Type 2 Error Question - How to calculate for a two tailed?












1












$begingroup$


The modulus of rupture (MOR) for a particular grade of pencil lead is known to have
a standard deviation of 250 psi. Process standards call for a target value of 6500 psi
for the true mean MOR. For each batch, an inspector tests a random sample of 16
leads. Management wishes to detect any change in the true mean MOR. (Assume normal
distribution.)



QUESTION : Find the probability of type II error of the test when the true mean MOR is 6400.



How do we solve for two tails? i am able to get 0.4821 but with 1 tail method.
6397-6400/62.5 = 0.045
than using normcdf on MATLAB, i got this value






probability statistics data-analysis hypothesis-testing







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    The modulus of rupture (MOR) for a particular grade of pencil lead is known to have
    a standard deviation of 250 psi. Process standards call for a target value of 6500 psi
    for the true mean MOR. For each batch, an inspector tests a random sample of 16
    leads. Management wishes to detect any change in the true mean MOR. (Assume normal
    distribution.)



    QUESTION : Find the probability of type II error of the test when the true mean MOR is 6400.



    How do we solve for two tails? i am able to get 0.4821 but with 1 tail method.
    6397-6400/62.5 = 0.045
    than using normcdf on MATLAB, i got this value






    probability statistics data-analysis hypothesis-testing







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      The modulus of rupture (MOR) for a particular grade of pencil lead is known to have
      a standard deviation of 250 psi. Process standards call for a target value of 6500 psi
      for the true mean MOR. For each batch, an inspector tests a random sample of 16
      leads. Management wishes to detect any change in the true mean MOR. (Assume normal
      distribution.)



      QUESTION : Find the probability of type II error of the test when the true mean MOR is 6400.



      How do we solve for two tails? i am able to get 0.4821 but with 1 tail method.
      6397-6400/62.5 = 0.045
      than using normcdf on MATLAB, i got this value






      probability statistics data-analysis hypothesis-testing







      share|cite|improve this question









      $endgroup$




      The modulus of rupture (MOR) for a particular grade of pencil lead is known to have
      a standard deviation of 250 psi. Process standards call for a target value of 6500 psi
      for the true mean MOR. For each batch, an inspector tests a random sample of 16
      leads. Management wishes to detect any change in the true mean MOR. (Assume normal
      distribution.)



      QUESTION : Find the probability of type II error of the test when the true mean MOR is 6400.



      How do we solve for two tails? i am able to get 0.4821 but with 1 tail method.
      6397-6400/62.5 = 0.045
      than using normcdf on MATLAB, i got this value






      probability statistics data-analysis hypothesis-testing




      probability statistics data-analysis hypothesis-testing






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Apr 27 '16 at 1:12









      user334902user334902

      115




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          $begingroup$

          You have not said what significance level you are using, but it seems that what you did for one tail used $5%$ and




          • standard error of the mean $dfrac{250}{sqrt{16}} = 62.5$

          • If the true population mean is $6500$ then there is a $5%$ probability that the the sample mean will be below $6500 + 62.5 Phi^{-1}(0.05) approx 6397$

          • If the true population mean is $6400$ then there is about a $Phileft(dfrac{6397-6400}{62.5}right) approx 48.21%$ probability that the the sample mean will be below $6397$


          You need to do something similar for two tails:




          • If the true population mean is $6500$ then there is a $2.5%$ probability that the the sample mean will be below $6500 + 62.5 Phi^{-1}(0.025) approx 6377.5$

          • If the true population mean is $6500$ then there is a $2.5%$ probability that the the sample mean will be above $6500 + 62.5 Phi^{-1}(0.975) approx 6622.5$

          • If the true population mean is $6400$ then there is about a $Phileft(dfrac{6377.5-6400}{62.5}right) approx 35.94%$ probability that the the sample mean will be below $6377.5$

          • If the true population mean is $6400$ then there is about a $1-Phileft(dfrac{6622.5-6400}{62.5}right) approx 0.02%$ probability that the the sample mean will be above $6622.5$

          • So the probability of a type II error of the test when the true mean is $6400$ is about $35.96%$






          share|cite|improve this answer









          $endgroup$













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            0












            $begingroup$

            You have not said what significance level you are using, but it seems that what you did for one tail used $5%$ and




            • standard error of the mean $dfrac{250}{sqrt{16}} = 62.5$

            • If the true population mean is $6500$ then there is a $5%$ probability that the the sample mean will be below $6500 + 62.5 Phi^{-1}(0.05) approx 6397$

            • If the true population mean is $6400$ then there is about a $Phileft(dfrac{6397-6400}{62.5}right) approx 48.21%$ probability that the the sample mean will be below $6397$


            You need to do something similar for two tails:




            • If the true population mean is $6500$ then there is a $2.5%$ probability that the the sample mean will be below $6500 + 62.5 Phi^{-1}(0.025) approx 6377.5$

