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Unformatted text preview: 1974.40) = 26.8
= 3190.4 n 20
2 = 77990016 ! (39488)( 39499) ˆ Y = !26.8 + 1.0139 X " (X
i =1 n i ! X ) = " X i2
2 () # & % " X i( $ ' ! n 20 = 77968254 ! (39488) 2 = 3146.8 b= 3190.4 = 1.0139 3146.8 Predicting Y from X
If a cadaver had an age estimated to be born in 1975 based on radioactivity, what does the regression line predict its year of birth to be?! r2 predicts the amount of variance in Y explained by the regression line
r2 is the “coefficient of determination: it is the square of the correlation coefficient r ˆ Y = !26.8 + 1.0139 X = !26.8 + 1.0139 (1975) = 1975.6 Caution: It is unwise to extrapolate beyond the range of the data.
Number of species of fish as predicted by the area of a desert pool If we were to extrapolate to ask how many species might be in a pool of 50000m2, we would guess about 20. More data on fish in desert pools Log transformed data: Testing hypotheses about regression
H 0: HA: =0 !0 Sums of squares for regression
# & Yi ( %! $ ' 2 = ! Yi " n
2 SSTotal SSregression = b ! ( X i " X )(Yi " Y ) SSresidual + SSregression = SSTotal
With n  2 degrees of freedom for the residual Radioactive teeth: Sums of squares
# & Yi ( %! $ ' 2 = ! Yi " n = 78011881 "
2 Smoking: Sums of squares
SSresidual = SSTotal ! SSregression SSTotal (39499) 2
20 = 3330.95 SSregression = b ! ( X i " X )(Yi " Y ) = (1.0139)( 3190.4 ) = 3234.6 df residual = 3330.95 ! 3234.6 = 96.35 = 20 ! 2 = 18 Calculating residual mean squares Standard error of a slope
SE b = MSresidual MSresidual = SSresidual / dfresidual
MSresidual = 96.35 = 5.35 18 " (X i ! X) 2 = 5.35 = 0.041 3146.8 b has a t distribution
Confidence interval for a slope: b ± t! [ 2 ],...
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This note was uploaded on 04/16/2010 for the course MATHEMATIC 1231 taught by Professor Driscoll during the Spring '10 term at Clayton College of Natural Health.
 Spring '10
 Driscoll
 Statistics, Correlation, Linear Regression, Variance

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