### Feb. 2, 2008

burt <- read.table("burt.txt", header = T)

attach(burt)

model<-lm(FostIQ~OwnIQ)

## Regression Lingo Alert

**Regress the outcome variable (Y) on the predictor variable**-We regress Y on X

coef(model)

(Intercept) OwnIQ

9.719491 0.907920

anova(model)

Analysis of Variance Table

Response: FostIQ

Df Sum Sq Mean Sq F value Pr(>F)

OwnIQ 1 9250.7 9250.7 169.42 < 2.2e-16 ***

Residuals 51 2784.7 54.6

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

## Fitted Model

-Drops error from the regression equation.

Yvariable-sub-i = (Intercept) + Slope*Xvariable-sub-i

(Regression Equation includes error so the Y variable is not an estimate but in the Fitted Model the Y is an estimate [

**needs the hat**])

library(lattice)

xyplot(FostIQ~OwnIQ,type=c("p","r"))

## R^2

R^2 = SSmodel/SStotal

R^2 = 9251/12035 = .769

*76.9% of difference in FostIQ is accounted for by difference in OwnIQ and 23.1% is not.

We cannot account for how the unexplained variation divides the variation between other systematic components, measurement error, and individual variation.

## Estimated Residual Variance

What is the variance of the mean estimates for the scores at each point on the line (

**REmember that the line represents the means of the potential distribution at any point on a line.**)

sigma-hat-

sigma-hat^2-sub-X|Y = SSresiduals/n

sigma-hat^2-sub-X|Y = SSresiduals/n - (parameters in equation)

sigma-hat^2-sub-X|Y = 2785/31 - 2

SD=Sqrt(sigma-hat^2-sub-X|Y = 2785/31 - 2)

sqrt(54.6)

[1] 7.389181

We are therefore sure to around 95% (2 SDs) that our predicted values will be within about 15 points either side of any particular point estimate.

9.72+.91*75

[1] 77.97

77.97-14.8

[1] 63.17

77.97+14.8

[1] 92.77

So we are sure that for an OwnIQ score of 75 we would expect the Fost IQ score to be somewhere between 63.1,92.8.