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BruceET
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A paired approach seems appropriate because of the way data were collected. A paired Wilcoxon test is essentially the same a one-sample Wilcoxon Signed Rank test on the differences.

Unfortunately, assuming no ties among the five differences, the smallest possible P-value $1/16 = 0.0625$ for a 2-sided test occurs when all differences have the same sign. (By chance alone there is 1 chance in 32 that all five differences would be positive and the same that tall five would be negative: $1/32+1/32 = 1/16.)$

Here are computations in R statistical software, based on some hypothetical data:

x1 = c(1,4,5,7,8)
x2 = c(12,18,17,20,23)
d = x2-x1; d
[1] 11 14 12 13 15

Identical P-values of $1/16$1/16$ for both versions of the Wilcoxon signed-rank test, using paired data:

wilcox.test(x1, x2, pair=T)

        Wilcoxon signed rank test

data:  x1 and x2
V = 0, p-value = 0.0625
alternative hypothesis: 
   true location shift is not equal to 0

wilcox.test(d)

        Wilcoxon signed rank test

data:  d
V = 15, p-value = 0.0625
alternative hypothesis: 
    true location is not equal to 0

Doing a 2-sided Mann-Whitney-Wilcoxon two-sample rank sum test, there are ten observations altogether, then if (as for my illustrative hypothetical data) all observations in one group are larger than any value in the other a smaller P-value is possible. Specifically, the P-value for a complete separation assuming the null hypothesis of no difference between groups is $2$ chances in ${10 \choose 5} = 252$ or $2/252 = 0.00794.$

wilcox.test(x1, x2)

        Wilcoxon rank sum test

data:  x1 and x2
W = 0, p-value = 0.007937
alternative hypothesis: 
   true location shift is not equal to 0

From what you say about your data, I suppose you have a complete separation. That provides clear evidence that something has produced a significant effect.

However, if 'treatment' runs always followed 'control' runs (or always preceded them), then you need to mention the that the difference might somehow be partially due to order of scanning in addition to your 'treatment'. It's a good idea to get statistical advice about experimental design, including randomization, before you begin to collect data.

A paired approach seems appropriate because of the way data were collected. A paired Wilcoxon test is a one-sample Wilcoxon Signed Rank test on the differences.

Unfortunately, assuming no ties among the five differences, the smallest possible P-value $1/16 = 0.0625$ for a 2-sided test occurs when all differences have the same sign. (By chance alone there is 1 chance in 32 that all five differences would be positive and the same that tall five would be negative: $1/32+1/32 = 1/16.)$

x1 = c(1,4,5,7,8)
x2 = c(12,18,17,20,23)
d = x2-x1; d
[1] 11 14 12 13 15

Identical P-values of $1/16 for both versions of the Wilcoxon signed-rank test, using paired data:

wilcox.test(x1, x2, pair=T)

        Wilcoxon signed rank test

data:  x1 and x2
V = 0, p-value = 0.0625
alternative hypothesis: 
   true location shift is not equal to 0

wilcox.test(d)

        Wilcoxon signed rank test

data:  d
V = 15, p-value = 0.0625
alternative hypothesis: 
    true location is not equal to 0

Doing a 2-sided Mann-Whitney-Wilcoxon two-sample rank sum test, there are ten observations altogether, then if (as for my illustrative hypothetical data) all observations in one group are larger than any value in the other a smaller P-value is possible. Specifically, the P-value for a complete separation assuming the null hypothesis of no difference between groups is $2$ chances in ${10 \choose 5} = 252$ or $2/252 = 0.00794.$

wilcox.test(x1, x2)

        Wilcoxon rank sum test

data:  x1 and x2
W = 0, p-value = 0.007937
alternative hypothesis: 
   true location shift is not equal to 0

From what you say about your data, I suppose you have a complete separation. That provides clear evidence that something has produced a significant effect.

However, if 'treatment' runs always followed 'control' runs (or always preceded them), then you need to mention the that the difference might somehow be partially due to order of scanning in addition to your 'treatment'. It's a good idea to get statistical advice about experimental design, including randomization, before you begin to collect data.

