What will happen if I use a nonparametric test with normally distributed data?
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I was asked if a type I error in the Shapiro-Wilk test would impact the main analysis and if the wrong test was used if it would matter or not if my data was normally distributed...
nonparametric
asked Nov 11 at 2:26
DaisyRiver
462
add a comment |
up vote
7
down vote
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I was asked if a type I error in the Shapiro-Wilk test would impact the main analysis and if the wrong test was used if it would matter or not if my data was normally distributed...
nonparametric
asked Nov 11 at 2:26
DaisyRiver
462
That makes sense thank you!
– DaisyRiver
Nov 11 at 2:56
@a_statistician - might want to expand that a little and post it as an answer, since it is!
– jbowman
Nov 11 at 3:02
@jbowman Thank you for your editing.
– user158565
Nov 11 at 3:28
add a comment |
up vote
7
down vote
favorite
up vote
7
down vote
favorite
I was asked if a type I error in the Shapiro-Wilk test would impact the main analysis and if the wrong test was used if it would matter or not if my data was normally distributed...
nonparametric
asked Nov 11 at 2:26
DaisyRiver
462
I was asked if a type I error in the Shapiro-Wilk test would impact the main analysis and if the wrong test was used if it would matter or not if my data was normally distributed...
nonparametric
nonparametric
asked Nov 11 at 2:26
DaisyRiver
462
asked Nov 11 at 2:26
DaisyRiver
462
asked Nov 11 at 2:26
DaisyRiver
462
asked Nov 11 at 2:26
DaisyRiver
462
asked Nov 11 at 2:26
DaisyRiver
462
462
That makes sense thank you!
– DaisyRiver
Nov 11 at 2:56
@a_statistician - might want to expand that a little and post it as an answer, since it is!
– jbowman
Nov 11 at 3:02
@jbowman Thank you for your editing.
– user158565
Nov 11 at 3:28
add a comment |
That makes sense thank you!
– DaisyRiver
Nov 11 at 2:56
@a_statistician - might want to expand that a little and post it as an answer, since it is!
– jbowman
Nov 11 at 3:02
@jbowman Thank you for your editing.
– user158565
Nov 11 at 3:28
That makes sense thank you!
– DaisyRiver
Nov 11 at 2:56
That makes sense thank you!
– DaisyRiver
Nov 11 at 2:56
@a_statistician - might want to expand that a little and post it as an answer, since it is!
– jbowman
Nov 11 at 3:02
@a_statistician - might want to expand that a little and post it as an answer, since it is!
– jbowman
Nov 11 at 3:02
@jbowman Thank you for your editing.
– user158565
Nov 11 at 3:28
@jbowman Thank you for your editing.
– user158565
Nov 11 at 3:28
add a comment |
2 Answers
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up vote
12
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In statistical analysis, if your data follow a parametric distribution, you should utilize the benefit of knowing the distribution, and employ the statistical methods based on that distribution.
But sometimes we do not know the distribution of the random variable, so the nonparametric statistical methods were developed to embrace the wide range of the distributions while sacrificing some efficiency.
Given you know the distribution of random variable and use the nonparametric statistical method, instead of parametric statistical methods based on knowing the distribution, it will be inefficient, i.e., the power of test will decrease, standard error will increase, and the confidence intervals will be wider than with the parametric method.
edited Nov 11 at 16:36
John Bentin
14516
answered Nov 11 at 3:11
user158565
4,4441316
add a comment |
up vote
12
down vote
If your data happened to be drawn from a normal population (and the other usual assumptions for an ordinary t-test apply), then the test works as it should (it's non-parametric, it's supposed to work). There's no drama on that score.
If you know enough that you're confident in assuming normality you may want to take advantage of that knowledge, but for many tests it doesn't help you a lot.
If you're doing one of the common location-tests (Wilcoxon signed rank test, Wilcoxon-Mann-Whitney test) you lose almost nothing (power-wise) in a test for a location shift by ignoring the normality. [You need one extra observation for every 21 observations to match the power of the most powerful test when all its assumptions hold.]
