How to test the equality of two Pearson correlation coefficients computed from the same sample? ...
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How to test the equality of two Pearson correlation coefficients computed from the same sample?
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Is there a reliable way to say if two Pearson correlations from the same sample (do not) differ significantly? More concrete, I calculated the correlation between a total score on a questionnaire and an other variable, and a subscore of the same questionnaire and the variable. The correlations are respectively .239 and .234, so they look quite similar to me. (The other two subscales did not significantly correlate with the variable). Could I use a fisher Z to check if the two correlations indeed do not significantly differ, or is the fact that they are not independent a problem?
hypothesis-testing correlation non-independent
edited yesterday
amoeba
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asked yesterday
ChaFoChaFo
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add a comment |
$begingroup$
Is there a reliable way to say if two Pearson correlations from the same sample (do not) differ significantly? More concrete, I calculated the correlation between a total score on a questionnaire and an other variable, and a subscore of the same questionnaire and the variable. The correlations are respectively .239 and .234, so they look quite similar to me. (The other two subscales did not significantly correlate with the variable). Could I use a fisher Z to check if the two correlations indeed do not significantly differ, or is the fact that they are not independent a problem?
hypothesis-testing correlation non-independent
edited yesterday
amoeba
62.3k15208267
asked yesterday
ChaFoChaFo
161
New contributor
ChaFo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Is there a reliable way to say if two Pearson correlations from the same sample (do not) differ significantly? More concrete, I calculated the correlation between a total score on a questionnaire and an other variable, and a subscore of the same questionnaire and the variable. The correlations are respectively .239 and .234, so they look quite similar to me. (The other two subscales did not significantly correlate with the variable). Could I use a fisher Z to check if the two correlations indeed do not significantly differ, or is the fact that they are not independent a problem?
hypothesis-testing correlation non-independent
edited yesterday
amoeba
62.3k15208267
asked yesterday
ChaFoChaFo
161
New contributor
ChaFo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Is there a reliable way to say if two Pearson correlations from the same sample (do not) differ significantly? More concrete, I calculated the correlation between a total score on a questionnaire and an other variable, and a subscore of the same questionnaire and the variable. The correlations are respectively .239 and .234, so they look quite similar to me. (The other two subscales did not significantly correlate with the variable). Could I use a fisher Z to check if the two correlations indeed do not significantly differ, or is the fact that they are not independent a problem?
hypothesis-testing correlation non-independent
hypothesis-testing correlation non-independent
edited yesterday
amoeba
62.3k15208267
asked yesterday
ChaFoChaFo
161
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ChaFo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited yesterday
amoeba
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asked yesterday
ChaFoChaFo
161
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edited yesterday
amoeba
62.3k15208267
edited yesterday
amoeba
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edited yesterday
amoeba
62.3k15208267
62.3k15208267
asked yesterday
ChaFoChaFo
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asked yesterday
ChaFoChaFo
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asked yesterday
ChaFoChaFo
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3 Answers
3
active
oldest
votes
$begingroup$
Firstly I would point out that these correlations are fairly low.
Second, have you plotted the data to investigate possible non-linear associations?
Third, I would say that common sense should dictate that correlations of 0.239 and 0.234 are essentially the same and searching for a test to confirm this, unless the sample size is absolutely enormous, is folly.
Fourth, you could calculate confidence intervals for both statistics, and if they do not overlap, then you can conclude that they are statistically significantly different. However, this would be invalid since the 2 samples are not independent. Moreover, as per my third point, even if you did have such an enormous sample and a test which validly concluded that a significant difference exists, I would find it hard to belive that the difference was practically significant.
answered yesterday
Robert LongRobert Long
11.9k22552
$endgroup$
$begingroup$
Thanks for your reply. I am indeed aware that the correlations are small. I have checked for non-linear associations. I also feel like the difference is not meaningfull, but I just wanted make I do everything in the best possible way. Thanks!
$endgroup$
– ChaFo
17 hours ago
$begingroup$
@ChaFo that's OK, but you don't always have to make a formal test, especially where it seems obvious that they are essentially the same. How many observations do you have?
$endgroup$
– Robert Long
16 hours ago
add a comment |
$begingroup$
Expanding on Robert Long's answer (+1 to Robert) I'd say that testing for a difference between these is folly, regardless of sample size. Look! Is 0.239 different from 0.234? Well, maybe it is. There are situations where a very small effect size is very important. If a plane crashes 1 in 1,000 flights, that's a big big problem. I can't think, offhand, of a situation where this tiny difference in correlations could be meaningful, but maybe there is one. Whether it is significant or not is not the point.
