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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



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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 2 hours ago







      Nakx

















      asked 3 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:
            $$
            p(H_0|D) = fracp(Dneg H_0)p(neg H_0)
            $$

            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 1 hour 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:
                      $$
                      p(H_0|D) = fracp(Dneg H_0)p(neg H_0)
                      $$

                      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:
                        $$
                        p(H_0|D) = fracp(Dneg H_0)p(neg H_0)
                        $$

                        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:
                          $$
                          p(H_0|D) = fracp(Dneg H_0)p(neg H_0)
                          $$

                          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:
                          $$
                          p(H_0|D) = fracp(Dneg H_0)p(neg H_0)
                          $$

                          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 1 hour ago









                          Ruben van BergenRuben van Bergen

                          4,0391924




                          4,0391924



























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                              Tom Holland Mục lục Đầu đời và giáo dục | Sự nghiệp | Cuộc sống cá nhân | Phim tham gia | Giải thưởng và đề cử | Chú thích | Liên kết ngoài | Trình đơn chuyển hướngProfile“Person Details for Thomas Stanley Holland, "England and Wales Birth Registration Index, 1837-2008" — FamilySearch.org”"Meet Tom Holland... the 16-year-old star of The Impossible""Schoolboy actor Tom Holland finds himself in Oscar contention for role in tsunami drama"“Naomi Watts on the Prince William and Harry's reaction to her film about the late Princess Diana”lưu trữ"Holland and Pflueger Are West End's Two New 'Billy Elliots'""I'm so envious of my son, the movie star! British writer Dominic Holland's spent 20 years trying to crack Hollywood - but he's been beaten to it by a very unlikely rival"“Richard and Margaret Povey of Jersey, Channel Islands, UK: Information about Thomas Stanley Holland”"Tom Holland to play Billy Elliot""New Billy Elliot leaving the garage"Billy Elliot the Musical - Tom Holland - Billy"A Tale of four Billys: Tom Holland""The Feel Good Factor""Thames Christian College schoolboys join Myleene Klass for The Feelgood Factor""Government launches £600,000 arts bursaries pilot""BILLY's Chapman, Holland, Gardner & Jackson-Keen Visit Prime Minister""Elton John 'blown away' by Billy Elliot fifth birthday" (video with John's interview and fragments of Holland's performance)"First News interviews Arrietty's Tom Holland"“33rd Critics' Circle Film Awards winners”“National Board of Review Current Awards”Bản gốc"Ron Howard Whaling Tale 'In The Heart Of The Sea' Casts Tom Holland"“'Spider-Man' Finds Tom Holland to Star as New Web-Slinger”lưu trữ“Captain America: Civil War (2016)”“Film Review: ‘Captain America: Civil War’”lưu trữ“‘Captain America: Civil War’ review: Choose your own avenger”lưu trữ“The Lost City of Z reviews”“Sony Pictures and Marvel Studios Find Their 'Spider-Man' Star and Director”“‘Mary Magdalene’, ‘Current War’ & ‘Wind River’ Get 2017 Release Dates From Weinstein”“Lionsgate Unleashing Daisy Ridley & Tom Holland Starrer ‘Chaos Walking’ In Cannes”“PTA's 'Master' Leads Chicago Film Critics Nominations, UPDATED: Houston and Indiana Critics Nominations”“Nominaciones Goya 2013 Telecinco Cinema – ENG”“Jameson Empire Film Awards: Martin Freeman wins best actor for performance in The Hobbit”“34th Annual Young Artist Awards”Bản gốc“Teen Choice Awards 2016—Captain America: Civil War Leads Second Wave of Nominations”“BAFTA Film Award Nominations: ‘La La Land’ Leads Race”“Saturn Awards Nominations 2017: 'Rogue One,' 'Walking Dead' Lead”Tom HollandTom HollandTom HollandTom Hollandmedia.gettyimages.comWorldCat Identities300279794no20130442900000 0004 0355 42791085670554170004732cb16706349t(data)XX5557367