Problem 1

The surface finish of metal parts made on $a=4$ machines is under investigation. Each machine can be run by one of $b=3$ operators. Because of the location of the machines, operators are specific to a particular machine. Therefore, a nested design with fixed factors is used. Each operator produces $n=2$ samples. The data is availabe on Blackboard as an Excel File.

Part (a)

Explain the difference between crossed factors and nested factors.

Factors $A$ and $B$ are crossed in an experimental design if the levels of $B$ are the same at each level of $A$.

Factor $B$ is nested within factor $A$ if the levels of $B$ are different for each of the levels of factor $A$.

Part (b)

Write the model for a nested design. Provide algebraic formulas for the estimates $\hat{\tau}i$and$\hat{\beta}){j(i)}$.

Given that $A$ and $B$ are fixed factors,

where ${\tau_i}$ ($\sum_i \tau_i = 0$) with $\df{A}=a-1$ and ${\beta_(j(i))}$ ($\sum_j \beta_{j(i)} = 0$ for $i=1,\ldots,a$) with $\df{B} = a(b-1)$ are fixed effects.

Estimators are given by

and

We see that level factors of $A$ are compared with ${\hat{\tau}_i}$ while level factors of $B$ are compared only with the same level of $A$, e.g., at the $i$-th level of $A$, level factors of $B$ are compared with ${\beta_j(i) }$.

Part (c)

Compute the $F_A$ statistic for testing factor $A$ effects, and the $F_{B(A)}$ statistic for testing nested factor $B$ effects. Compute the $p$-values. Provide an overall interpretation, stated in the context of the problem.

For fixed factor $A$ and fixed factor $B$ nested in $A$,

and

where \begin{align*} \ms{A} &= \frac{b n \sum_{i=1}^{a} \hat{\tau}i^2 }{a-1},\ \ms{B(A)} &= \frac{n \sum{i=1}^{a} \sum_{j=1}^{b} \hat{\beta}{j(i)}^2}{a(b-1)},\ \ms{E} &= \frac{\sum{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n} (Y_{i j k} - \bar{Y}_{i j \cdot})}{a b (n-1)}. \end{align*}

Under the model $\tau_1 = \cdots \tau_a = 0$,

and under $\beta_{j(i)} = 0$ for all $i,j$,

We perform these computations in R with:

library("readxl")
data = read_excel("handout9data.xlsx")
A = as.factor(na.omit(data$mchine))
B = as.factor(na.omit(data$operator))
y = na.omit(data$surface)

# use contrasts to define parameter restrictions for the fixed effects in the
# model
contrasts(A)=contr.sum
contrasts(B)=contr.sum

# to fit a model with a nested fixed effect, we use the / notation within the
# aov command.
nested.mod = aov(y ~ A/B)
summary(nested.mod) # compute F statistics for factor effects
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##             Df Sum Sq Mean Sq F value   Pr(>F)    
## A            3   3618  1205.9  14.271 0.000291 ***
## A:B          8   2818   352.2   4.168 0.013408 *  
## Residuals   12   1014    84.5                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
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We see that $F_A = 14.271$ ($p$-value $= .000$) and $F_{B(A)} = 4.168$ ($p$-value $= .013$).

Interpretation

The experiment finds that the machine has an effect on surface finish. Also, the experiment finds that the operators within a machine has an effect on surface finish.

Part (d)

Compute estimates for each of the effect parameters. Identify which machine performs best, and which operator performs best on each machine. (Higher scores of response are preferred.)

Comparing machines