Algebraic transformation — where is my mistake?Find least squares regression lineVariance of Least Squares...

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Algebraic transformation — where is my mistake?


Find least squares regression lineVariance of Least Squares EstimatorEstimating $beta_o$ and $beta_1$ with Weighted Least Squares with Logit linkSimple linear regression and sum of squared errorsCalculation of variance beta hatDifference in fitting models using least squaresLinear regression propertyLeast squares assuming functional form of solutionHow can I prove $hatbeta_0$ and $hatbeta_1$ are linear in $hat Y_i$?Derivation of the ordinary least squares estimator β1 and the sampling distribution?













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$begingroup$


I tried to find the estimators of $hat{beta_1}$ and $hat{beta_0}$ via the least-squares method algebraically. Somehow I seem to have messed up.
Can you tell me where?



My Calculations.










share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    I tried to find the estimators of $hat{beta_1}$ and $hat{beta_0}$ via the least-squares method algebraically. Somehow I seem to have messed up.
    Can you tell me where?



    My Calculations.










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I tried to find the estimators of $hat{beta_1}$ and $hat{beta_0}$ via the least-squares method algebraically. Somehow I seem to have messed up.
      Can you tell me where?



      My Calculations.










      share|cite|improve this question









      $endgroup$




      I tried to find the estimators of $hat{beta_1}$ and $hat{beta_0}$ via the least-squares method algebraically. Somehow I seem to have messed up.
      Can you tell me where?



      My Calculations.







      transformation least-squares linear-regression






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Feb 22 at 18:05









      Jürgen ErhardtJürgen Erhardt

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          $begingroup$

          Your calculations seem fine, you should only continue rearranging the terms and using the fact that from the basic definition of a mean, we have:



          $$ sum x_i = nbar x = sum bar x .$$






          share|cite|improve this answer









          $endgroup$













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            0












            $begingroup$

            Your calculations seem fine, you should only continue rearranging the terms and using the fact that from the basic definition of a mean, we have:



            $$ sum x_i = nbar x = sum bar x .$$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Your calculations seem fine, you should only continue rearranging the terms and using the fact that from the basic definition of a mean, we have:



              $$ sum x_i = nbar x = sum bar x .$$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Your calculations seem fine, you should only continue rearranging the terms and using the fact that from the basic definition of a mean, we have:



                $$ sum x_i = nbar x = sum bar x .$$






                share|cite|improve this answer









                $endgroup$



                Your calculations seem fine, you should only continue rearranging the terms and using the fact that from the basic definition of a mean, we have:



                $$ sum x_i = nbar x = sum bar x .$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered yesterday









                pdbpdb

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