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CHAPTER 9

THE ONE-WAY MODEL

9.4.

(a)

(8 points)) The table is

Analysis of Variance
Source	DF	Sum of Squares	Mean Square	F Stat	Prob > F
Treatment	6	1.32	0.22	3.38	0.0207
Blocks	3	2.28	0.76	11.69	0.0002
Error	18	1.17	0.065
C Total	27	4.77

(b)

(6 points)) There are seven treatments.

(c)

(6 points)) There are four blocks.

9.6.

(a)

(5 points)) Randomize the order of the runs.

(b)

(10 points)) Means model: $Y_{ij}=\mu_i+\epsilon_{ij}$ , or effects model: $Y_{ij}=\mu+\tau_i+\epsilon_{ij}$ .For the means model the estimates of the means are:

$\begin{displaymath} \hat{\mu}_1=600.21,~\hat{\mu}_2=913.64,~\hat{\mu}_3=1523.20,~ \hat{\mu}_4=478.31.\end{displaymath}$

(c)

(10 points)) Plots of the studentized residuals (Figure 1), reveal one possible outlier. Their distribution has a short upper tail compared with the normal distribution.

**Figure 1:** *Studentized residual plots, 9.6(c)*
$\begin{figure} \centerline{ \psfig {file=exsoln9a.eps,height=3in,width=5in} } \vspace{2ex}\end{figure}$

(d)

(10 points)) The ANOVA table is

**Figure 1:** *Studentized residual plots, 9.6(c)*
Analysis of Variance
Source	DF	Sum of Squares	Mean Square	F Stat	Prob > F
Model	3	$1.96\times 10^6$	$6.54\times 10^5$	50	0.0001
Error	8	$1.05\times 10^5$	13089
C Total	11	$2.07\times 10^6$

The p-value of the F test is $\leq 0.0001$ , indicating a highly significant difference in means. The means are ordered: 3>2>1>4. Level 0.95 Tukey intervals are

Comparison	Interval
3-2	(310.42,908.71)
3-1	(623.84,1222.14)
3-4	(745.74,1344.04)
2-1	(14.28,612.57)
2-4	(136.18,734.47)
1-4	(-177.25,421.05)

From this we see that the mean of 3 is significantly larger than the other means, and the mean of 2 is significantly larger than that of 1 or 4. There are no differences among the other means.

(e)

(10 points)) The report should state:

(i): A one-way means model fit reasonably well (one possible outlier).
(ii): The F test revealed a significant difference in means for different configurations.
(iv): A Tukey multiple comparison test showed that the mean force under configuration 3 exceeded that under the other three configurations; that the force under configuration 2 exceeded that under configurations 1 and 4; and that the force under configurations 1 and 4 are not significantly different.

9.10.

(10 points)) The one-way model

$\begin{displaymath} RMS_{ij}=\mu+SHAPE_i+\epsilon_{ij},\end{displaymath}$

was fit. The MSE was 0.007, compared with 0.004 for the full data model in Exercise 9-9. Without the reduction in variation due to cams, the F test is no longer significant (p=0.1497).

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Joseph D Petruccelli
12/14/1999