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I have multiple rows of observation periods [OP] (of varying durations) showing the average food intake rate of the first quarter of the OP compared to the last quarter of the OP. See the table below.

OP Intake.Q1 Intake.Q4
1 6 6
2 19.5 5
3 5.1 2.5
4 4.7 2.1
5 10.6 5.4

I want to find out if there is a significant difference between the first and the last quarter of each OP.

If I run a Wilcoxon test in R:

wilcox.test(IntakeQ1, IntakeQ4, alternative=("two.sided"), paired=TRUE)

My question is: How will R analyse this data, because I need it to compare the time points in each OP then give me a final result of whether intake rate significantly changes during observation period.

So will it compare 6 & 6, then 19.5 & 5, then 5.1 and 2.5, etc. then give me a final output, or will it aggregate everything in Intake.Q1 and compare to everything in Intake.Q4?

Thank you.

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Yes, since a X and Y are provided and paired=TRUE option is used.

From help:

"If both x and y are given and paired is TRUE, a Wilcoxon signed rank test of the null that the distribution of of x - y (in the paired two sample case) is symmetric about mu is performed."

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  • $\begingroup$ I'm sorry, just to clarify, it will analyse row by row? $\endgroup$ Feb 19 '21 at 17:36
  • $\begingroup$ Yes, it will pair the first x with the first y, the second x with the second y, etc... $\endgroup$
    – Dave2e
    Feb 19 '21 at 18:29
  • $\begingroup$ Nice idea to quote R help. (+1) $\endgroup$
    – BruceET
    Feb 20 '21 at 1:34
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Putting your data into R as vectors q1 and q4:

q1 = c(6, 19.5, 5.1, 4.7, 10.6)
q4 = c(6,  5,   2.5, 2.1,  5.4)

Your paired Wilcoxon test gives the following results:

wilcox.test(q1, q4, pair=T)  # 2-sided is default

        Wilcoxon signed rank test with 
        continuity correction

data:  q1 and q4
V = 10, p-value = 0.1003
alternative hypothesis: 
 true location shift is not equal to 0

Warning message:
In wilcox.test.default(q1, q2, pair = T) :
  cannot compute exact p-value with zeroes

Now find the five paired differences.

d = q1 - q4;  d
[1]  0.0 14.5  2.6  2.6  5.2

wilcox.test(d)  # 2-sided test with 0 null is default

        Wilcoxon signed rank test 
        with continuity correction

data:  d
V = 10, p-value = 0.1003
alternative hypothesis: 
  true location is not equal to 0

Warning message:
In wilcox.test.default(d) : 
  cannot compute exact p-value with zeroes

The result is exactly the same as for the 'paired' test. The paired test begins by finding the paired differences and then doing the one-sample Wilcoxon signed-rank test on the differences (instead of the two-sample Wilcoxon rank sum test on the two samples, as if indepencent).

Notes: (1) Additional evidence.

  • Taken separately, there are no ties in your two columns of data. So you might have wondered about the Warning message for the 'paired' test.

  • However, upon taking differences, you do have a a $0$ difference and a tie (at 2.6) for two out of the five differences.

(2) Neither the $0$ nor the tie is the reason one fails to get significant results at the 5% level--for just the five subjects you list as examples.

The following simulation repeatedly jitters the data very slightly to avoid $0$ and to break ties. None of the jittered data give significant results at the 5% level:

pv = replicate(10^4, wilcox.test(
                   d+runif(5, -.001,.001))$p.val)
summary(pv)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
0.06250 0.06250 0.06250 0.09354 0.12500 0.12500 

Also, with only five subjects the smallest possible P-value for a one-sided test is $1/32$ and the smallest for a 2-sided test is $1/16 = 0.0625,$ achieved as the min in the simulation.

(3) The data seem to be numerical (instead of merely ordinal). Even if we assume differences are normal (and with only $n=5$ of them, there's no use trying to test that), a paired t test comes nowhere near significance at the 5% level, partly because of a large variance.

t.test(d)$p.val
[1] 0.1191081
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