What to bootstrap for hypothesis testingExplaining to laypeople why bootstrapping works Bootstrap vs. permutation hypothesis testingHypothesis testing: small timeseries changesWhy the data should be resampled under null hypothesis in bootstrap hypothesis testing?lmer() parametric bootstrap testing for fixed effectsBootstrap hypothesis testing with small sample sizesIs this popular approach to Bootstrap hypothesis testing correct?Hypothesis testing using the non-parametric bootstrapBootstrap hypothesis testing of equality of distributionsWhat is the difference between bootstrap hypothesis testing/permutation test and traditional hypothesis testing?Hypothesis testing for percentage

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What to bootstrap for hypothesis testing


Explaining to laypeople why bootstrapping works Bootstrap vs. permutation hypothesis testingHypothesis testing: small timeseries changesWhy the data should be resampled under null hypothesis in bootstrap hypothesis testing?lmer() parametric bootstrap testing for fixed effectsBootstrap hypothesis testing with small sample sizesIs this popular approach to Bootstrap hypothesis testing correct?Hypothesis testing using the non-parametric bootstrapBootstrap hypothesis testing of equality of distributionsWhat is the difference between bootstrap hypothesis testing/permutation test and traditional hypothesis testing?Hypothesis testing for percentage






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








1












$begingroup$


I have a small question about the concept behind hypothesis testing using bootstrap. Assume that I need to evaluate two independent population mean differences: population a and population b. My doubt is the following:




  1. Should I apply bootstrap on a single population, and check the difference of the mean after that?



    Mean[BOOT(a)-BOOT(b)]



  2. Alternatively, should I compute di difference: Mean(a)-Mean(b) and then apply bootstrap?



    BOOT[Mean(a)-Mean(b)]


I used this code by using the second approach:



set.seed(123)
a <- rnorm(100)
b <- rnorm(100)
hist(a)
hist(b)

c = a-b
hist(c)

boot_1 = function(R,dati_oss)
n = length(dati_oss)
media_boot = vector("numeric",R)
for(i in 1:R)
ind = sample(1:n,replace=T)
media_boot[i] = mean(dati_oss[ind])

return(media_boot)


res=boot_1(500000,c)

hist(res)

stat = matrix(c(mean(c), mean(res), mean(res)-mean(c), sqrt(var(res)),
as.vector(quantile(res, c(0.025,0.975)))), 1, 6)
colnames(stat) = c("Observed", "Mean-boot", "Bias", "SE", "0.95LCI", "0.95UCI")
row.names(stat) = c("Mean")
stat









share|cite|improve this question











$endgroup$











  • $begingroup$
    Did you try running your code? I get an error. Also, your code doesn't actually match your method 2.
    $endgroup$
    – gung
    8 hours ago










  • $begingroup$
    @gung where did you get the error?
    $endgroup$
    – an.dr.ea
    8 hours ago










  • $begingroup$
    After running res=boot_1(500000,c).
    $endgroup$
    – gung
    8 hours ago










  • $begingroup$
    @MichaelM Thanks, I just wanted to know if approach 1 or 2 is correct.
    $endgroup$
    – an.dr.ea
    8 hours ago










  • $begingroup$
    @gung I don't have this problem, try with fewer reps res = boot_1 (5000, c)
    $endgroup$
    – an.dr.ea
    8 hours ago

















1












$begingroup$


I have a small question about the concept behind hypothesis testing using bootstrap. Assume that I need to evaluate two independent population mean differences: population a and population b. My doubt is the following:




  1. Should I apply bootstrap on a single population, and check the difference of the mean after that?



    Mean[BOOT(a)-BOOT(b)]



  2. Alternatively, should I compute di difference: Mean(a)-Mean(b) and then apply bootstrap?



    BOOT[Mean(a)-Mean(b)]