            • If the true population mean is $6500$ then there is a $2.5%$ probability that the the sample mean will be above $6500 + 62.5 Phi^{-1}(0.975) approx 6622.5$

            • If the true population mean is $6400$ then there is about a $Phileft(dfrac{6377.5-6400}{62.5}right) approx 35.94%$ probability that the the sample mean will be below $6377.5$

            • If the true population mean is $6400$ then there is about a $1-Phileft(dfrac{6622.5-6400}{62.5}right) approx 0.02%$ probability that the the sample mean will be above $6622.5$

            • So the probability of a type II error of the test when the true mean is $6400$ is about $35.96%$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              You have not said what significance level you are using, but it seems that what you did for one tail used $5%$ and




              • standard error of the mean $dfrac{250}{sqrt{16}} = 62.5$

              • If the true population mean is $6500$ then there is a $5%$ probability that the the sample mean will be below $6500 + 62.5 Phi^{-1}(0.05) approx 6397$

              • If the true population mean is $6400$ then there is about a $Phileft(dfrac{6397-6400}{62.5}right) approx 48.21%$ probability that the the sample mean will be below $6397$


              You need to do something similar for two tails:




              • If the true population mean is $6500$ then there is a $2.5%$ probability that the the sample mean will be below $6500 + 62.5 Phi^{-1}(0.025) approx 6377.5$

              • If the true population mean is $6500$ then there is a $2.5%$ probability that the the sample mean will be above $6500 + 62.5 Phi^{-1}(0.975) approx 6622.5$

              • If the true population mean is $6400$ then there is about a $Phileft(dfrac{6377.5-6400}{62.5}right) approx 35.94%$ probability that the the sample mean will be below $6377.5$

              • If the true population mean is $6400$ then there is about a $1-Phileft(dfrac{6622.5-6400}{62.5}right) approx 0.02%$ probability that the the sample mean will be above $6622.5$

              • So the probability of a type II error of the test when the true mean is $6400$ is about $35.96%$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                You have not said what significance level you are using, but it seems that what you did for one tail used $5%$ and




                • standard error of the mean $dfrac{250}{sqrt{16}} = 62.5$

                • If the true population mean is $6500$ then there is a $5%$ probability that the the sample mean will be below $6500 + 62.5 Phi^{-1}(0.05) approx 6397$

                • If the true population mean is $6400$ then there is about a $Phileft(dfrac{6397-6400}{62.5}right) approx 48.21%$ probability that the the sample mean will be below $6397$


                You need to do something similar for two tails:




                • If the true population mean is $6500$ then there is a $2.5%$ probability that the the sample mean will be below $6500 + 62.5 Phi^{-1}(0.025) approx 6377.5$

                • If the true population mean is $6500$ then there is a $2.5%$ probability that the the sample mean will be above $6500 + 62.5 Phi^{-1}(0.975) approx 6622.5$

                • If the true population mean is $6400$ then there is about a $Phileft(dfrac{6377.5-6400}{62.5}right) approx 35.94%$ probability that the the sample mean will be below $6377.5$

                • If the true population mean is $6400$ then there is about a $1-Phileft(dfrac{6622.5-6400}{62.5}right) approx 0.02%$ probability that the the sample mean will be above $6622.5$

                • So the probability of a type II error of the test when the true mean is $6400$ is about $35.96%$






                share|cite|improve this answer









                $endgroup$



                You have not said what significance level you are using, but it seems that what you did for one tail used $5%$ and




                • standard error of the mean $dfrac{250}{sqrt{16}} = 62.5$

                • If the true population mean is $6500$ then there is a $5%$ probability that the the sample mean will be below $6500 + 62.5 Phi^{-1}(0.05) approx 6397$

                • If the true population mean is $6400$ then there is about a $Phileft(dfrac{6397-6400}{62.5}right) approx 48.21%$ probability that the the sample mean will be below $6397$


                You need to do something similar for two tails:




                • If the true population mean is $6500$ then there is a $2.5%$ probability that the the sample mean will be below $6500 + 62.5 Phi^{-1}(0.025) approx 6377.5$

                • If the true population mean is $6500$ then there is a $2.5%$ probability that the the sample mean will be above $6500 + 62.5 Phi^{-1}(0.975) approx 6622.5$

                • If the true population mean is $6400$ then there is about a $Phileft(dfrac{6377.5-6400}{62.5}right) approx 35.94%$ probability that the the sample mean will be below $6377.5$

                • If the true population mean is $6400$ then there is about a $1-Phileft(dfrac{6622.5-6400}{62.5}right) approx 0.02%$ probability that the the sample mean will be above $6622.5$

                • So the probability of a type II error of the test when the true mean is $6400$ is about $35.96%$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Apr 28 '16 at 7:59









                HenryHenry

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                101k481168






























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