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Why is my p-value correlated to difference between means in two sample tests?



The 2019 Stack Overflow Developer Survey Results Are InIs it possible to use a two sample $t$ test here?Mann-Whitney null hypothesis under unequal varianceDoes statistically insignificant difference of means imply equality of means?Evaluating close calls with the Wilcon Sum Rank test two sided vs. one sidedTest for systematic difference between two samplesHow to adjust p-value to reject null hypothesis from sample size in Mann Whitney U test?In distribution tests, why do we assume that any distribution is true unless proven otherwise?Calculating the p-value of two independent counts?Mann–Whitney U test shows there is a difference between two sample sets, how do I know which sample set is better?Two sample t-test to show equality of the two means



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








2












$begingroup$


A colleague has recently made the claim that a large p-value was not more support for the null hypothesis than a low one. Of course, this is also what I learned (uniform distribution under the null hypothesis, we can only reject the null hypothesis...). But when I simulate two random normal distributions (100 samples in each group) in R, my p-value is correlated to the difference (averaged over 30 repetitions) between the two means (with for example a T test or a Mann & Whitney test).



Why is my p-value, above the threshold of 0.05, correlated to the difference between the means of my two groups?



enter image description here



With 1000 repetitions for each x (difference between means/2) value.
enter image description here



My R code in case this is just a silly mistake.



pvaluetot<-NULL
xtot<-NULL
seqx<-seq(0,5,0.01)
for (x in seqx)
ptemp<-NULL
pmean<-NULL
a<-0

repeat
a<-a+1
pop1<-rnorm(100,0+x,2)
pop2<-rnorm(100,0-x,2)
pvalue<-t.test(pop1,pop2)$p.value

ptemp<-c(ptemp,pvalue)
#print(ptemp)
if (a==30)
break


pmean<-mean(ptemp)
pvaluetot<-c(pvaluetot,pmean)
xtot<-c(xtot,x)
print(x)


pvaluetot
xtot
plot(pvaluetot,xtot)









share|cite|improve this question











$endgroup$


















    2












    $begingroup$


    A colleague has recently made the claim that a large p-value was not more support for the null hypothesis than a low one. Of course, this is also what I learned (uniform distribution under the null hypothesis, we can only reject the null hypothesis...). But when I simulate two random normal distributions (100 samples in each group) in R, my p-value is correlated to the difference (averaged over 30 repetitions) between the two means (with for example a T test or a Mann & Whitney test).



    Why is my p-value, above the threshold of 0.05, correlated to the difference between the means of my two groups?



    enter image description here



    With 1000 repetitions for each x (difference between means/2) value.
    enter image description here



    My R code in case this is just a silly mistake.



    pvaluetot<-NULL
    xtot<-NULL
    seqx<-seq(0,5,0.01)
    for (x in seqx)
    ptemp<-NULL
    pmean<-NULL
    a<-0

    repeat
    a<-a+1
    pop1<-rnorm(100,0+x,2)
    pop2<-rnorm(100,0-x,2)
    pvalue<-t.test(pop1,pop2)$p.value

    ptemp<-c(ptemp,pvalue)
    #print(ptemp)
    if (a==30)
    break


    pmean<-mean(ptemp)
    pvaluetot<-c(pvaluetot,pmean)
    xtot<-c(xtot,x)
    print(x)


    pvaluetot
    xtot
    plot(pvaluetot,xtot)









    share|cite|improve this question











    $endgroup$














      2












      2








      2





      $begingroup$


      A colleague has recently made the claim that a large p-value was not more support for the null hypothesis than a low one. Of course, this is also what I learned (uniform distribution under the null hypothesis, we can only reject the null hypothesis...). But when I simulate two random normal distributions (100 samples in each group) in R, my p-value is correlated to the difference (averaged over 30 repetitions) between the two means (with for example a T test or a Mann & Whitney test).



