Paired Samples Mann Whitney Wilcoxon Test not working
up vote
2
down vote
favorite
I am trying to perform paired Mann Whitney U tests. My samples for each test are very different and by the t.test do produce different results, but the results are identical for the MW test. By contrast the default settings for the MW test work just fine and do give different results. Any ideas? Attached is code which contains a sample of my overall data and the test function. The t.test is included to show that different results are produced by the test:
x1 <- c(26.33323, 26.69508, 26.25390, 18.78399, 24.11386, 23.94950,
23.77843, 21.09932, 17.71425, 19.03429, 20.01796, 19.86626, 12.84303)
x2 <- c(20.82535, 20.27921, 17.99138, 10.40184, 23.23184, 22.56530,
18.69153, 18.33580, 13.83343, 18.22934, 15.21738, 10.07495, 9.93721)
y1 <- c(169.73751, 134.85579, 122.62475, 67.87308, 110.10757, 125.72300,
133.87937, 135.56772, 79.41600, 96.92930, 97.92528, 68.62409, 40.21653)
y2 <- c(92.88698, 54.23404, 51.58410, 21.72830, 72.02835, 70.74432,
69.52055, 89.59934, 49.08684, 79.98573, 50.58707, 22.80362, 22.49185)
wilcox.test(x1, x2, paired = TRUE)
wilcox.test(y1, y2, paired = TRUE)
t.test(x1, x2, paired = TRUE)
t.test(y1, y2, paired = TRUE)
r
asked Nov 19 at 18:37
Joe Flannery Sutherland
133
add a comment |
up vote
2
down vote
favorite
I am trying to perform paired Mann Whitney U tests. My samples for each test are very different and by the t.test do produce different results, but the results are identical for the MW test. By contrast the default settings for the MW test work just fine and do give different results. Any ideas? Attached is code which contains a sample of my overall data and the test function. The t.test is included to show that different results are produced by the test:
x1 <- c(26.33323, 26.69508, 26.25390, 18.78399, 24.11386, 23.94950,
23.77843, 21.09932, 17.71425, 19.03429, 20.01796, 19.86626, 12.84303)
x2 <- c(20.82535, 20.27921, 17.99138, 10.40184, 23.23184, 22.56530,
18.69153, 18.33580, 13.83343, 18.22934, 15.21738, 10.07495, 9.93721)
y1 <- c(169.73751, 134.85579, 122.62475, 67.87308, 110.10757, 125.72300,
133.87937, 135.56772, 79.41600, 96.92930, 97.92528, 68.62409, 40.21653)
y2 <- c(92.88698, 54.23404, 51.58410, 21.72830, 72.02835, 70.74432,
69.52055, 89.59934, 49.08684, 79.98573, 50.58707, 22.80362, 22.49185)
wilcox.test(x1, x2, paired = TRUE)
wilcox.test(y1, y2, paired = TRUE)
t.test(x1, x2, paired = TRUE)
t.test(y1, y2, paired = TRUE)
r
asked Nov 19 at 18:37
Joe Flannery Sutherland
133
Welcome to SO, this isn't really a minimal example. Can you edit and provide just the vectors you made and the code required to reproduce the error? Asking people to download files is sketchy at best and should only be a last resort sharing option.
– Nate
Nov 19 at 19:15
1
Hi Nate. Thanks for the advice about posting. I will certainly follow it in future, and hopefully here! I have amended my post accordingly.
– Joe Flannery Sutherland
Nov 19 at 19:30
the R functionwilcox.test()is called different tests under the hood so you are getting different results. This is described in the details section of?wilcox.test. Whenxandyare given withpaired = TRUEa Wilcoxon signed rank test is performed test if the two samples are from the same distribution. Whenpaired = FALSEa Wilcoxon rank sum test is performed testing if the difference in means between the two samples is equal tomu(which defaults to 0).
– Nate
Nov 19 at 20:04
So they are very similar but employ different formulas, if you want a deeper dive into the math CrossValidated is the place for you!