Also, the dependence will surely be a problem. If you really wanted to see something like this, I'd find a third correlation: The correlation between the test after removing the subtest. Then you can compare that to the correlation with the subtest.
Finally, it's unclear to me what you are trying to show, but I think you are trying to show that these are not different. In that case, the usual null hypothesis tests are inappropriate. You should be looking at tests of equivalence (if, in fact you want to look at significance at all).
answered yesterday
Peter Flom♦Peter Flom
77.5k12109218
$endgroup$
1
$begingroup$
Excellent points, Peter (+1)
$endgroup$
– Robert Long
yesterday
1
$begingroup$
Peter Flom, the population perspective in epidemiology says, in effect, that a tiny change in risk—one that is so small as to be effectively inconsequential clinically—is a big deal if it is multiplied across an entire population. Changing someone's risk of stroke by 1 in 10,000 per year is kinda meh. Changing 10,000,000 people's risk of stroke by 1 in 10,000 is a change of a 1,000 strokes per year: a big deal. See Rose, G. (1985). Sick individuals and sick populations. International Journal of Epidemiology, 14(1), 32–28.
$endgroup$
– Alexis
yesterday
$begingroup$
Of course, Pearson's correlation coefficient alone isn't likely to be the most used measure of contrasts in risk, but I think small associations can matter.
$endgroup$
– Alexis
yesterday
add a comment |
$begingroup$
Yes, it is possible to perform a significance test using the Fisher transform. This also depends on $N$, the number of samples used to compute the Pearson correlations. This blog post describes the method in more detail, and provides R code for it.
answered yesterday
BaiBai
101
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2
$begingroup$
Your reference is inappropriate for comparing correlation coefficients that share data, as is the case here. The OP points out that "the fact they are not independent" is the problem.
$endgroup$
– whuber♦
yesterday
1
$begingroup$
Yes, I see. OP's situation involves overlap between the two datasets, but is not a case of paired data. Therefore, my answer is inappropriate.
$endgroup$
– Bai
yesterday
$begingroup$
Actually, it sounds like the data are triples: that's what makes it possible to compute more than one correlation coefficient.
$endgroup$
– whuber♦
yesterday
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Firstly I would point out that these correlations are fairly low.
Second, have you plotted the data to investigate possible non-linear associations?
Third, I would say that common sense should dictate that correlations of 0.239 and 0.234 are essentially the same and searching for a test to confirm this, unless the sample size is absolutely enormous, is folly.
Fourth, you could calculate confidence intervals for both statistics, and if they do not overlap, then you can conclude that they are statistically significantly different. However, this would be invalid since the 2 samples are not independent. Moreover, as per my third point, even if you did have such an enormous sample and a test which validly concluded that a significant difference exists, I would find it hard to belive that the difference was practically significant.
answered yesterday
Robert LongRobert Long
11.9k22552
$endgroup$
$begingroup$
Thanks for your reply. I am indeed aware that the correlations are small. I have checked for non-linear associations. I also feel like the difference is not meaningfull, but I just wanted make I do everything in the best possible way. Thanks!
$endgroup$
– ChaFo
17 hours ago
$begingroup$
@ChaFo that's OK, but you don't always have to make a formal test, especially where it seems obvious that they are essentially the same. How many observations do you have?
$endgroup$
– Robert Long
16 hours ago
add a comment |
$begingroup$
Firstly I would point out that these correlations are fairly low.
Second, have you plotted the data to investigate possible non-linear associations?
Third, I would say that common sense should dictate that correlations of 0.239 and 0.234 are essentially the same and searching for a test to confirm this, unless the sample size is absolutely enormous, is folly.
Fourth, you could calculate confidence intervals for both statistics, and if they do not overlap, then you can conclude that they are statistically significantly different. However, this would be invalid since the 2 samples are not independent. Moreover, as per my third point, even if you did have such an enormous sample and a test which validly concluded that a significant difference exists, I would find it hard to belive that the difference was practically significant.
answered yesterday
Robert LongRobert Long
11.9k22552
$endgroup$
$begingroup$
Thanks for your reply. I am indeed aware that the correlations are small. I have checked for non-linear associations. I also feel like the difference is not meaningfull, but I just wanted make I do everything in the best possible way. Thanks!
