Paired T Test Hypothesis Interpretation











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




























    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


























      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited yesterday

























      asked yesterday









      AxGryndr

      1125




      1125






















          1 Answer
          1






          active

          oldest

          votes

















          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











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














           

          draft saved


          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2999266%2fpaired-t-test-hypothesis-interpretation%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          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










          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

          61.8k65997




          61.8k65997












          • 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


















           

          draft saved


          draft discarded



















































           


          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2999266%2fpaired-t-test-hypothesis-interpretation%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          Popular posts from this blog

          AnyDesk - Fatal Program Failure

          QoS: MAC-Priority for clients behind a repeater

          Актюбинская область