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Math Help - Coefficient of Determination

  1. #1
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    Coefficient of Determination

    Im stuck on a question where I must find the coefficient of determination and then interpret the resulting value.
    The value of the correlation coefficient is 0.916

    Help is appreciated
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  2. #2
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    Quote Originally Posted by lumberjackjim View Post
    Im stuck on a question where I must find the coefficient of determination and then interpret the resulting value.
    The value of the correlation coefficient is 0.916

    Help is appreciated
    The correlation coefficient is r = 0.916.

    The coefficient of determination is r^2 = (0.916)^2 = 0.839056.

    So is this a good linear fit?

    -Dan
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  3. #3
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    thanks a lot, I was unaware of how incredibly easy that is. this next one seems even more confusing to me though

    Regression Analysis: Larvae versus Stumps
    The regression equation is
    Larvae = - 1.29 + 11.9 Stumps

    Predictor Coef SE Coef T P
    Constant -1.286 2.853 -0.45 0.657
    Stumps 11.894 1.136 10.47 0.000

    S = 6.41939 R-Sq = 83.9% R-Sq(adj) = 83.1%

    Analysis of Variance
    Source DF SS MS F P
    Regression 1 4514.4 4514.4 109.55 0.000
    Residual Error 21 865.4 41.2
    Total 22 5379.8

    Unusual Observations
    Obs Stumps Larvae Fit SE Fit Residual St Resid
    10 5.00 56.00 58.18 3.43 -2.18 -0.40 X
    X denotes an observation whose X value gives it large influence.

    How does the value of the Coefficient of Determination included in this
    output compare with that in the previous question?
    Based on the regression line obtained, predict the number of larvae clusters for a circular plot in which there were four stumps.

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  4. #4
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    Quote Originally Posted by lumberjackjim View Post
    thanks a lot, I was unaware of how incredibly easy that is. this next one seems even more confusing to me though

    Regression Analysis: Larvae versus Stumps
    The regression equation is
    Larvae = - 1.29 + 11.9 Stumps

    Predictor Coef SE Coef T P
    Constant -1.286 2.853 -0.45 0.657
    Stumps 11.894 1.136 10.47 0.000

    S = 6.41939 R-Sq = 83.9% R-Sq(adj) = 83.1%

    Analysis of Variance
    Source DF SS MS F P
    Regression 1 4514.4 4514.4 109.55 0.000
    Residual Error 21 865.4 41.2
    Total 22 5379.8

    Unusual Observations
    Obs Stumps Larvae Fit SE Fit Residual St Resid
    10 5.00 56.00 58.18 3.43 -2.18 -0.40 X
    X denotes an observation whose X value gives it large influence.

    How does the value of the Coefficient of Determination included in this
    output compare with that in the previous question?
    Based on the regression line obtained, predict the number of larvae clusters for a circular plot in which there were four stumps.

    The R^2 value is the same as in your first question to the precission it is quoted
    in this question 83.9% is 0.839. The corrected R^2 is little different.

    The prediction is obtained by plugging 4 in for the number of stumps in the
    regression equation:

    Larvae = - 1.29 + 11.9 Stumps = -1.29 + 47.6 = 46.31

    RonL
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  5. #5
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    thanks alot
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