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what is the residual for the equation 6.4+0.93x for x=60 and y=59

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The residual is the difference between the data point and what the predictor equation predicts. Compare y=59 to y=6.4_0.93(60).
So i get that the answer is 3.2, but would it be positive or negative
I think negative to show that the point is below the predictor line.

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It would be negative. The E(error/residual) = Y (actual value) - y (predicted value/estimated value) so E(residual) = 59 - [6.4 + .93(60)] = -3.2

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