$endgroup$
– ChaFo
17 hours ago
$begingroup$
@ChaFo that's OK, but you don't always have to make a formal test, especially where it seems obvious that they are essentially the same. How many observations do you have?
$endgroup$
– Robert Long
16 hours ago
add a comment |
$begingroup$
Firstly I would point out that these correlations are fairly low.
Second, have you plotted the data to investigate possible non-linear associations?
Third, I would say that common sense should dictate that correlations of 0.239 and 0.234 are essentially the same and searching for a test to confirm this, unless the sample size is absolutely enormous, is folly.
Fourth, you could calculate confidence intervals for both statistics, and if they do not overlap, then you can conclude that they are statistically significantly different. However, this would be invalid since the 2 samples are not independent. Moreover, as per my third point, even if you did have such an enormous sample and a test which validly concluded that a significant difference exists, I would find it hard to belive that the difference was practically significant.
answered yesterday
Robert LongRobert Long
11.9k22552
$endgroup$
Firstly I would point out that these correlations are fairly low.
Second, have you plotted the data to investigate possible non-linear associations?
Third, I would say that common sense should dictate that correlations of 0.239 and 0.234 are essentially the same and searching for a test to confirm this, unless the sample size is absolutely enormous, is folly.
Fourth, you could calculate confidence intervals for both statistics, and if they do not overlap, then you can conclude that they are statistically significantly different. However, this would be invalid since the 2 samples are not independent. Moreover, as per my third point, even if you did have such an enormous sample and a test which validly concluded that a significant difference exists, I would find it hard to belive that the difference was practically significant.
answered yesterday
Robert LongRobert Long
11.9k22552
answered yesterday
Robert LongRobert Long
11.9k22552
answered yesterday
Robert LongRobert Long
11.9k22552
answered yesterday
Robert LongRobert Long
11.9k22552
11.9k22552
$begingroup$
Thanks for your reply. I am indeed aware that the correlations are small. I have checked for non-linear associations. I also feel like the difference is not meaningfull, but I just wanted make I do everything in the best possible way. Thanks!
$endgroup$
– ChaFo
17 hours ago
$begingroup$
@ChaFo that's OK, but you don't always have to make a formal test, especially where it seems obvious that they are essentially the same. How many observations do you have?
$endgroup$
– Robert Long
16 hours ago
add a comment |
$begingroup$
Thanks for your reply. I am indeed aware that the correlations are small. I have checked for non-linear associations. I also feel like the difference is not meaningfull, but I just wanted make I do everything in the best possible way. Thanks!
$endgroup$
– ChaFo
17 hours ago
$begingroup$
@ChaFo that's OK, but you don't always have to make a formal test, especially where it seems obvious that they are essentially the same. How many observations do you have?
$endgroup$
– Robert Long
16 hours ago
$begingroup$
Thanks for your reply. I am indeed aware that the correlations are small. I have checked for non-linear associations. I also feel like the difference is not meaningfull, but I just wanted make I do everything in the best possible way. Thanks!
$endgroup$
– ChaFo
17 hours ago
$begingroup$
Thanks for your reply. I am indeed aware that the correlations are small. I have checked for non-linear associations. I also feel like the difference is not meaningfull, but I just wanted make I do everything in the best possible way. Thanks!
$endgroup$
– ChaFo
17 hours ago
$begingroup$
@ChaFo that's OK, but you don't always have to make a formal test, especially where it seems obvious that they are essentially the same. How many observations do you have?
$endgroup$
– Robert Long
16 hours ago
$begingroup$
@ChaFo that's OK, but you don't always have to make a formal test, especially where it seems obvious that they are essentially the same. How many observations do you have?
$endgroup$
– Robert Long
16 hours ago
add a comment |
$begingroup$
Expanding on Robert Long's answer (+1 to Robert) I'd say that testing for a difference between these is folly, regardless of sample size. Look! Is 0.239 different from 0.234? Well, maybe it is. There are situations where a very small effect size is very important. If a plane crashes 1 in 1,000 flights, that's a big big problem. I can't think, offhand, of a situation where this tiny difference in correlations could be meaningful, but maybe there is one. Whether it is significant or not is not the point.
Also, the dependence will surely be a problem. If you really wanted to see something like this, I'd find a third correlation: The correlation between the test after removing the subtest. Then you can compare that to the correlation with the subtest.