      Why is my p-value, above the threshold of 0.05, correlated to the difference between the means of my two groups?



      enter image description here



      With 1000 repetitions for each x (difference between means/2) value.
      enter image description here



      My R code in case this is just a silly mistake.



      pvaluetot<-NULL
      xtot<-NULL
      seqx<-seq(0,5,0.01)
      for (x in seqx)
      ptemp<-NULL
      pmean<-NULL
      a<-0

      repeat
      a<-a+1
      pop1<-rnorm(100,0+x,2)
      pop2<-rnorm(100,0-x,2)
      pvalue<-t.test(pop1,pop2)$p.value

      ptemp<-c(ptemp,pvalue)
      #print(ptemp)
      if (a==30)
      break


      pmean<-mean(ptemp)
      pvaluetot<-c(pvaluetot,pmean)
      xtot<-c(xtot,x)
      print(x)


      pvaluetot
      xtot
      plot(pvaluetot,xtot)









      share|cite|improve this question











      $endgroup$




      A colleague has recently made the claim that a large p-value was not more support for the null hypothesis than a low one. Of course, this is also what I learned (uniform distribution under the null hypothesis, we can only reject the null hypothesis...). But when I simulate two random normal distributions (100 samples in each group) in R, my p-value is correlated to the difference (averaged over 30 repetitions) between the two means (with for example a T test or a Mann & Whitney test).



      Why is my p-value, above the threshold of 0.05, correlated to the difference between the means of my two groups?



      enter image description here



      With 1000 repetitions for each x (difference between means/2) value.
      enter image description here



      My R code in case this is just a silly mistake.



      pvaluetot<-NULL
      xtot<-NULL
      seqx<-seq(0,5,0.01)
      for (x in seqx)
      ptemp<-NULL
      pmean<-NULL
      a<-0

      repeat
      a<-a+1
      pop1<-rnorm(100,0+x,2)
      pop2<-rnorm(100,0-x,2)
      pvalue<-t.test(pop1,pop2)$p.value

      ptemp<-c(ptemp,pvalue)
      #print(ptemp)
      if (a==30)
      break


      pmean<-mean(ptemp)
      pvaluetot<-c(pvaluetot,pmean)
      xtot<-c(xtot,x)
      print(x)


      pvaluetot
      xtot
      plot(pvaluetot,xtot)






      hypothesis-testing statistical-significance p-value effect-size






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 1 hour ago







      Nakx

















      asked 2 hours ago









      NakxNakx

      324115




      324115




















          2 Answers
          2






          active

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          2












          $begingroup$

          Why would you expect anything else? You don't need a simulation to know this is going to happen. Look at the formula for the t-statistic:
          $t = fracbarx_1 - barx_2 sqrt fracs^2_1n_1 + fracs^2_2n_2 $



          Obviously if you increase the true difference of means you expect $barx_1 - barx_2$ will be larger. You are holding the variance and sample size constant, so the t-statistic must be larger and thus the p-value smaller.



          I think you are confusing a philosophical rule about hypothesis testing with a mathematical fact. If the null hypothesis is true, you would expect a higher p-value. This has to be true in order for hypothesis testing to make any sense.






          share|cite|improve this answer









          $endgroup$




















            0












            $begingroup$

            As you said, the p-value is uniformly distributed under the null hypothesis. That is, if the null hypothesis is really true, then upon repeated experiments we expect to find a fully random, flat distribution of p-values between [0, 1]. Consequently, a frequentist p-value says nothing about how likely the null hypothesis is to be true, since any p-value is equally probable under the null.



            What you're looking at is the distribution of p-values under an alternative hypothesis. Depending on the formulation of this hypothesis, the resulting p-values can have any non-Uniform, positively skewed distribution between [0, 1]. But this doesn't tell you anything about the probability of the null. The reason is that the p-value expresses the probability of the evidence under the null hypothesis, i.e. $p(D|H_0)$, whereas you want to know $p(H_0|D)$. These two are related by Bayes' rule:
            $$
            p(H_0|D) = fracH_0)p(H_0)p(D
            $$

            This means that in order to calculate the probability you're interested in, you need to know and take into account the prior probability of the null being true ($p(H_0)$), the prior probability of the null being false ($p(neg H_0)$) and the probability of the data given that the null is false ($p(D|neg H_0)$). This is the purview of Bayesian, rather than frequentist statistics.