I used this code by using the second approach:



set.seed(123)
a <- rnorm(100)
b <- rnorm(100)
hist(a)
hist(b)

c = a-b
hist(c)

boot_1 = function(R,dati_oss)
n = length(dati_oss)
media_boot = vector("numeric",R)
for(i in 1:R)
ind = sample(1:n,replace=T)
media_boot[i] = mean(dati_oss[ind])

return(media_boot)


res=boot_1(500000,c)

hist(res)

stat = matrix(c(mean(c), mean(res), mean(res)-mean(c), sqrt(var(res)),
as.vector(quantile(res, c(0.025,0.975)))), 1, 6)
colnames(stat) = c("Observed", "Mean-boot", "Bias", "SE", "0.95LCI", "0.95UCI")
row.names(stat) = c("Mean")
stat









share|cite|improve this question











$endgroup$











  • $begingroup$
    Did you try running your code? I get an error. Also, your code doesn't actually match your method 2.
    $endgroup$
    – gung
    8 hours ago










  • $begingroup$
    @gung where did you get the error?
    $endgroup$
    – an.dr.ea
    8 hours ago










  • $begingroup$
    After running res=boot_1(500000,c).
    $endgroup$
    – gung
    8 hours ago










  • $begingroup$
    @MichaelM Thanks, I just wanted to know if approach 1 or 2 is correct.
    $endgroup$
    – an.dr.ea
    8 hours ago










  • $begingroup$
    @gung I don't have this problem, try with fewer reps res = boot_1 (5000, c)
    $endgroup$
    – an.dr.ea
    8 hours ago













1












1








1





$begingroup$


I have a small question about the concept behind hypothesis testing using bootstrap. Assume that I need to evaluate two independent population mean differences: population a and population b. My doubt is the following:




  1. Should I apply bootstrap on a single population, and check the difference of the mean after that?



    Mean[BOOT(a)-BOOT(b)]



  2. Alternatively, should I compute di difference: Mean(a)-Mean(b) and then apply bootstrap?



    BOOT[Mean(a)-Mean(b)]


I used this code by using the second approach:



set.seed(123)
a <- rnorm(100)
b <- rnorm(100)
hist(a)
hist(b)

c = a-b
hist(c)

boot_1 = function(R,dati_oss)
n = length(dati_oss)
media_boot = vector("numeric",R)
for(i in 1:R)
ind = sample(1:n,replace=T)
media_boot[i] = mean(dati_oss[ind])

return(media_boot)


res=boot_1(500000,c)

hist(res)

stat = matrix(c(mean(c), mean(res), mean(res)-mean(c), sqrt(var(res)),
as.vector(quantile(res, c(0.025,0.975)))), 1, 6)
colnames(stat) = c("Observed", "Mean-boot", "Bias", "SE", "0.95LCI", "0.95UCI")
row.names(stat) = c("Mean")
stat









share|cite|improve this question











$endgroup$




I have a small question about the concept behind hypothesis testing using bootstrap. Assume that I need to evaluate two independent population mean differences: population a and population b. My doubt is the following:




  1. Should I apply bootstrap on a single population, and check the difference of the mean after that?



    Mean[BOOT(a)-BOOT(b)]



  2. Alternatively, should I compute di difference: Mean(a)-Mean(b) and then apply bootstrap?



    BOOT[Mean(a)-Mean(b)]


I used this code by using the second approach:



set.seed(123)
a <- rnorm(100)
b <- rnorm(100)
hist(a)
hist(b)

c = a-b
hist(c)

boot_1 = function(R,dati_oss)
n = length(dati_oss)
media_boot = vector("numeric",R)
for(i in 1:R)
ind = sample(1:n,replace=T)
media_boot[i] = mean(dati_oss[ind])

return(media_boot)


res=boot_1(500000,c)

hist(res)

stat = matrix(c(mean(c), mean(res), mean(res)-mean(c), sqrt(var(res)),
as.vector(quantile(res, c(0.025,0.975)))), 1, 6)
colnames(stat) = c("Observed", "Mean-boot", "Bias", "SE", "0.95LCI", "0.95UCI")
row.names(stat) = c("Mean")
stat






r hypothesis-testing bootstrap






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 8 hours ago









gung

110k34271542




110k34271542










asked 9 hours ago









an.dr.eaan.dr.ea

214




214











  • $begingroup$
    Did you try running your code? I get an error. Also, your code doesn't actually match your method 2.
    $endgroup$
    – gung
    8 hours ago










  • $begingroup$
    @gung where did you get the error?
    $endgroup$
    – an.dr.ea
    8 hours ago