Finally, it's unclear to me what you are trying to show, but I think you are trying to show that these are not different. In that case, the usual null hypothesis tests are inappropriate. You should be looking at tests of equivalence (if, in fact you want to look at significance at all).
answered yesterday
Peter Flom♦Peter Flom
77.5k12109218
$endgroup$
1
$begingroup$
Excellent points, Peter (+1)
$endgroup$
– Robert Long
yesterday
1
$begingroup$
Peter Flom, the population perspective in epidemiology says, in effect, that a tiny change in risk—one that is so small as to be effectively inconsequential clinically—is a big deal if it is multiplied across an entire population. Changing someone's risk of stroke by 1 in 10,000 per year is kinda meh. Changing 10,000,000 people's risk of stroke by 1 in 10,000 is a change of a 1,000 strokes per year: a big deal. See Rose, G. (1985). Sick individuals and sick populations. International Journal of Epidemiology, 14(1), 32–28.
$endgroup$
– Alexis
yesterday
$begingroup$
Of course, Pearson's correlation coefficient alone isn't likely to be the most used measure of contrasts in risk, but I think small associations can matter.
$endgroup$
– Alexis
yesterday
add a comment |
$begingroup$
Expanding on Robert Long's answer (+1 to Robert) I'd say that testing for a difference between these is folly, regardless of sample size. Look! Is 0.239 different from 0.234? Well, maybe it is. There are situations where a very small effect size is very important. If a plane crashes 1 in 1,000 flights, that's a big big problem. I can't think, offhand, of a situation where this tiny difference in correlations could be meaningful, but maybe there is one. Whether it is significant or not is not the point.
Also, the dependence will surely be a problem. If you really wanted to see something like this, I'd find a third correlation: The correlation between the test after removing the subtest. Then you can compare that to the correlation with the subtest.
Finally, it's unclear to me what you are trying to show, but I think you are trying to show that these are not different. In that case, the usual null hypothesis tests are inappropriate. You should be looking at tests of equivalence (if, in fact you want to look at significance at all).
answered yesterday
Peter Flom♦Peter Flom
77.5k12109218
$endgroup$
1
$begingroup$
Excellent points, Peter (+1)
$endgroup$
– Robert Long
yesterday
1
$begingroup$
Peter Flom, the population perspective in epidemiology says, in effect, that a tiny change in risk—one that is so small as to be effectively inconsequential clinically—is a big deal if it is multiplied across an entire population. Changing someone's risk of stroke by 1 in 10,000 per year is kinda meh. Changing 10,000,000 people's risk of stroke by 1 in 10,000 is a change of a 1,000 strokes per year: a big deal. See Rose, G. (1985). Sick individuals and sick populations. International Journal of Epidemiology, 14(1), 32–28.
$endgroup$
– Alexis
yesterday
$begingroup$
Of course, Pearson's correlation coefficient alone isn't likely to be the most used measure of contrasts in risk, but I think small associations can matter.
$endgroup$
– Alexis
yesterday
add a comment |
$begingroup$
Expanding on Robert Long's answer (+1 to Robert) I'd say that testing for a difference between these is folly, regardless of sample size. Look! Is 0.239 different from 0.234? Well, maybe it is. There are situations where a very small effect size is very important. If a plane crashes 1 in 1,000 flights, that's a big big problem. I can't think, offhand, of a situation where this tiny difference in correlations could be meaningful, but maybe there is one. Whether it is significant or not is not the point.
Also, the dependence will surely be a problem. If you really wanted to see something like this, I'd find a third correlation: The correlation between the test after removing the subtest. Then you can compare that to the correlation with the subtest.
Finally, it's unclear to me what you are trying to show, but I think you are trying to show that these are not different. In that case, the usual null hypothesis tests are inappropriate. You should be looking at tests of equivalence (if, in fact you want to look at significance at all).
answered yesterday
Peter Flom♦Peter Flom
77.5k12109218
$endgroup$
Expanding on Robert Long's answer (+1 to Robert) I'd say that testing for a difference between these is folly, regardless of sample size. Look! Is 0.239 different from 0.234? Well, maybe it is. There are situations where a very small effect size is very important. If a plane crashes 1 in 1,000 flights, that's a big big problem. I can't think, offhand, of a situation where this tiny difference in correlations could be meaningful, but maybe there is one. Whether it is significant or not is not the point.
Also, the dependence will surely be a problem. If you really wanted to see something like this, I'd find a third correlation: The correlation between the test after removing the subtest. Then you can compare that to the correlation with the subtest.
