Mean difference (95% CI) between paired samples

Question

I have some results from a study pre- and post-intervention. It's for an underpowered feasibility study, hence I am not performing formal hypothesis testing but I'd like to calculate mean difference (with 95% CI) as per CONSORT 2010 guidelines.

My question is, what is the best approach for this? I am struggling for a couple of reasons:

1. I don't believe my data is normally distributed (it is blood test results and survey scores)
2. My data is paired, i.e., it's the same group of participants pre- and post.

And example of data is as follows:

Pre-intervention	Post-intervention
37.14	33.21
34.69	31.25
65.52	43.73
40.56	38.26
41.32	41.72
39.34	38.14
43.95	41.07
44.26	39.49
35.28	28.50
37.12	32.40
37.82	35.87
34.71	33.79
34.08	30.49
33.08	36.29
36.89	33.18
41.38	37.70
39.29	33.86
41.62	39.21
39.36	35.43

Answer 1

Here is an R solution applying the $t$-test for paired data.

x = scan()
1: 37.14    33.21
3: 34.69    31.25
5: 65.52    43.73
7: 40.56    38.26
9: 41.32    41.72
11: 39.34   38.14
13: 43.95   41.07
15: 44.26   39.49
17: 35.28   28.50
19: 37.12   32.40
21: 37.82   35.87
23: 34.71   33.79
25: 34.08   30.49
27: 33.08   36.29
29: 36.89   33.18
31: 41.38   37.70
33: 39.29   33.86
35: 41.62   39.21
37: 39.36   35.43

xmat <- matrix(x, ncol=2, byrow = TRUE)
t.test(xmat[,1], xmat[,2], paired = TRUE)

    Paired t-test

data:  xmat[, 1] and xmat[, 2]
t = 3.4757, df = 18, p-value = 0.002698
alternative hypothesis: true mean difference is not equal to 0
95 percent confidence interval:
 1.536795 6.233732
sample estimates:
mean difference 
       3.885263 

And here is another solution based on the Wilcoxon signed-rank test for two paired samples, aka the Mann-Whitney test:

wilcox.test(xmat[,1], xmat[,2], paired = TRUE, conf.int = TRUE,correct = TRUE, )

    Wilcoxon signed rank test with continuity correction

data:  xmat[, 1] and xmat[, 2]
V = 181, p-value = 0.0005795
alternative hypothesis: true location shift is not equal to 0
95 percent confidence interval:
 2.179984 4.350064
sample estimates:
(pseudo)median 
      3.314607 

Warning messages:
1: In wilcox.test.default(xmat[, 1], xmat[, 2], paired = TRUE, conf.int = TRUE,  :
  cannot compute exact p-value with ties
2: In wilcox.test.default(xmat[, 1], xmat[, 2], paired = TRUE, conf.int = TRUE,  :
  cannot compute exact confidence interval with ties

The two warning messages you see, as they say it, are due to the presence of ties for which the exact sampling distribution doesn't hold and we have to resort to asymptotic results.

Comment. The two confidence intervals are quite different although I'm not sure if this difference is relevant from a clinical point of view. Nevertheless, presumably this difference may be to the presence of a slight departure from normality shown by QQ-plot applied to the difference between the two variables.

qqnorm(xmat[,1]-xmat[,2])
qqline(xmat[,1]-xmat[,2])