  • $begingroup$
    After running res=boot_1(500000,c).
    $endgroup$
    – gung
    8 hours ago










  • $begingroup$
    @MichaelM Thanks, I just wanted to know if approach 1 or 2 is correct.
    $endgroup$
    – an.dr.ea
    8 hours ago










  • $begingroup$
    @gung I don't have this problem, try with fewer reps res = boot_1 (5000, c)
    $endgroup$
    – an.dr.ea
    8 hours ago
















  • $begingroup$
    Did you try running your code? I get an error. Also, your code doesn't actually match your method 2.
    $endgroup$
    – gung
    8 hours ago










  • $begingroup$
    @gung where did you get the error?
    $endgroup$
    – an.dr.ea
    8 hours ago










  • $begingroup$
    After running res=boot_1(500000,c).
    $endgroup$
    – gung
    8 hours ago










  • $begingroup$
    @MichaelM Thanks, I just wanted to know if approach 1 or 2 is correct.
    $endgroup$
    – an.dr.ea
    8 hours ago










  • $begingroup$
    @gung I don't have this problem, try with fewer reps res = boot_1 (5000, c)
    $endgroup$
    – an.dr.ea
    8 hours ago















$begingroup$
Did you try running your code? I get an error. Also, your code doesn't actually match your method 2.
$endgroup$
– gung
8 hours ago




$begingroup$
Did you try running your code? I get an error. Also, your code doesn't actually match your method 2.
$endgroup$
– gung
8 hours ago












$begingroup$
@gung where did you get the error?
$endgroup$
– an.dr.ea
8 hours ago




$begingroup$
@gung where did you get the error?
$endgroup$
– an.dr.ea
8 hours ago












$begingroup$
After running res=boot_1(500000,c).
$endgroup$
– gung
8 hours ago




$begingroup$
After running res=boot_1(500000,c).
$endgroup$
– gung
8 hours ago












$begingroup$
@MichaelM Thanks, I just wanted to know if approach 1 or 2 is correct.
$endgroup$
– an.dr.ea
8 hours ago




$begingroup$
@MichaelM Thanks, I just wanted to know if approach 1 or 2 is correct.
$endgroup$
– an.dr.ea
8 hours ago












$begingroup$
@gung I don't have this problem, try with fewer reps res = boot_1 (5000, c)
$endgroup$
– an.dr.ea
8 hours ago




$begingroup$
@gung I don't have this problem, try with fewer reps res = boot_1 (5000, c)
$endgroup$
– an.dr.ea
8 hours ago










2 Answers
2






active

oldest

votes


















2












$begingroup$

The basic principle to apply, quoting @MichaelChernick, is: "Sampling with replacement behaves on the original sample the way the original sample behaves on a population."



Think about how you analyzed the original sample. You took the sample, calculated the mean of each of the 2 groups, and determined the difference between their mean values to get an estimate of the a-b difference.



So you proceed similarly with each bootstrapped resample: resample from the original sample, calculate the mean of each group, and determine the difference between the means of the two groups as represented in the resample. Do this a large number of times to estimate the distribution of a-b differences. Compare the mean of the bootstrapped a-b differences against the a-b difference found in the original sample to estimate the bias in the original a-b difference.



Note that the way you design the resampling might depend on the original study design. If you had two independent populations from which you took samples, then the resampling should proceed comparably, within each of the populations. If you sampled from a mixed population in which individual cases were labeled as belonging to population a versus b, then you should resample from a pool of all the cases in the original sample.






share|cite|improve this answer









$endgroup$




















    2












    $begingroup$

    Without going into the coding of it, consider what happens if you calculate Mean(a)-Mean(b) first (proposal 2). You calculate Mean(a) and get a single number, then you calculate Mean(b) and get a single number. Then you take your two numbers, and calculate Mean(a)-Mean(b) to get a single value for your difference. No matter how many times you sample/bootstrap this single value, you will get the same number. Try it out!



    Whereas if you take multiple samples of your population A and population B, then calculate the difference of means of your samples (proposal 1) you will get slightly different combinations nearly every time you sample (assuming you don't re-set the random seed at the wrong point in your code), so you will get a range of values for the difference.