– Nate
Nov 19 at 20:04
Thanks for the clarification, I didn't realise they were separate kinds of tests. My tests have to be paired, however, due to the nature of the samples. I'm still unsure as to why each pair of vectors, which are different to each other and the other pairs, all produce the same result with paired = TRUE (U = 91, p = 0.0002441). Something must still be going wrong somewhere right?
– Joe Flannery Sutherland
Nov 19 at 23:11
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am trying to perform paired Mann Whitney U tests. My samples for each test are very different and by the t.test do produce different results, but the results are identical for the MW test. By contrast the default settings for the MW test work just fine and do give different results. Any ideas? Attached is code which contains a sample of my overall data and the test function. The t.test is included to show that different results are produced by the test:
x1 <- c(26.33323, 26.69508, 26.25390, 18.78399, 24.11386, 23.94950,
23.77843, 21.09932, 17.71425, 19.03429, 20.01796, 19.86626, 12.84303)
x2 <- c(20.82535, 20.27921, 17.99138, 10.40184, 23.23184, 22.56530,
18.69153, 18.33580, 13.83343, 18.22934, 15.21738, 10.07495, 9.93721)
y1 <- c(169.73751, 134.85579, 122.62475, 67.87308, 110.10757, 125.72300,
133.87937, 135.56772, 79.41600, 96.92930, 97.92528, 68.62409, 40.21653)
y2 <- c(92.88698, 54.23404, 51.58410, 21.72830, 72.02835, 70.74432,
69.52055, 89.59934, 49.08684, 79.98573, 50.58707, 22.80362, 22.49185)
wilcox.test(x1, x2, paired = TRUE)
wilcox.test(y1, y2, paired = TRUE)
t.test(x1, x2, paired = TRUE)
t.test(y1, y2, paired = TRUE)
r
asked Nov 19 at 18:37
Joe Flannery Sutherland
133
I am trying to perform paired Mann Whitney U tests. My samples for each test are very different and by the t.test do produce different results, but the results are identical for the MW test. By contrast the default settings for the MW test work just fine and do give different results. Any ideas? Attached is code which contains a sample of my overall data and the test function. The t.test is included to show that different results are produced by the test:
x1 <- c(26.33323, 26.69508, 26.25390, 18.78399, 24.11386, 23.94950,
23.77843, 21.09932, 17.71425, 19.03429, 20.01796, 19.86626, 12.84303)
x2 <- c(20.82535, 20.27921, 17.99138, 10.40184, 23.23184, 22.56530,
18.69153, 18.33580, 13.83343, 18.22934, 15.21738, 10.07495, 9.93721)
y1 <- c(169.73751, 134.85579, 122.62475, 67.87308, 110.10757, 125.72300,
133.87937, 135.56772, 79.41600, 96.92930, 97.92528, 68.62409, 40.21653)
y2 <- c(92.88698, 54.23404, 51.58410, 21.72830, 72.02835, 70.74432,
69.52055, 89.59934, 49.08684, 79.98573, 50.58707, 22.80362, 22.49185)
wilcox.test(x1, x2, paired = TRUE)
wilcox.test(y1, y2, paired = TRUE)
t.test(x1, x2, paired = TRUE)
t.test(y1, y2, paired = TRUE)
r
r
asked Nov 19 at 18:37
Joe Flannery Sutherland
133
asked Nov 19 at 18:37
Joe Flannery Sutherland
133
edited Nov 19 at 19:33
asked Nov 19 at 18:37
Joe Flannery Sutherland
133
asked Nov 19 at 18:37
Joe Flannery Sutherland
133
asked Nov 19 at 18:37
Joe Flannery Sutherland
133
133
Welcome to SO, this isn't really a minimal example. Can you edit and provide just the vectors you made and the code required to reproduce the error? Asking people to download files is sketchy at best and should only be a last resort sharing option.
– Nate
Nov 19 at 19:15
1
Hi Nate. Thanks for the advice about posting. I will certainly follow it in future, and hopefully here! I have amended my post accordingly.