If you're dealing with some other tests is may matter a bit more (though some may matter even less). One example where it makes a somewhat bigger difference is using a Friedman test compared to the corresponding ANOVA test in a randomized blocks design.
answered Nov 11 at 6:15
Glen_b♦
207k22395732
Aren't those tests are only location shift tests if the distributes have the same shape (normal here) and the same variance?
– Alexis
Nov 11 at 17:37
1
@Alexis The assumption under the null for a permutation test is exchangeability (so permuting of labels/signs etc as needed doesn't change the distribution of the statistic); typically the slightly stronger "independent identically distributed" is assumed for H0, giving convenient significance level calculations for rank tests. Combined with a location-shift alternative (which is already specified in my answer), this gives the identify of shape and spread you mention. Perhaps I need to write down algebraically what "location shift alternative" means algebraically so people can see it's covered
– Glen_b♦
Nov 11 at 23:08
1
If we're not willing to specify a location shift alternative (or at the least to say "that's what we're interested in seeing the power against), I'd say we shouldn't be holding up a t-test as a basis of comparison in the first place, since it would also be impacted if we didn't have a location-shift alternative. If we abandon the location-shift part of the comparison, until we then specify a sequence of alternatives we're interested in, we're left with a question that specifies too little to compute power on.
– Glen_b♦
Nov 11 at 23:10
1
@Alexis Actually, on reflection I think I'll include both some algebra and more explanation; it's important to be clear what is implied by the circumstances already given in the question and answer, unfortunately I can't do it right now. Thanks for your comment, it makes it clear I don't explain enough here.
– Glen_b♦
Nov 11 at 23:23
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
12
down vote
In statistical analysis, if your data follow a parametric distribution, you should utilize the benefit of knowing the distribution, and employ the statistical methods based on that distribution.
But sometimes we do not know the distribution of the random variable, so the nonparametric statistical methods were developed to embrace the wide range of the distributions while sacrificing some efficiency.
Given you know the distribution of random variable and use the nonparametric statistical method, instead of parametric statistical methods based on knowing the distribution, it will be inefficient, i.e., the power of test will decrease, standard error will increase, and the confidence intervals will be wider than with the parametric method.
edited Nov 11 at 16:36
John Bentin
14516
answered Nov 11 at 3:11
user158565
4,4441316
add a comment |
up vote
12
down vote
In statistical analysis, if your data follow a parametric distribution, you should utilize the benefit of knowing the distribution, and employ the statistical methods based on that distribution.
But sometimes we do not know the distribution of the random variable, so the nonparametric statistical methods were developed to embrace the wide range of the distributions while sacrificing some efficiency.
Given you know the distribution of random variable and use the nonparametric statistical method, instead of parametric statistical methods based on knowing the distribution, it will be inefficient, i.e., the power of test will decrease, standard error will increase, and the confidence intervals will be wider than with the parametric method.
edited Nov 11 at 16:36
John Bentin
14516
answered Nov 11 at 3:11
user158565
4,4441316
add a comment |
up vote
12
down vote
up vote
12
down vote
In statistical analysis, if your data follow a parametric distribution, you should utilize the benefit of knowing the distribution, and employ the statistical methods based on that distribution.
But sometimes we do not know the distribution of the random variable, so the nonparametric statistical methods were developed to embrace the wide range of the distributions while sacrificing some efficiency.
Given you know the distribution of random variable and use the nonparametric statistical method, instead of parametric statistical methods based on knowing the distribution, it will be inefficient, i.e., the power of test will decrease, standard error will increase, and the confidence intervals will be wider than with the parametric method.
edited Nov 11 at 16:36
John Bentin
14516
answered Nov 11 at 3:11
user158565
4,4441316
In statistical analysis, if your data follow a parametric distribution, you should utilize the benefit of knowing the distribution, and employ the statistical methods based on that distribution.
But sometimes we do not know the distribution of the random variable, so the nonparametric statistical methods were developed to embrace the wide range of the distributions while sacrificing some efficiency.
