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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

          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









          $endgroup$




















            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

              oldest

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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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