Solution to Exercise 11.18.

  Results of Minitab Two way ANOVA:

Two-way Analysis of Variance


Analysis of Variance for Yield   
Source        DF        SS        MS        F        P
Formulat       1   2253.44   2253.44   376.27    0.000
Speed          2    230.81    115.41    19.27    0.000
Interaction    2     18.58      9.29     1.55    0.252
Error         12     71.87      5.99
Total         17   2574.70

                       Individual 95% CI
Formulat        Mean   --------+---------+---------+---------+---
1             187.03                                    (-*--)
2             164.66    (-*--)
                       --------+---------+---------+---------+---
                          168.00    175.00    182.00    189.00

                       Individual 95% CI
Speed           Mean   --------+---------+---------+---------+---
60             177.8                           (------*------)
70             170.8   (------*-------)
80             178.9                              (------*-------)
                       --------+---------+---------+---------+---
                           171.0     174.0     177.0     180.0

(a) The interaction was not significant (p = 0.252), so I guess the answer is, ``No, there does not appear to be an interaction ...''

(b) Bot of the main effects are significant (p = 0.000 in both cases), so the yield does appear to depend on both speed and formulation.

(c) The main effects:

Factor Level Symbol Estimate

$$\bar{x}_{1 \cdot \cdot}$$ Formulation 1 187.03

$$\bar{x}_{2 \cdot \cdot}$$ Formulation 2 164.66

$$\bar{x}_{\cdot 1 \cdot}$$ Speed 60 177.8

$$\bar{x}_{\cdot 2 \cdot}$$ Speed 70 170.8

$$\bar{x}_{\cdot 3 \cdot}$$ Speed 80 178.9

(d) I asked to store the fitted values and residuals, so they appear in the printout of the data set below:

Yield Speed Formulation RESI1 FITS1

189.7 60 1 0.23333 189.467

188.6 60 1 -0.86667 189.467

190.1 60 1 0.63333 189.467

185.1 70 1 4.50000 180.600

179.4 70 1 -1.20000 180.600

177.3 70 1 -3.30000 180.600

189.0 80 1 -2.03333 191.033

193.0 80 1 1.96667 191.033

191.1 80 1 0.06667 191.033

165.1 60 2 -1.10000 166.200

165.9 60 2 -0.30000 166.200

167.6 60 2 1.40000 166.200

161.7 70 2 0.66667 161.033

159.8 70 2 -1.23333 161.033

161.6 70 2 0.56667 161.033

163.3 80 2 -3.43333 166.733

166.6 80 2 -0.13333 166.733

170.3 80 2 3.56667 166.733

(e) The normal probability plot appears here. It looks pretty good. However, the plot of residuals vs. fitted values which appears here shows evidence that there is not a constant variance.