Given you know the distribution of random variable and use the nonparametric statistical method, instead of parametric statistical methods based on knowing the distribution, it will be inefficient, i.e., the power of test will decrease, standard error will increase, and the confidence intervals will be wider than with the parametric method.
edited Nov 11 at 16:36
John Bentin
14516
answered Nov 11 at 3:11
user158565
4,4441316
edited Nov 11 at 16:36
John Bentin
14516
edited Nov 11 at 16:36
John Bentin
14516
edited Nov 11 at 16:36
John Bentin
14516
14516
answered Nov 11 at 3:11
user158565
4,4441316
answered Nov 11 at 3:11
user158565
4,4441316
answered Nov 11 at 3:11
user158565
4,4441316
4,4441316
add a comment |
add a comment |
up vote
12
down vote
If your data happened to be drawn from a normal population (and the other usual assumptions for an ordinary t-test apply), then the test works as it should (it's non-parametric, it's supposed to work). There's no drama on that score.
If you know enough that you're confident in assuming normality you may want to take advantage of that knowledge, but for many tests it doesn't help you a lot.
If you're doing one of the common location-tests (Wilcoxon signed rank test, Wilcoxon-Mann-Whitney test) you lose almost nothing (power-wise) in a test for a location shift by ignoring the normality. [You need one extra observation for every 21 observations to match the power of the most powerful test when all its assumptions hold.]
If you're dealing with some other tests is may matter a bit more (though some may matter even less). One example where it makes a somewhat bigger difference is using a Friedman test compared to the corresponding ANOVA test in a randomized blocks design.
answered Nov 11 at 6:15
Glen_b♦
207k22395732
Aren't those tests are only location shift tests if the distributes have the same shape (normal here) and the same variance?
– Alexis
Nov 11 at 17:37
1
@Alexis The assumption under the null for a permutation test is exchangeability (so permuting of labels/signs etc as needed doesn't change the distribution of the statistic); typically the slightly stronger "independent identically distributed" is assumed for H0, giving convenient significance level calculations for rank tests. Combined with a location-shift alternative (which is already specified in my answer), this gives the identify of shape and spread you mention. Perhaps I need to write down algebraically what "location shift alternative" means algebraically so people can see it's covered
– Glen_b♦
Nov 11 at 23:08
1
If we're not willing to specify a location shift alternative (or at the least to say "that's what we're interested in seeing the power against), I'd say we shouldn't be holding up a t-test as a basis of comparison in the first place, since it would also be impacted if we didn't have a location-shift alternative. If we abandon the location-shift part of the comparison, until we then specify a sequence of alternatives we're interested in, we're left with a question that specifies too little to compute power on.
– Glen_b♦
Nov 11 at 23:10
1
@Alexis Actually, on reflection I think I'll include both some algebra and more explanation; it's important to be clear what is implied by the circumstances already given in the question and answer, unfortunately I can't do it right now. Thanks for your comment, it makes it clear I don't explain enough here.
– Glen_b♦
Nov 11 at 23:23
add a comment |
up vote
12
down vote
If your data happened to be drawn from a normal population (and the other usual assumptions for an ordinary t-test apply), then the test works as it should (it's non-parametric, it's supposed to work). There's no drama on that score.
If you know enough that you're confident in assuming normality you may want to take advantage of that knowledge, but for many tests it doesn't help you a lot.
If you're doing one of the common location-tests (Wilcoxon signed rank test, Wilcoxon-Mann-Whitney test) you lose almost nothing (power-wise) in a test for a location shift by ignoring the normality. [You need one extra observation for every 21 observations to match the power of the most powerful test when all its assumptions hold.]
If you're dealing with some other tests is may matter a bit more (though some may matter even less). One example where it makes a somewhat bigger difference is using a Friedman test compared to the corresponding ANOVA test in a randomized blocks design.
answered Nov 11 at 6:15
Glen_b♦
207k22395732
Aren't those tests are only location shift tests if the distributes have the same shape (normal here) and the same variance?
– Alexis
Nov 11 at 17:37
1
@Alexis The assumption under the null for a permutation test is exchangeability (so permuting of labels/signs etc as needed doesn't change the distribution of the statistic); typically the slightly stronger "independent identically distributed" is assumed for H0, giving convenient significance level calculations for rank tests. Combined with a location-shift alternative (which is already specified in my answer), this gives the identify of shape and spread you mention. Perhaps I need to write down algebraically what "location shift alternative" means algebraically so people can see it's covered
– Glen_b♦
Nov 11 at 23:08
1
If we're not willing to specify a location shift alternative (or at the least to say "that's what we're interested in seeing the power against), I'd say we shouldn't be holding up a t-test as a basis of comparison in the first place, since it would also be impacted if we didn't have a location-shift alternative. If we abandon the location-shift part of the comparison, until we then specify a sequence of alternatives we're interested in, we're left with a question that specifies too little to compute power on.
