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Thread: Using linear regression information to make a prediction

  1. #1
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    Using linear regression information to make a prediction

    A regression was run to determine if there is a relationship between hours of TV watched per day (x) and number of situps a person can do (y).

    The results of the regression were:
    y=ax+b a=-0.864 b=37.082 r2=0.850084 r=-0.922
    Use this to predict the number of situps a person who watches 4.5 hours of TV can do (to one decimal place.

    I plugged in 4.5 into the equation and got 40.9. The answer is wrong. What am I doing wrong?
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  2. #2
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    Re: Using linear regression information to make a prediction

    Hey DasRabbit.

    Here is the results using R:

    > -0.864*4.5 +37.082
    [1] 33.194
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  3. #3
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    Re: Using linear regression information to make a prediction

    hey I am having a similar problem and have scoured for someone to show me how it's done. I do not get where the r's com into the equation.
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  4. #4
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    Re: Using linear regression information to make a prediction

    The "r" does not come into the equation at all!

    DasRabbit wrote
    The results of the regression were:
    y=ax+b a=-0.864 b=37.082 r2=0.850084 r=-0.922
    That is y= -0.864x+ 37.082, and then wrote
    Use this to predict the number of situps a person who watches 4.5 hours of TV can do (to one decimal place.

    I plugged in 4.5 into the equation and got 40.9. The answer is wrong. What am I doing wrong?
    And he simply did the arithmetic wrong y(4.5)= -0.864(4.5)+ 37.082= -3.888+ 37.082= 33.194.

    It looks like he forgot the "-" on a= -0.865. If it were +0.865(4.5)+ 37.082= 3.888+ 37.082, then it would be 40.97.

    A "linear regression" replaces some unknown function with a linear approximation. The "r" gives an estimate of how accurate that liner approximation is likely to be.
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