– Joe Flannery Sutherland
Nov 19 at 19:30
the R functionwilcox.test()is called different tests under the hood so you are getting different results. This is described in the details section of?wilcox.test. Whenxandyare given withpaired = TRUEa Wilcoxon signed rank test is performed test if the two samples are from the same distribution. Whenpaired = FALSEa Wilcoxon rank sum test is performed testing if the difference in means between the two samples is equal tomu(which defaults to 0).
– Nate
Nov 19 at 20:04
So they are very similar but employ different formulas, if you want a deeper dive into the math CrossValidated is the place for you!
– Nate
Nov 19 at 20:04
Thanks for the clarification, I didn't realise they were separate kinds of tests. My tests have to be paired, however, due to the nature of the samples. I'm still unsure as to why each pair of vectors, which are different to each other and the other pairs, all produce the same result with paired = TRUE (U = 91, p = 0.0002441). Something must still be going wrong somewhere right?
– Joe Flannery Sutherland
Nov 19 at 23:11
add a comment |
Welcome to SO, this isn't really a minimal example. Can you edit and provide just the vectors you made and the code required to reproduce the error? Asking people to download files is sketchy at best and should only be a last resort sharing option.
– Nate
Nov 19 at 19:15
1
Hi Nate. Thanks for the advice about posting. I will certainly follow it in future, and hopefully here! I have amended my post accordingly.
– Joe Flannery Sutherland
Nov 19 at 19:30
the R functionwilcox.test()is called different tests under the hood so you are getting different results. This is described in the details section of?wilcox.test. Whenxandyare given withpaired = TRUEa Wilcoxon signed rank test is performed test if the two samples are from the same distribution. Whenpaired = FALSEa Wilcoxon rank sum test is performed testing if the difference in means between the two samples is equal tomu(which defaults to 0).
– Nate
Nov 19 at 20:04
So they are very similar but employ different formulas, if you want a deeper dive into the math CrossValidated is the place for you!
– Nate
Nov 19 at 20:04
Thanks for the clarification, I didn't realise they were separate kinds of tests. My tests have to be paired, however, due to the nature of the samples. I'm still unsure as to why each pair of vectors, which are different to each other and the other pairs, all produce the same result with paired = TRUE (U = 91, p = 0.0002441). Something must still be going wrong somewhere right?
– Joe Flannery Sutherland
Nov 19 at 23:11
Welcome to SO, this isn't really a minimal example. Can you edit and provide just the vectors you made and the code required to reproduce the error? Asking people to download files is sketchy at best and should only be a last resort sharing option.
– Nate
Nov 19 at 19:15
Welcome to SO, this isn't really a minimal example. Can you edit and provide just the vectors you made and the code required to reproduce the error? Asking people to download files is sketchy at best and should only be a last resort sharing option.
– Nate
Nov 19 at 19:15
1
1
Hi Nate. Thanks for the advice about posting. I will certainly follow it in future, and hopefully here! I have amended my post accordingly.
– Joe Flannery Sutherland
Nov 19 at 19:30
Hi Nate. Thanks for the advice about posting. I will certainly follow it in future, and hopefully here! I have amended my post accordingly.
– Joe Flannery Sutherland
Nov 19 at 19:30
the R function
wilcox.test() is called different tests under the hood so you are getting different results. This is described in the details section of ?wilcox.test. When x and y are given with paired = TRUE a Wilcoxon signed rank test is performed test if the two samples are from the same distribution. When paired = FALSE a Wilcoxon rank sum test is performed testing if the difference in means between the two samples is equal to mu (which defaults to 0).– Nate
Nov 19 at 20:04
the R function
wilcox.test() is called different tests under the hood so you are getting different results. This is described in the details section of ?wilcox.test. When x and y are given with paired = TRUE a Wilcoxon signed rank test is performed test if the two samples are from the same distribution. When paired = FALSE a Wilcoxon rank sum test is performed testing if the difference in means between the two samples is equal to mu (which defaults to 0).– Nate
Nov 19 at 20:04
So they are very similar but employ different formulas, if you want a deeper dive into the math CrossValidated is the place for you!
