Question

Oehlert provides data from a small experiment with n = 16 observations on baking packaged cake mixes. Two factors, X1 = baking time in minutes and X2 = baking temperature in degrees F, were varied in the experiment. The response Y was the average palatability score of four cakes baked at a given combination of (X1, X2), with higher values desirable. We fit the full second model Model I: Y; = Bo + B1211 + B22:2 +B112 + B2222 Model II: Y; = Bo + B22:2 + B2222 Model III: Y; = Bo + B1 Xil + B22:2 + B112? + B22x3 + B122:1012 ## Test for model II and model III

Answer 1

(a)

ô = 0.7289 MSR = mean sum of squares due to residual = 0.72892 = 0.5313 with degrees of freedom (df) = 16 – 5= 11. SSR= = sum of squares due to residuals 11 * MSR= 11 * 0.5313 = 5.8443 SSR R= 0.8125 = 1 - SSTO 5.8443/(1 – 0.8125) = 31.1696. SSTO

ANOVA:

Source	SS	df	MS	F
Regression	31.1696-5.8443=25.3253	4	25.3253/4=6.3313	6.3313/0.5313=11.9166
Residual	5.8443	15-4=11	5.8443/11=0.5313
Total	31.1696	16-1=15

(b) Value of F statistic=(14.4380-2.7525)/(5-2))/(14.4380/(16-6))=2.6979

P-value=P(F>2.6979|F~F_3,10)= 0.1023(R code: round(1-pf(2.6979,3,10),4)).

Since P-value>0.01, hence X₁, X₁² and X₁X₂ are not significant.

(c) AIC=n*log(SSR/n)+2p

BIC=n*log(SSR)+(p-n)log n

where, n=no. of observations, p=no. of explanatory variables.

Model 2:

SSR=14.4380, n=16, p=2 then AIC=16*log(14.4380/16)+2*2=2.3564, BIC=16*log(14.4380)+(2-16)*log(16)=3.9016.

Model 3:

SSR=2.7525, n=16, p=5 then AIC=16*log(2.7525/16)+2*5=-18.1613, BIC=16*log(2.7525)+(5-16)*log(16)=-14.2983.

Model 2 is used.