            As for the correlation you observed: as I said above the p-values will be positively skewed under the alternative hypothesis. How skewed depends what that alternative hypothesis is. In the case of a two-sample t-test, the more you increase the difference between your population means, the more skewed the p-values will become. This reflects the fact that you're making your samples increasingly more different from what is plausible under the null, and so by definition the resulting p-values (reflecting the probability of the data under the null) must decrease.






            share|cite|improve this answer









            $endgroup$













              Your Answer





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              2 Answers
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              2 Answers
              2






              active

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              active

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              active

              oldest

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              2












              $begingroup$

              Why would you expect anything else? You don't need a simulation to know this is going to happen. Look at the formula for the t-statistic:
              $t = fracbarx_1 - barx_2 sqrt fracs^2_1n_1 + fracs^2_2n_2 $



              Obviously if you increase the true difference of means you expect $barx_1 - barx_2$ will be larger. You are holding the variance and sample size constant, so the t-statistic must be larger and thus the p-value smaller.



              I think you are confusing a philosophical rule about hypothesis testing with a mathematical fact. If the null hypothesis is true, you would expect a higher p-value. This has to be true in order for hypothesis testing to make any sense.






              share|cite|improve this answer









              $endgroup$

















                2












                $begingroup$

                Why would you expect anything else? You don't need a simulation to know this is going to happen. Look at the formula for the t-statistic:
                $t = fracbarx_1 - barx_2 sqrt fracs^2_1n_1 + fracs^2_2n_2 $



                Obviously if you increase the true difference of means you expect $barx_1 - barx_2$ will be larger. You are holding the variance and sample size constant, so the t-statistic must be larger and thus the p-value smaller.



                I think you are confusing a philosophical rule about hypothesis testing with a mathematical fact. If the null hypothesis is true, you would expect a higher p-value. This has to be true in order for hypothesis testing to make any sense.






                share|cite|improve this answer









                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  Why would you expect anything else? You don't need a simulation to know this is going to happen. Look at the formula for the t-statistic:
                  $t = fracbarx_1 - barx_2 sqrt fracs^2_1n_1 + fracs^2_2n_2 $



                  Obviously if you increase the true difference of means you expect $barx_1 - barx_2$ will be larger. You are holding the variance and sample size constant, so the t-statistic must be larger and thus the p-value smaller.



                  I think you are confusing a philosophical rule about hypothesis testing with a mathematical fact. If the null hypothesis is true, you would expect a higher p-value. This has to be true in order for hypothesis testing to make any sense.






                  share|cite|improve this answer









                  $endgroup$



                  Why would you expect anything else? You don't need a simulation to know this is going to happen. Look at the formula for the t-statistic:
                  $t = fracbarx_1 - barx_2 sqrt fracs^2_1n_1 + fracs^2_2n_2 $



                  Obviously if you increase the true difference of means you expect $barx_1 - barx_2$ will be larger. You are holding the variance and sample size constant, so the t-statistic must be larger and thus the p-value smaller.



                  I think you are confusing a philosophical rule about hypothesis testing with a mathematical fact. If the null hypothesis is true, you would expect a higher p-value. This has to be true in order for hypothesis testing to make any sense.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 33 mins ago









                  Matt PMatt P

                  1163




                  1163























                      0












                      $begingroup$

                      As you said, the p-value is uniformly distributed under the null hypothesis. That is, if the null hypothesis is really true, then upon repeated experiments we expect to find a fully random, flat distribution of p-values between [0, 1]. Consequently, a frequentist p-value says nothing about how likely the null hypothesis is to be true, since any p-value is equally probable under the null.



                      What you're looking at is the distribution of p-values under an alternative hypothesis. Depending on the formulation of this hypothesis, the resulting p-values can have any non-Uniform, positively skewed distribution between [0, 1]. But this doesn't tell you anything about the probability of the null. The reason is that the p-value expresses the probability of the evidence under the null hypothesis, i.e. $p(D|H_0)$, whereas you want to know $p(H_0|D)$. These two are related by Bayes' rule:
                      $$
                      p(H_0|D) = fracH_0)p(H_0)p(D
                      $$

                      This means that in order to calculate the probability you're interested in, you need to know and take into account the prior probability of the null being true ($p(H_0)$), the prior probability of the null being false ($p(neg H_0)$) and the probability of the data given that the null is false ($p(D|neg H_0)$). This is the purview of Bayesian, rather than frequentist statistics.