– Nate
Nov 19 at 20:04
So they are very similar but employ different formulas, if you want a deeper dive into the math CrossValidated is the place for you!
– Nate
Nov 19 at 20:04
Thanks for the clarification, I didn't realise they were separate kinds of tests. My tests have to be paired, however, due to the nature of the samples. I'm still unsure as to why each pair of vectors, which are different to each other and the other pairs, all produce the same result with paired = TRUE (U = 91, p = 0.0002441). Something must still be going wrong somewhere right?
– Joe Flannery Sutherland
Nov 19 at 23:11
Thanks for the clarification, I didn't realise they were separate kinds of tests. My tests have to be paired, however, due to the nature of the samples. I'm still unsure as to why each pair of vectors, which are different to each other and the other pairs, all produce the same result with paired = TRUE (U = 91, p = 0.0002441). Something must still be going wrong somewhere right?
– Joe Flannery Sutherland
Nov 19 at 23:11
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
No the test is performing as expected. Maybe seeing the arithmatic in R of wilcox.test will help.
From the Wikipedia page example here are the steps of wilcox.text(x, y, paired = TRUE) or a Wilcoxon signed rank test:
# calculate pair-wise differences
x_diffs <- x1 - x2
y_diffs <- y1 - y2
# rank them
x_ranks <- rank(x_diffs)
y_ranks <- rank(y_diffs)
# adjust ranks by the original sign of each difference
x_ranks_signed <- rank(x_diffs) * sign(x_diffs)
y_ranks_signed <- rank(y_diffs) * sign(y_diffs)
# sum them for the test statistic
sum(x_ranks_signed)
sum(y_ranks_signed)
In your example both of the 2 values are smaller than the 1 values so no ranks get multiplied by -1, since all of the deltas are positive numbers. Since both x and y are the same length 13, you correctly get the same test statistic 91 or sum(1:13) for both tests.
To get a different test statistic you would need to introduce some negative delta values, ie making x3 bigger than x1.
x3 <- c(28, 29, 17.99138, 10.40184, 23.23184, 22.56530,
18.69153, 18.33580, 13.83343, 18.22934, 15.21738, 10.07495, 9.93721)
wilcox.test(x1, x2, paired = T) # same old test statistic and p-value
Wilcoxon signed rank test
data: x1 and x2
V = 91, p-value = 0.0002441
alternative hypothesis: true location shift is not equal to 0
wilcox.test(x1, x3, paired = T) # new negative deltas >> new test stat and new p-value
Wilcoxon signed rank test
data: x1 and x3
V = 82, p-value = 0.008057
alternative hypothesis: true location shift is not equal to 0
answered Nov 20 at 16:19
Nate
5,86711528
Ah I see! Thanks for taking me through the test process
– Joe Flannery Sutherland
Nov 20 at 16:48
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
No the test is performing as expected. Maybe seeing the arithmatic in R of wilcox.test will help.
From the Wikipedia page example here are the steps of wilcox.text(x, y, paired = TRUE) or a Wilcoxon signed rank test:
# calculate pair-wise differences
x_diffs <- x1 - x2
y_diffs <- y1 - y2
# rank them
x_ranks <- rank(x_diffs)
y_ranks <- rank(y_diffs)
# adjust ranks by the original sign of each difference
x_ranks_signed <- rank(x_diffs) * sign(x_diffs)
y_ranks_signed <- rank(y_diffs) * sign(y_diffs)
# sum them for the test statistic
sum(x_ranks_signed)
sum(y_ranks_signed)
In your example both of the 2 values are smaller than the 1 values so no ranks get multiplied by -1, since all of the deltas are positive numbers. Since both x and y are the same length 13, you correctly get the same test statistic 91 or sum(1:13) for both tests.
