal to 0
98.3 percent confidence interval:
-1.3099358 -0.5300047
sample estimates:
difference in location
-0.9199606


Summarizing, we see that A and B are significantly different according to the Bonferroni criterion [CI $$(-1.02, -0.12)$$]; B and C are not significantly different [CI includes 0]; A and C are highly significantly different [CI $$(-1.31 -0.53)].$$

Note: Data were simulated as follows:

set.seed(918); n = 50
A = 1+round(4*rbeta(n, 2, 2),2)
B = 1+round(4*rbeta(n, 3, 2),2)
C = 1+round(4*rbeta(n, 3.5, 2),2)
X = c(A,B,C);  gp=rep(1:3, each=n)

• Thanks for your help - I managed to get this working and it fits with what I need. Very much appreciate it. – ashamc Oct 19 '18 at 5:15
• So to please clarify your answer further, if these tests showed p<0.0167 BUT the CI includes 0, do I then report these as non-significant? – ashamc Nov 3 '18 at 4:33
• P-value < 5% seems inconsistent with 95% CI that includes 0. Something seems wrong with that. Sometimes with software implementations of nonparametric procedures the approximation for the test and the aprx for the CI don't match exactly. But 0.017 is much below 0.05, so it's hard to believe a 95% CI including 0. What software are you using? Can you show relevant input and output? // Are you using Bonferroni CI's to control family error rate. If so, with K-W, its possible overall test rejects, but that doesn't mean all pairs are signif different. – BruceET Nov 3 '18 at 5:04
• I have edited the original question to provide more information – ashamc Nov 3 '18 at 5:18

The Kruskal-Wallis test (generalization of the Wilcoxon-Mann-Whitney test) is a special case of the proportional odds (PO) ordinal logistic regression model. You can use the the PO model to get any contrast you need, and even to get simultaneous confidence intervals. These effects are on the odds ratio scale. The R rms package orm, contrast.rms, and summary.rms functions make these easy to do.