                      As for the correlation you observed: as I said above the p-values will be positively skewed under the alternative hypothesis. How skewed depends what that alternative hypothesis is. In the case of a two-sample t-test, the more you increase the difference between your population means, the more skewed the p-values will become. This reflects the fact that you're making your samples increasingly more different from what is plausible under the null, and so by definition the resulting p-values (reflecting the probability of the data under the null) must decrease.






                      share|cite|improve this answer









                      $endgroup$

















                        0












                        $begingroup$

                        As you said, the p-value is uniformly distributed under the null hypothesis. That is, if the null hypothesis is really true, then upon repeated experiments we expect to find a fully random, flat distribution of p-values between [0, 1]. Consequently, a frequentist p-value says nothing about how likely the null hypothesis is to be true, since any p-value is equally probable under the null.



                        What you're looking at is the distribution of p-values under an alternative hypothesis. Depending on the formulation of this hypothesis, the resulting p-values can have any non-Uniform, positively skewed distribution between [0, 1]. But this doesn't tell you anything about the probability of the null. The reason is that the p-value expresses the probability of the evidence under the null hypothesis, i.e. $p(D|H_0)$, whereas you want to know $p(H_0|D)$. These two are related by Bayes' rule:
                        $$
                        p(H_0|D) = fracH_0)p(H_0)p(D
                        $$

                        This means that in order to calculate the probability you're interested in, you need to know and take into account the prior probability of the null being true ($p(H_0)$), the prior probability of the null being false ($p(neg H_0)$) and the probability of the data given that the null is false ($p(D|neg H_0)$). This is the purview of Bayesian, rather than frequentist statistics.



                        As for the correlation you observed: as I said above the p-values will be positively skewed under the alternative hypothesis. How skewed depends what that alternative hypothesis is. In the case of a two-sample t-test, the more you increase the difference between your population means, the more skewed the p-values will become. This reflects the fact that you're making your samples increasingly more different from what is plausible under the null, and so by definition the resulting p-values (reflecting the probability of the data under the null) must decrease.






                        share|cite|improve this answer









                        $endgroup$















                          0












                          0








                          0





                          $begingroup$

                          As you said, the p-value is uniformly distributed under the null hypothesis. That is, if the null hypothesis is really true, then upon repeated experiments we expect to find a fully random, flat distribution of p-values between [0, 1]. Consequently, a frequentist p-value says nothing about how likely the null hypothesis is to be true, since any p-value is equally probable under the null.



                          What you're looking at is the distribution of p-values under an alternative hypothesis. Depending on the formulation of this hypothesis, the resulting p-values can have any non-Uniform, positively skewed distribution between [0, 1]. But this doesn't tell you anything about the probability of the null. The reason is that the p-value expresses the probability of the evidence under the null hypothesis, i.e. $p(D|H_0)$, whereas you want to know $p(H_0|D)$. These two are related by Bayes' rule:
                          $$
                          p(H_0|D) = fracH_0)p(H_0)p(D
                          $$

                          This means that in order to calculate the probability you're interested in, you need to know and take into account the prior probability of the null being true ($p(H_0)$), the prior probability of the null being false ($p(neg H_0)$) and the probability of the data given that the null is false ($p(D|neg H_0)$). This is the purview of Bayesian, rather than frequentist statistics.



                          As for the correlation you observed: as I said above the p-values will be positively skewed under the alternative hypothesis. How skewed depends what that alternative hypothesis is. In the case of a two-sample t-test, the more you increase the difference between your population means, the more skewed the p-values will become. This reflects the fact that you're making your samples increasingly more different from what is plausible under the null, and so by definition the resulting p-values (reflecting the probability of the data under the null) must decrease.






                          share|cite|improve this answer









                          $endgroup$



                          As you said, the p-value is uniformly distributed under the null hypothesis. That is, if the null hypothesis is really true, then upon repeated experiments we expect to find a fully random, flat distribution of p-values between [0, 1]. Consequently, a frequentist p-value says nothing about how likely the null hypothesis is to be true, since any p-value is equally probable under the null.



                          What you're looking at is the distribution of p-values under an alternative hypothesis. Depending on the formulation of this hypothesis, the resulting p-values can have any non-Uniform, positively skewed distribution between [0, 1]. But this doesn't tell you anything about the probability of the null. The reason is that the p-value expresses the probability of the evidence under the null hypothesis, i.e. $p(D|H_0)$, whereas you want to know $p(H_0|D)$. These two are related by Bayes' rule:
                          $$
                          p(H_0|D) = fracH_0)p(H_0)p(D
                          $$

                          This means that in order to calculate the probability you're interested in, you need to know and take into account the prior probability of the null being true ($p(H_0)$), the prior probability of the null being false ($p(neg H_0)$) and the probability of the data given that the null is false ($p(D|neg H_0)$). This is the purview of Bayesian, rather than frequentist statistics.