– Glen_b♦
Nov 11 at 23:10
1
@Alexis Actually, on reflection I think I'll include both some algebra and more explanation; it's important to be clear what is implied by the circumstances already given in the question and answer, unfortunately I can't do it right now. Thanks for your comment, it makes it clear I don't explain enough here.
– Glen_b♦
Nov 11 at 23:23
add a comment |
up vote
12
down vote
up vote
12
down vote
If your data happened to be drawn from a normal population (and the other usual assumptions for an ordinary t-test apply), then the test works as it should (it's non-parametric, it's supposed to work). There's no drama on that score.
If you know enough that you're confident in assuming normality you may want to take advantage of that knowledge, but for many tests it doesn't help you a lot.
If you're doing one of the common location-tests (Wilcoxon signed rank test, Wilcoxon-Mann-Whitney test) you lose almost nothing (power-wise) in a test for a location shift by ignoring the normality. [You need one extra observation for every 21 observations to match the power of the most powerful test when all its assumptions hold.]
If you're dealing with some other tests is may matter a bit more (though some may matter even less). One example where it makes a somewhat bigger difference is using a Friedman test compared to the corresponding ANOVA test in a randomized blocks design.
answered Nov 11 at 6:15
Glen_b♦
207k22395732
If your data happened to be drawn from a normal population (and the other usual assumptions for an ordinary t-test apply), then the test works as it should (it's non-parametric, it's supposed to work). There's no drama on that score.
If you know enough that you're confident in assuming normality you may want to take advantage of that knowledge, but for many tests it doesn't help you a lot.
If you're doing one of the common location-tests (Wilcoxon signed rank test, Wilcoxon-Mann-Whitney test) you lose almost nothing (power-wise) in a test for a location shift by ignoring the normality. [You need one extra observation for every 21 observations to match the power of the most powerful test when all its assumptions hold.]
If you're dealing with some other tests is may matter a bit more (though some may matter even less). One example where it makes a somewhat bigger difference is using a Friedman test compared to the corresponding ANOVA test in a randomized blocks design.
answered Nov 11 at 6:15
Glen_b♦
207k22395732
edited Nov 11 at 23:16
answered Nov 11 at 6:15
Glen_b♦
207k22395732
answered Nov 11 at 6:15
Glen_b♦
207k22395732
answered Nov 11 at 6:15
Glen_b♦
207k22395732
207k22395732
Aren't those tests are only location shift tests if the distributes have the same shape (normal here) and the same variance?
– Alexis
Nov 11 at 17:37
1
@Alexis The assumption under the null for a permutation test is exchangeability (so permuting of labels/signs etc as needed doesn't change the distribution of the statistic); typically the slightly stronger "independent identically distributed" is assumed for H0, giving convenient significance level calculations for rank tests. Combined with a location-shift alternative (which is already specified in my answer), this gives the identify of shape and spread you mention. Perhaps I need to write down algebraically what "location shift alternative" means algebraically so people can see it's covered
– Glen_b♦
Nov 11 at 23:08
1
If we're not willing to specify a location shift alternative (or at the least to say "that's what we're interested in seeing the power against), I'd say we shouldn't be holding up a t-test as a basis of comparison in the first place, since it would also be impacted if we didn't have a location-shift alternative. If we abandon the location-shift part of the comparison, until we then specify a sequence of alternatives we're interested in, we're left with a question that specifies too little to compute power on.
– Glen_b♦
Nov 11 at 23:10
1
@Alexis Actually, on reflection I think I'll include both some algebra and more explanation; it's important to be clear what is implied by the circumstances already given in the question and answer, unfortunately I can't do it right now. Thanks for your comment, it makes it clear I don't explain enough here.
– Glen_b♦
Nov 11 at 23:23
add a comment |
Aren't those tests are only location shift tests if the distributes have the same shape (normal here) and the same variance?