    I would consider proposal 1 to be a form of bootstrapping, but I wouldn't say the same about proposal 2!






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      The basic principle to apply, quoting @MichaelChernick, is: "Sampling with replacement behaves on the original sample the way the original sample behaves on a population."



      Think about how you analyzed the original sample. You took the sample, calculated the mean of each of the 2 groups, and determined the difference between their mean values to get an estimate of the a-b difference.



      So you proceed similarly with each bootstrapped resample: resample from the original sample, calculate the mean of each group, and determine the difference between the means of the two groups as represented in the resample. Do this a large number of times to estimate the distribution of a-b differences. Compare the mean of the bootstrapped a-b differences against the a-b difference found in the original sample to estimate the bias in the original a-b difference.



      Note that the way you design the resampling might depend on the original study design. If you had two independent populations from which you took samples, then the resampling should proceed comparably, within each of the populations. If you sampled from a mixed population in which individual cases were labeled as belonging to population a versus b, then you should resample from a pool of all the cases in the original sample.






      share|cite|improve this answer









      $endgroup$

















        2












        $begingroup$

        The basic principle to apply, quoting @MichaelChernick, is: "Sampling with replacement behaves on the original sample the way the original sample behaves on a population."



        Think about how you analyzed the original sample. You took the sample, calculated the mean of each of the 2 groups, and determined the difference between their mean values to get an estimate of the a-b difference.



        So you proceed similarly with each bootstrapped resample: resample from the original sample, calculate the mean of each group, and determine the difference between the means of the two groups as represented in the resample. Do this a large number of times to estimate the distribution of a-b differences. Compare the mean of the bootstrapped a-b differences against the a-b difference found in the original sample to estimate the bias in the original a-b difference.



        Note that the way you design the resampling might depend on the original study design. If you had two independent populations from which you took samples, then the resampling should proceed comparably, within each of the populations. If you sampled from a mixed population in which individual cases were labeled as belonging to population a versus b, then you should resample from a pool of all the cases in the original sample.






        share|cite|improve this answer









        $endgroup$















          2












          2








          2





          $begingroup$

          The basic principle to apply, quoting @MichaelChernick, is: "Sampling with replacement behaves on the original sample the way the original sample behaves on a population."



          Think about how you analyzed the original sample. You took the sample, calculated the mean of each of the 2 groups, and determined the difference between their mean values to get an estimate of the a-b difference.



          So you proceed similarly with each bootstrapped resample: resample from the original sample, calculate the mean of each group, and determine the difference between the means of the two groups as represented in the resample. Do this a large number of times to estimate the distribution of a-b differences. Compare the mean of the bootstrapped a-b differences against the a-b difference found in the original sample to estimate the bias in the original a-b difference.



          Note that the way you design the resampling might depend on the original study design. If you had two independent populations from which you took samples, then the resampling should proceed comparably, within each of the populations. If you sampled from a mixed population in which individual cases were labeled as belonging to population a versus b, then you should resample from a pool of all the cases in the original sample.






          share|cite|improve this answer









          $endgroup$



          The basic principle to apply, quoting @MichaelChernick, is: "Sampling with replacement behaves on the original sample the way the original sample behaves on a population."



          Think about how you analyzed the original sample. You took the sample, calculated the mean of each of the 2 groups, and determined the difference between their mean values to get an estimate of the a-b difference.



          So you proceed similarly with each bootstrapped resample: resample from the original sample, calculate the mean of each group, and determine the difference between the means of the two groups as represented in the resample. Do this a large number of times to estimate the distribution of a-b differences. Compare the mean of the bootstrapped a-b differences against the a-b difference found in the original sample to estimate the bias in the original a-b difference.



          Note that the way you design the resampling might depend on the original study design. If you had two independent populations from which you took samples, then the resampling should proceed comparably, within each of the populations. If you sampled from a mixed population in which individual cases were labeled as belonging to population a versus b, then you should resample from a pool of all the cases in the original sample.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 6 hours ago









          EdMEdM

          23.8k234102




          23.8k234102























              2












              $begingroup$

              Without going into the coding of it, consider what happens if you calculate Mean(a)-Mean(b) first (proposal 2). You calculate Mean(a) and get a single number, then you calculate Mean(b) and get a single number. Then you take your two numbers, and calculate Mean(a)-Mean(b) to get a single value for your difference. No matter how many times you sample/bootstrap this single value, you will get the same number. Try it out!