Finally, it's unclear to me what you are trying to show, but I think you are trying to show that these are not different. In that case, the usual null hypothesis tests are inappropriate. You should be looking at tests of equivalence (if, in fact you want to look at significance at all).
answered yesterday
Peter Flom♦Peter Flom
77.5k12109218
answered yesterday
Peter Flom♦Peter Flom
77.5k12109218
answered yesterday
Peter Flom♦Peter Flom
77.5k12109218
answered yesterday
Peter Flom♦Peter Flom
77.5k12109218
77.5k12109218
1
$begingroup$
Excellent points, Peter (+1)
$endgroup$
– Robert Long
yesterday
1
$begingroup$
Peter Flom, the population perspective in epidemiology says, in effect, that a tiny change in risk—one that is so small as to be effectively inconsequential clinically—is a big deal if it is multiplied across an entire population. Changing someone's risk of stroke by 1 in 10,000 per year is kinda meh. Changing 10,000,000 people's risk of stroke by 1 in 10,000 is a change of a 1,000 strokes per year: a big deal. See Rose, G. (1985). Sick individuals and sick populations. International Journal of Epidemiology, 14(1), 32–28.
$endgroup$
– Alexis
yesterday
$begingroup$
Of course, Pearson's correlation coefficient alone isn't likely to be the most used measure of contrasts in risk, but I think small associations can matter.
$endgroup$
– Alexis
yesterday
add a comment |
1
$begingroup$
Excellent points, Peter (+1)
$endgroup$
– Robert Long
yesterday
1
$begingroup$
Peter Flom, the population perspective in epidemiology says, in effect, that a tiny change in risk—one that is so small as to be effectively inconsequential clinically—is a big deal if it is multiplied across an entire population. Changing someone's risk of stroke by 1 in 10,000 per year is kinda meh. Changing 10,000,000 people's risk of stroke by 1 in 10,000 is a change of a 1,000 strokes per year: a big deal. See Rose, G. (1985). Sick individuals and sick populations. International Journal of Epidemiology, 14(1), 32–28.
$endgroup$
– Alexis
yesterday
$begingroup$
Of course, Pearson's correlation coefficient alone isn't likely to be the most used measure of contrasts in risk, but I think small associations can matter.
$endgroup$
– Alexis
yesterday
1
1
$begingroup$
Excellent points, Peter (+1)
$endgroup$
– Robert Long
yesterday
$begingroup$
Excellent points, Peter (+1)
$endgroup$
– Robert Long
yesterday
1
1
$begingroup$
Peter Flom, the population perspective in epidemiology says, in effect, that a tiny change in risk—one that is so small as to be effectively inconsequential clinically—is a big deal if it is multiplied across an entire population. Changing someone's risk of stroke by 1 in 10,000 per year is kinda meh. Changing 10,000,000 people's risk of stroke by 1 in 10,000 is a change of a 1,000 strokes per year: a big deal. See Rose, G. (1985). Sick individuals and sick populations. International Journal of Epidemiology, 14(1), 32–28.
$endgroup$
– Alexis
yesterday
$begingroup$
Peter Flom, the population perspective in epidemiology says, in effect, that a tiny change in risk—one that is so small as to be effectively inconsequential clinically—is a big deal if it is multiplied across an entire population. Changing someone's risk of stroke by 1 in 10,000 per year is kinda meh. Changing 10,000,000 people's risk of stroke by 1 in 10,000 is a change of a 1,000 strokes per year: a big deal. See Rose, G. (1985). Sick individuals and sick populations. International Journal of Epidemiology, 14(1), 32–28.
$endgroup$
– Alexis
yesterday
$begingroup$
Of course, Pearson's correlation coefficient alone isn't likely to be the most used measure of contrasts in risk, but I think small associations can matter.
$endgroup$
– Alexis
yesterday
$begingroup$
Of course, Pearson's correlation coefficient alone isn't likely to be the most used measure of contrasts in risk, but I think small associations can matter.
$endgroup$
– Alexis
yesterday
add a comment |
$begingroup$
Yes, it is possible to perform a significance test using the Fisher transform. This also depends on $N$, the number of samples used to compute the Pearson correlations. This blog post describes the method in more detail, and provides R code for it.
answered yesterday
BaiBai
101
New contributor
Bai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
Your reference is inappropriate for comparing correlation coefficients that share data, as is the case here. The OP points out that "the fact they are not independent" is the problem.