– Alexis
Nov 11 at 17:37
1
@Alexis The assumption under the null for a permutation test is exchangeability (so permuting of labels/signs etc as needed doesn't change the distribution of the statistic); typically the slightly stronger "independent identically distributed" is assumed for H0, giving convenient significance level calculations for rank tests. Combined with a location-shift alternative (which is already specified in my answer), this gives the identify of shape and spread you mention. Perhaps I need to write down algebraically what "location shift alternative" means algebraically so people can see it's covered
– Glen_b♦
Nov 11 at 23:08
1
If we're not willing to specify a location shift alternative (or at the least to say "that's what we're interested in seeing the power against), I'd say we shouldn't be holding up a t-test as a basis of comparison in the first place, since it would also be impacted if we didn't have a location-shift alternative. If we abandon the location-shift part of the comparison, until we then specify a sequence of alternatives we're interested in, we're left with a question that specifies too little to compute power on.
– Glen_b♦
Nov 11 at 23:10
1
@Alexis Actually, on reflection I think I'll include both some algebra and more explanation; it's important to be clear what is implied by the circumstances already given in the question and answer, unfortunately I can't do it right now. Thanks for your comment, it makes it clear I don't explain enough here.
– Glen_b♦
Nov 11 at 23:23
Aren't those tests are only location shift tests if the distributes have the same shape (normal here) and the same variance?
– Alexis
Nov 11 at 17:37
Aren't those tests are only location shift tests if the distributes have the same shape (normal here) and the same variance?
– Alexis
Nov 11 at 17:37
1
1
@Alexis The assumption under the null for a permutation test is exchangeability (so permuting of labels/signs etc as needed doesn't change the distribution of the statistic); typically the slightly stronger "independent identically distributed" is assumed for H0, giving convenient significance level calculations for rank tests. Combined with a location-shift alternative (which is already specified in my answer), this gives the identify of shape and spread you mention. Perhaps I need to write down algebraically what "location shift alternative" means algebraically so people can see it's covered
– Glen_b♦
Nov 11 at 23:08
@Alexis The assumption under the null for a permutation test is exchangeability (so permuting of labels/signs etc as needed doesn't change the distribution of the statistic); typically the slightly stronger "independent identically distributed" is assumed for H0, giving convenient significance level calculations for rank tests. Combined with a location-shift alternative (which is already specified in my answer), this gives the identify of shape and spread you mention. Perhaps I need to write down algebraically what "location shift alternative" means algebraically so people can see it's covered
– Glen_b♦
Nov 11 at 23:08
1
1
If we're not willing to specify a location shift alternative (or at the least to say "that's what we're interested in seeing the power against), I'd say we shouldn't be holding up a t-test as a basis of comparison in the first place, since it would also be impacted if we didn't have a location-shift alternative. If we abandon the location-shift part of the comparison, until we then specify a sequence of alternatives we're interested in, we're left with a question that specifies too little to compute power on.
– Glen_b♦
Nov 11 at 23:10
If we're not willing to specify a location shift alternative (or at the least to say "that's what we're interested in seeing the power against), I'd say we shouldn't be holding up a t-test as a basis of comparison in the first place, since it would also be impacted if we didn't have a location-shift alternative. If we abandon the location-shift part of the comparison, until we then specify a sequence of alternatives we're interested in, we're left with a question that specifies too little to compute power on.
– Glen_b♦
Nov 11 at 23:10
1
1
@Alexis Actually, on reflection I think I'll include both some algebra and more explanation; it's important to be clear what is implied by the circumstances already given in the question and answer, unfortunately I can't do it right now. Thanks for your comment, it makes it clear I don't explain enough here.
– Glen_b♦
Nov 11 at 23:23
@Alexis Actually, on reflection I think I'll include both some algebra and more explanation; it's important to be clear what is implied by the circumstances already given in the question and answer, unfortunately I can't do it right now. Thanks for your comment, it makes it clear I don't explain enough here.
– Glen_b♦
Nov 11 at 23:23
add a comment |
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That makes sense thank you!
– DaisyRiver
Nov 11 at 2:56
@a_statistician - might want to expand that a little and post it as an answer, since it is!
– jbowman
Nov 11 at 3:02
@jbowman Thank you for your editing.
– user158565
Nov 11 at 3:28