              Whereas if you take multiple samples of your population A and population B, then calculate the difference of means of your samples (proposal 1) you will get slightly different combinations nearly every time you sample (assuming you don't re-set the random seed at the wrong point in your code), so you will get a range of values for the difference.



              I would consider proposal 1 to be a form of bootstrapping, but I wouldn't say the same about proposal 2!






              share|cite|improve this answer









              $endgroup$

















                2












                $begingroup$

                Without going into the coding of it, consider what happens if you calculate Mean(a)-Mean(b) first (proposal 2). You calculate Mean(a) and get a single number, then you calculate Mean(b) and get a single number. Then you take your two numbers, and calculate Mean(a)-Mean(b) to get a single value for your difference. No matter how many times you sample/bootstrap this single value, you will get the same number. Try it out!



                Whereas if you take multiple samples of your population A and population B, then calculate the difference of means of your samples (proposal 1) you will get slightly different combinations nearly every time you sample (assuming you don't re-set the random seed at the wrong point in your code), so you will get a range of values for the difference.



                I would consider proposal 1 to be a form of bootstrapping, but I wouldn't say the same about proposal 2!






                share|cite|improve this answer









                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  Without going into the coding of it, consider what happens if you calculate Mean(a)-Mean(b) first (proposal 2). You calculate Mean(a) and get a single number, then you calculate Mean(b) and get a single number. Then you take your two numbers, and calculate Mean(a)-Mean(b) to get a single value for your difference. No matter how many times you sample/bootstrap this single value, you will get the same number. Try it out!



                  Whereas if you take multiple samples of your population A and population B, then calculate the difference of means of your samples (proposal 1) you will get slightly different combinations nearly every time you sample (assuming you don't re-set the random seed at the wrong point in your code), so you will get a range of values for the difference.



                  I would consider proposal 1 to be a form of bootstrapping, but I wouldn't say the same about proposal 2!






                  share|cite|improve this answer









                  $endgroup$



                  Without going into the coding of it, consider what happens if you calculate Mean(a)-Mean(b) first (proposal 2). You calculate Mean(a) and get a single number, then you calculate Mean(b) and get a single number. Then you take your two numbers, and calculate Mean(a)-Mean(b) to get a single value for your difference. No matter how many times you sample/bootstrap this single value, you will get the same number. Try it out!



                  Whereas if you take multiple samples of your population A and population B, then calculate the difference of means of your samples (proposal 1) you will get slightly different combinations nearly every time you sample (assuming you don't re-set the random seed at the wrong point in your code), so you will get a range of values for the difference.



                  I would consider proposal 1 to be a form of bootstrapping, but I wouldn't say the same about proposal 2!







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 6 hours ago









                  IzyIzy

                  337212




                  337212



























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                      François Viète Contents Biography Work and thought Bibliography See also Notes Further reading External links Navigation menup. 21Google Bookspp. 75–77Google BooksDe thou (from University of Saint Andrews)ArchivedGoogle BooksGoogle BooksGoogle BooksGoogle booksGoogle Bookscc-parthenay.frL'histoire universelle (fr)Universal History (en)ArchivedAdsabs.harvard.eduPagesperso-orange.frArchive.orgChikara Sasaki. Descartes' mathematical thought p.259Google BooksGoogle BooksGoogle Bookspp. 152 and onwardGoogle BooksGoogle BooksScribd.comGoogle Books1257-7979Google BooksGoogle BooksGoogle BooksGoogle BooksGoogle BooksGoogle BooksGallica.bnf.frGoogle BooksGoogle Books"François Viète"Francois Viète: Father of Modern Algebraic NotationThe Lawyer and the GamblerAbout TarporleySite de Jean-Paul GuichardL'algèbre nouvelle"About the Harmonicon"cb120511976(data)1188044800000 0001 0913 5903n82164680ola2013766880073431702w6vt1sb70287374827140948071409480