Output for Problem 4.40 in B2 Book

> wafer.lm=lm(FAILTIME~TEMP+I(TEMP^2),data=WAFER)
> summary(wafer.lm)

Call:
lm(formula = FAILTIME ~ TEMP + I(TEMP^2), data = WAFER)

Residuals:
     Min       1Q   Median       3Q      Max 
-1260.49  -475.70   -15.57   528.45  1131.69 

Coefficients:
              Estimate Std. Error t value Pr(>|t|)    
(Intercept) 154242.914  21868.474   7.053 1.03e-06 ***
TEMP         -1908.850    303.664  -6.286 4.92e-06 ***
I(TEMP^2)        5.929      1.048   5.659 1.86e-05 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 688.1 on 19 degrees of freedom
Multiple R-squared:  0.9415, Adjusted R-squared:  0.9354 
F-statistic: 152.9 on 2 and 19 DF,  p-value: 1.937e-12


Using Estimation Equation for Temp= 140 and 150.
Note 140^2 =  19600 
Note 150^2 =  22500

(Failtime | temp = 140) = 154242.914 - 1908.850(140) + 5.929 (19600)
                        = 3211.191

R Command to compute:↔
predict(wafer.lm,data.frame(TEMP=c(140,150)))
       1        2 

3211.191 1316.629