Paired T Test Hypothesis Interpretation











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With a paired T test I understand the first part of the hypothesis:



$H0: mu_d = 0$



The alternative hypothesis I get a little mixed up and I am trying to figure out when to appropriately use the alternatives.



If the alternative hypothesis is:



$H1: mu_dneq0$



I think this means we are just using the paired T test to see if there was a change between test 1 and test 2.



If the alternative hypothesis is:



$H1: mu_d>0$



Is this setting up that the alternative hypothesis is checking that the results of test 2 were lower than that result of test one?



If so then by extension an alternative hypothesis of:



$H1: mu _{d} < 0$



This is checking that test 2 had higher results then test 1?



Any insight would be greatly appreciated.










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    up vote
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    down vote

    favorite












    With a paired T test I understand the first part of the hypothesis:



    $H0: mu_d = 0$



    The alternative hypothesis I get a little mixed up and I am trying to figure out when to appropriately use the alternatives.



    If the alternative hypothesis is:



    $H1: mu_dneq0$



    I think this means we are just using the paired T test to see if there was a change between test 1 and test 2.



    If the alternative hypothesis is:



    $H1: mu_d>0$



    Is this setting up that the alternative hypothesis is checking that the results of test 2 were lower than that result of test one?



    If so then by extension an alternative hypothesis of:



    $H1: mu _{d} < 0$



    This is checking that test 2 had higher results then test 1?



    Any insight would be greatly appreciated.










    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      With a paired T test I understand the first part of the hypothesis:



      $H0: mu_d = 0$



      The alternative hypothesis I get a little mixed up and I am trying to figure out when to appropriately use the alternatives.



      If the alternative hypothesis is:



      $H1: mu_dneq0$



      I think this means we are just using the paired T test to see if there was a change between test 1 and test 2.



      If the alternative hypothesis is:



      $H1: mu_d>0$



      Is this setting up that the alternative hypothesis is checking that the results of test 2 were lower than that result of test one?



      If so then by extension an alternative hypothesis of:



      $H1: mu _{d} < 0$



      This is checking that test 2 had higher results then test 1?



      Any insight would be greatly appreciated.










      share|cite|improve this question















      With a paired T test I understand the first part of the hypothesis:



      $H0: mu_d = 0$



      The alternative hypothesis I get a little mixed up and I am trying to figure out when to appropriately use the alternatives.



      If the alternative hypothesis is:



      $H1: mu_dneq0$



      I think this means we are just using the paired T test to see if there was a change between test 1 and test 2.



      If the alternative hypothesis is:



      $H1: mu_d>0$



      Is this setting up that the alternative hypothesis is checking that the results of test 2 were lower than that result of test one?



      If so then by extension an alternative hypothesis of:



      $H1: mu _{d} < 0$



      This is checking that test 2 had higher results then test 1?



      Any insight would be greatly appreciated.







      hypothesis-testing






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          That depends entirely on how the test statistic is specified, or equivalently, how $mu_d$ relates to the means of each group. If you define $mu_d = mu_2 - mu_1$, that is, the mean of the second tests minus the mean of the first tests, then $mu_d > 0$ implies $mu_2 > mu_1$; conversely, if $mu_d = mu_1 - mu_2$, then $mu_d > 0$ implies $mu_2 < mu_1$. Correspondingly, if you use the first definition, your test statistic must be constructed as $$T = frac{bar x_2 - bar x_1}{s/sqrt{n}},$$ whereas if using the second definition of $mu_d$, the order of the difference must also be interchanged.






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          • Ah, that makes sense. Thank you!
            – AxGryndr
            yesterday











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          That depends entirely on how the test statistic is specified, or equivalently, how $mu_d$ relates to the means of each group. If you define $mu_d = mu_2 - mu_1$, that is, the mean of the second tests minus the mean of the first tests, then $mu_d > 0$ implies $mu_2 > mu_1$; conversely, if $mu_d = mu_1 - mu_2$, then $mu_d > 0$ implies $mu_2 < mu_1$. Correspondingly, if you use the first definition, your test statistic must be constructed as $$T = frac{bar x_2 - bar x_1}{s/sqrt{n}},$$ whereas if using the second definition of $mu_d$, the order of the difference must also be interchanged.






          share|cite|improve this answer





















          • Ah, that makes sense. Thank you!
            – AxGryndr
            yesterday















          up vote
          0
          down vote



          accepted










          That depends entirely on how the test statistic is specified, or equivalently, how $mu_d$ relates to the means of each group. If you define $mu_d = mu_2 - mu_1$, that is, the mean of the second tests minus the mean of the first tests, then $mu_d > 0$ implies $mu_2 > mu_1$; conversely, if $mu_d = mu_1 - mu_2$, then $mu_d > 0$ implies $mu_2 < mu_1$. Correspondingly, if you use the first definition, your test statistic must be constructed as $$T = frac{bar x_2 - bar x_1}{s/sqrt{n}},$$ whereas if using the second definition of $mu_d$, the order of the difference must also be interchanged.






          share|cite|improve this answer





















          • Ah, that makes sense. Thank you!
            – AxGryndr
            yesterday













          up vote
          0
          down vote



          accepted







          up vote
          0
          down vote



          accepted






          That depends entirely on how the test statistic is specified, or equivalently, how $mu_d$ relates to the means of each group. If you define $mu_d = mu_2 - mu_1$, that is, the mean of the second tests minus the mean of the first tests, then $mu_d > 0$ implies $mu_2 > mu_1$; conversely, if $mu_d = mu_1 - mu_2$, then $mu_d > 0$ implies $mu_2 < mu_1$. Correspondingly, if you use the first definition, your test statistic must be constructed as $$T = frac{bar x_2 - bar x_1}{s/sqrt{n}},$$ whereas if using the second definition of $mu_d$, the order of the difference must also be interchanged.






          share|cite|improve this answer












          That depends entirely on how the test statistic is specified, or equivalently, how $mu_d$ relates to the means of each group. If you define $mu_d = mu_2 - mu_1$, that is, the mean of the second tests minus the mean of the first tests, then $mu_d > 0$ implies $mu_2 > mu_1$; conversely, if $mu_d = mu_1 - mu_2$, then $mu_d > 0$ implies $mu_2 < mu_1$. Correspondingly, if you use the first definition, your test statistic must be constructed as $$T = frac{bar x_2 - bar x_1}{s/sqrt{n}},$$ whereas if using the second definition of $mu_d$, the order of the difference must also be interchanged.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered yesterday









          heropup

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          • Ah, that makes sense. Thank you!
            – AxGryndr
            yesterday


















          • Ah, that makes sense. Thank you!
            – AxGryndr
            yesterday
















          Ah, that makes sense. Thank you!
          – AxGryndr
          yesterday




          Ah, that makes sense. Thank you!
          – AxGryndr
          yesterday


















           

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