A paired approach seems appropriate because of the way data were collected. A paired Wilcoxon test is essentially the same a one-sample Wilcoxon Signed Rank test on the differences.

Unfortunately, assuming no ties among the five differences, the smallest possible P-value $1/16 = 0.0625$ for a 2-sided test occurs when all differences have the same sign. (By chance alone there is 1 chance in 32 that all five differences would be positive and the same that tall five would be negative: $1/32+1/32 = 1/16.)$

Here are computations in R statistical software, based on some hypothetical data:

x1 = c(1,4,5,7,8)
x2 = c(12,18,17,20,23)
d = x2-x1; d
[1] 11 14 12 13 15

Identical P-values of $1/16$ for both versions of the Wilcoxon signed-rank test, using paired data:

wilcox.test(x1, x2, pair=T)

        Wilcoxon signed rank test

data:  x1 and x2
V = 0, p-value = 0.0625
alternative hypothesis: 
   true location shift is not equal to 0

wilcox.test(d)

        Wilcoxon signed rank test

data:  d
V = 15, p-value = 0.0625
alternative hypothesis: 
    true location is not equal to 0

Doing a 2-sided Mann-Whitney-Wilcoxon two-sample rank sum test, there are ten observations altogether, then if (as for my illustrative hypothetical data) all observations in one group are larger than any value in the other a smaller P-value is possible. Specifically, the P-value for a complete separation assuming the null hypothesis of no difference between groups is $2$ chances in ${10 \choose 5} = 252$ or $2/252 = 0.00794.$

wilcox.test(x1, x2)

        Wilcoxon rank sum test

data:  x1 and x2
W = 0, p-value = 0.007937
alternative hypothesis: 
   true location shift is not equal to 0

From what you say about your data, I suppose you have a complete separation. That provides clear evidence that something has produced a significant effect.

However, if 'treatment' runs always followed 'control' runs (or always preceded them), then you need to mention the that the difference might somehow be partially due to order of scanning in addition to your 'treatment'. It's a good idea to get statistical advice about experimental design, including randomization, before you begin to collect data.

Source Link
BruceET
  • 57.6k
  • 2
  • 36
  • 94

A paired approach seems appropriate because of the way data were collected. A paired Wilcoxon test is a one-sample Wilcoxon Signed Rank test on the differences.

Unfortunately, assuming no ties among the five differences, the smallest possible P-value $1/16 = 0.0625$ for a 2-sided test occurs when all differences have the same sign. (By chance alone there is 1 chance in 32 that all five differences would be positive and the same that tall five would be negative: $1/32+1/32 = 1/16.)$

x1 = c(1,4,5,7,8)
x2 = c(12,18,17,20,23)
d = x2-x1; d
[1] 11 14 12 13 15

Identical P-values of $1/16 for both versions of the Wilcoxon signed-rank test, using paired data:

wilcox.test(x1, x2, pair=T)

        Wilcoxon signed rank test

data:  x1 and x2
V = 0, p-value = 0.0625
alternative hypothesis: 
   true location shift is not equal to 0

wilcox.test(d)

        Wilcoxon signed rank test

data:  d
V = 15, p-value = 0.0625
alternative hypothesis: 
    true location is not equal to 0

Doing a 2-sided Mann-Whitney-Wilcoxon two-sample rank sum test, there are ten observations altogether, then if (as for my illustrative hypothetical data) all observations in one group are larger than any value in the other a smaller P-value is possible. Specifically, the P-value for a complete separation assuming the null hypothesis of no difference between groups is $2$ chances in ${10 \choose 5} = 252$ or $2/252 = 0.00794.$

wilcox.test(x1, x2)

        Wilcoxon rank sum test

data:  x1 and x2
W = 0, p-value = 0.007937
alternative hypothesis: 
   true location shift is not equal to 0

From what you say about your data, I suppose you have a complete separation. That provides clear evidence that something has produced a significant effect.

However, if 'treatment' runs always followed 'control' runs (or always preceded them), then you need to mention the that the difference might somehow be partially due to order of scanning in addition to your 'treatment'. It's a good idea to get statistical advice about experimental design, including randomization, before you begin to collect data.