                          As for the correlation you observed: as I said above the p-values will be positively skewed under the alternative hypothesis. How skewed depends what that alternative hypothesis is. In the case of a two-sample t-test, the more you increase the difference between your population means, the more skewed the p-values will become. This reflects the fact that you're making your samples increasingly more different from what is plausible under the null, and so by definition the resulting p-values (reflecting the probability of the data under the null) must decrease.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 22 mins ago









                          Ruben van BergenRuben van Bergen

                          4,0391924




                          4,0391924



























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                              Mortes em março de 2019 Referências Menu de navegação«Zhores Alferov, Nobel de Física bielorrusso, morre aos 88 anos - Ciência»«Fallece Rafael Torija, o bispo emérito de Ciudad Real»«Peter Hurford dies at 88»«Keith Flint, vocalista do The Prodigy, morre aos 49 anos»«Luke Perry, ator de 'Barrados no baile' e 'Riverdale', morre aos 52 anos»«Former Rangers and Scotland captain Eric Caldow dies, aged 84»«Morreu, aos 61 anos, a antiga lenda do wrestling King Kong Bundy»«Fallece el actor y director teatral Abraham Stavans»«In Memoriam Guillaume Faye»«Sidney Sheinberg, a Force Behind Universal and Spielberg, Is Dead at 84»«Carmine Persico, Colombo Crime Family Boss, Is Dead at 85»«Dirigent Michael Gielen gestorben»«Ciclista tricampeã mundial e prata na Rio 2016 é encontrada morta em casa aos 23 anos»«Pagan Community Notes: Raven Grimassi dies, Indianapolis pop-up event cancelled, Circle Sanctuary announces new podcast, and more!»«Hal Blaine, Wrecking Crew Drummer, Dies at 90»«Morre Coutinho, que editou dupla lendária com Pelé no Santos»«Cantor Demétrius, ídolo da Jovem Guarda, morre em SP»«Ex-presidente do Vasco, Eurico Miranda morre no Rio de Janeiro»«Bronze no Mundial de basquete de 1971, Laís Elena morre aos 76 anos»«Diretor de Corridas da F1, Charlie Whiting morre aos 66 anos às vésperas do GP da Austrália»«Morreu o cardeal Danneels, da Bélgica»«Morreu o cartoonista Augusto Cid»«Morreu a atriz Maria Isabel de Lizandra, de "Vale Tudo" e novelas da Tupi»«WS Merwin, prize-winning poet of nature, dies at 91»«Atriz Márcia Real morre em São Paulo aos 88 anos»«Mauritanie: décès de l'ancien président Mohamed Mahmoud ould Louly»«Morreu Dick Dale, o rei da surf guitar e de "Pulp Fiction"»«Falleció Víctor Genes»«João Carlos Marinho, autor de 'O Gênio do Crime', morre em SP»«Legendary Horror Director and SFX Artist John Carl Buechler Dies at 66»«Morre em Salvador a religiosa Makota Valdina»«مرگ بازیکن‌ سابق نساجی بر اثر سقوط سنگ در مازندران»«Domingos Oliveira morre no Rio»«Morre Airton Ravagniani, ex-São Paulo, Fla, Vasco, Grêmio e Sport - Notícias»«Morre o escritor Flavio Moreira da Costa»«Larry Cohen, Writer-Director of 'It's Alive' and 'Hell Up in Harlem,' Dies at 77»«Scott Walker, experimental singer-songwriter, dead at 76»«Joseph Pilato, Day of the Dead Star and Horror Favorite, Dies at 70»«Sheffield United set to pay tribute to legendary goalkeeper Ted Burgin who has died at 91»«Morre Rafael Henzel, sobrevivente de acidente aéreo da Chapecoense»«Morre Valery Bykovsky, um dos primeiros cosmonautas da União Soviética»«Agnès Varda, cineasta da Nouvelle Vague, morre aos 90 anos»«Agnès Varda, cineasta francesa, morre aos 90 anos»«Tania Mallet, James Bond Actress and Helen Mirren's Cousin, Dies at 77»e