$endgroup$
– whuber♦
yesterday
1
$begingroup$
Yes, I see. OP's situation involves overlap between the two datasets, but is not a case of paired data. Therefore, my answer is inappropriate.
$endgroup$
– Bai
yesterday
$begingroup$
Actually, it sounds like the data are triples: that's what makes it possible to compute more than one correlation coefficient.
$endgroup$
– whuber♦
yesterday
add a comment |
$begingroup$
Yes, it is possible to perform a significance test using the Fisher transform. This also depends on $N$, the number of samples used to compute the Pearson correlations. This blog post describes the method in more detail, and provides R code for it.
answered yesterday
BaiBai
101
New contributor
Bai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
2
$begingroup$
Your reference is inappropriate for comparing correlation coefficients that share data, as is the case here. The OP points out that "the fact they are not independent" is the problem.
$endgroup$
– whuber♦
yesterday
1
$begingroup$
Yes, I see. OP's situation involves overlap between the two datasets, but is not a case of paired data. Therefore, my answer is inappropriate.
$endgroup$
– Bai
yesterday
$begingroup$
Actually, it sounds like the data are triples: that's what makes it possible to compute more than one correlation coefficient.
$endgroup$
– whuber♦
yesterday
add a comment |
$begingroup$
Yes, it is possible to perform a significance test using the Fisher transform. This also depends on $N$, the number of samples used to compute the Pearson correlations. This blog post describes the method in more detail, and provides R code for it.
answered yesterday
BaiBai
101
New contributor
Bai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Yes, it is possible to perform a significance test using the Fisher transform. This also depends on $N$, the number of samples used to compute the Pearson correlations. This blog post describes the method in more detail, and provides R code for it.
answered yesterday
BaiBai
101
New contributor
Bai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
BaiBai
101
New contributor
Bai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
BaiBai
101
answered yesterday
BaiBai
101
101
New contributor
Bai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Bai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Bai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
Your reference is inappropriate for comparing correlation coefficients that share data, as is the case here. The OP points out that "the fact they are not independent" is the problem.
$endgroup$
– whuber♦
yesterday
1
$begingroup$
Yes, I see. OP's situation involves overlap between the two datasets, but is not a case of paired data. Therefore, my answer is inappropriate.
$endgroup$
– Bai
yesterday
$begingroup$
Actually, it sounds like the data are triples: that's what makes it possible to compute more than one correlation coefficient.
$endgroup$
– whuber♦
yesterday
add a comment |
2
$begingroup$
Your reference is inappropriate for comparing correlation coefficients that share data, as is the case here. The OP points out that "the fact they are not independent" is the problem.
$endgroup$
– whuber♦
yesterday
1
$begingroup$
Yes, I see. OP's situation involves overlap between the two datasets, but is not a case of paired data. Therefore, my answer is inappropriate.
$endgroup$
– Bai
yesterday
$begingroup$
Actually, it sounds like the data are triples: that's what makes it possible to compute more than one correlation coefficient.
$endgroup$
– whuber♦
yesterday
2
2
$begingroup$
Your reference is inappropriate for comparing correlation coefficients that share data, as is the case here. The OP points out that "the fact they are not independent" is the problem.
$endgroup$
– whuber♦
yesterday
$begingroup$
Your reference is inappropriate for comparing correlation coefficients that share data, as is the case here. The OP points out that "the fact they are not independent" is the problem.
$endgroup$
– whuber♦
yesterday
1
1
$begingroup$
Yes, I see. OP's situation involves overlap between the two datasets, but is not a case of paired data. Therefore, my answer is inappropriate.
$endgroup$
– Bai
yesterday
$begingroup$
Yes, I see. OP's situation involves overlap between the two datasets, but is not a case of paired data. Therefore, my answer is inappropriate.
$endgroup$
– Bai
yesterday
$begingroup$
Actually, it sounds like the data are triples: that's what makes it possible to compute more than one correlation coefficient.
$endgroup$
– whuber♦
yesterday
$begingroup$
Actually, it sounds like the data are triples: that's what makes it possible to compute more than one correlation coefficient.
$endgroup$
– whuber♦
yesterday
add a comment |
ChaFo is a new contributor. Be nice, and check out our Code of Conduct.
ChaFo is a new contributor. Be nice, and check out our Code of Conduct.
ChaFo is a new contributor. Be nice, and check out our Code of Conduct.
ChaFo is a new contributor. Be nice, and check out our Code of Conduct.
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