To get a different test statistic you would need to introduce some negative delta values, ie making x3 bigger than x1.
x3 <- c(28, 29, 17.99138, 10.40184, 23.23184, 22.56530,
18.69153, 18.33580, 13.83343, 18.22934, 15.21738, 10.07495, 9.93721)
wilcox.test(x1, x2, paired = T) # same old test statistic and p-value
Wilcoxon signed rank test
data: x1 and x2
V = 91, p-value = 0.0002441
alternative hypothesis: true location shift is not equal to 0
wilcox.test(x1, x3, paired = T) # new negative deltas >> new test stat and new p-value
Wilcoxon signed rank test
data: x1 and x3
V = 82, p-value = 0.008057
alternative hypothesis: true location shift is not equal to 0
answered Nov 20 at 16:19
Nate
5,86711528
Ah I see! Thanks for taking me through the test process
– Joe Flannery Sutherland
Nov 20 at 16:48
add a comment |
up vote
0
down vote
accepted
No the test is performing as expected. Maybe seeing the arithmatic in R of wilcox.test will help.
From the Wikipedia page example here are the steps of wilcox.text(x, y, paired = TRUE) or a Wilcoxon signed rank test:
# calculate pair-wise differences
x_diffs <- x1 - x2
y_diffs <- y1 - y2
# rank them
x_ranks <- rank(x_diffs)
y_ranks <- rank(y_diffs)
# adjust ranks by the original sign of each difference
x_ranks_signed <- rank(x_diffs) * sign(x_diffs)
y_ranks_signed <- rank(y_diffs) * sign(y_diffs)
# sum them for the test statistic
sum(x_ranks_signed)
sum(y_ranks_signed)
In your example both of the 2 values are smaller than the 1 values so no ranks get multiplied by -1, since all of the deltas are positive numbers. Since both x and y are the same length 13, you correctly get the same test statistic 91 or sum(1:13) for both tests.
To get a different test statistic you would need to introduce some negative delta values, ie making x3 bigger than x1.
x3 <- c(28, 29, 17.99138, 10.40184, 23.23184, 22.56530,
18.69153, 18.33580, 13.83343, 18.22934, 15.21738, 10.07495, 9.93721)
wilcox.test(x1, x2, paired = T) # same old test statistic and p-value
Wilcoxon signed rank test
data: x1 and x2
V = 91, p-value = 0.0002441
alternative hypothesis: true location shift is not equal to 0
wilcox.test(x1, x3, paired = T) # new negative deltas >> new test stat and new p-value
Wilcoxon signed rank test
data: x1 and x3
V = 82, p-value = 0.008057
alternative hypothesis: true location shift is not equal to 0
answered Nov 20 at 16:19
Nate
5,86711528
Ah I see! Thanks for taking me through the test process
– Joe Flannery Sutherland
Nov 20 at 16:48
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
No the test is performing as expected. Maybe seeing the arithmatic in R of wilcox.test will help.
From the Wikipedia page example here are the steps of wilcox.text(x, y, paired = TRUE) or a Wilcoxon signed rank test:
# calculate pair-wise differences
x_diffs <- x1 - x2
y_diffs <- y1 - y2
# rank them
x_ranks <- rank(x_diffs)
y_ranks <- rank(y_diffs)
# adjust ranks by the original sign of each difference
x_ranks_signed <- rank(x_diffs) * sign(x_diffs)
y_ranks_signed <- rank(y_diffs) * sign(y_diffs)
# sum them for the test statistic
sum(x_ranks_signed)
sum(y_ranks_signed)
In your example both of the 2 values are smaller than the 1 values so no ranks get multiplied by -1, since all of the deltas are positive numbers. Since both x and y are the same length 13, you correctly get the same test statistic 91 or sum(1:13) for both tests.
To get a different test statistic you would need to introduce some negative delta values, ie making x3 bigger than x1.
x3 <- c(28, 29, 17.99138, 10.40184, 23.23184, 22.56530,
18.69153, 18.33580, 13.83343, 18.22934, 15.21738, 10.07495, 9.93721)
wilcox.test(x1, x2, paired = T) # same old test statistic and p-value
Wilcoxon signed rank test
data: x1 and x2
V = 91, p-value = 0.0002441
alternative hypothesis: true location shift is not equal to 0
wilcox.test(x1, x3, paired = T) # new negative deltas >> new test stat and new p-value
Wilcoxon signed rank test
data: x1 and x3
V = 82, p-value = 0.008057
alternative hypothesis: true location shift is not equal to 0
answered Nov 20 at 16:19
Nate
5,86711528
No the test is performing as expected. Maybe seeing the arithmatic in R of wilcox.test will help.
From the Wikipedia page example here are the steps of wilcox.text(x, y, paired = TRUE) or a Wilcoxon signed rank test:
# calculate pair-wise differences
x_diffs <- x1 - x2
y_diffs <- y1 - y2
# rank them
x_ranks <- rank(x_diffs)
y_ranks <- rank(y_diffs)
# adjust ranks by the original sign of each difference
x_ranks_signed <- rank(x_diffs) * sign(x_diffs)
y_ranks_signed <- rank(y_diffs) * sign(y_diffs)
# sum them for the test statistic
sum(x_ranks_signed)
sum(y_ranks_signed)
In your example both of the 2 values are smaller than the 1 values so no ranks get multiplied by -1, since all of the deltas are positive numbers. Since both x and y are the same length 13, you correctly get the same test statistic 91 or sum(1:13) for both tests.
To get a different test statistic you would need to introduce some negative delta values, ie making x3 bigger than x1.
x3 <- c(28, 29, 17.99138, 10.40184, 23.23184, 22.56530,
18.69153, 18.33580, 13.83343, 18.22934, 15.21738, 10.07495, 9.93721)
wilcox.test(x1, x2, paired = T) # same old test statistic and p-value
Wilcoxon signed rank test
data: x1 and x2
V = 91, p-value = 0.0002441
alternative hypothesis: true location shift is not equal to 0
wilcox.test(x1, x3, paired = T) # new negative deltas >> new test stat and new p-value
Wilcoxon signed rank test
data: x1 and x3
V = 82, p-value = 0.008057
alternative hypothesis: true location shift is not equal to 0
answered Nov 20 at 16:19
Nate
5,86711528
answered Nov 20 at 16:19
Nate
5,86711528
answered Nov 20 at 16:19
Nate
5,86711528
answered Nov 20 at 16:19
Nate
5,86711528
5,86711528
Ah I see! Thanks for taking me through the test process
– Joe Flannery Sutherland
Nov 20 at 16:48
add a comment |
Ah I see! Thanks for taking me through the test process
– Joe Flannery Sutherland
Nov 20 at 16:48
Ah I see! Thanks for taking me through the test process
– Joe Flannery Sutherland
Nov 20 at 16:48
Ah I see! Thanks for taking me through the test process
– Joe Flannery Sutherland
Nov 20 at 16:48
add a comment |
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Welcome to SO, this isn't really a minimal example. Can you edit and provide just the vectors you made and the code required to reproduce the error? Asking people to download files is sketchy at best and should only be a last resort sharing option.
– Nate
Nov 19 at 19:15
1
Hi Nate. Thanks for the advice about posting. I will certainly follow it in future, and hopefully here! I have amended my post accordingly.
– Joe Flannery Sutherland
Nov 19 at 19:30
the R function
wilcox.test()is called different tests under the hood so you are getting different results. This is described in the details section of?wilcox.test. Whenxandyare given withpaired = TRUEa Wilcoxon signed rank test is performed test if the two samples are from the same distribution. Whenpaired = FALSEa Wilcoxon rank sum test is performed testing if the difference in means between the two samples is equal tomu(which defaults to 0).– Nate
Nov 19 at 20:04
So they are very similar but employ different formulas, if you want a deeper dive into the math CrossValidated is the place for you!
– Nate
Nov 19 at 20:04
Thanks for the clarification, I didn't realise they were separate kinds of tests. My tests have to be paired, however, due to the nature of the samples. I'm still unsure as to why each pair of vectors, which are different to each other and the other pairs, all produce the same result with paired = TRUE (U = 91, p = 0.0002441). Something must still be going wrong somewhere right?
– Joe Flannery Sutherland
Nov 19 at 23:11