The surface finish of metal parts made on four randomly selected machines is being studied. An experiment is conducted in which each machine is run by three different operators and two specimens from each operator are collected and tested. Because of the location of the machines, different operators are used on each machine, and the operators are chosen at random. The data are shown in the table below. The machines and operators are denoted by M and O, respectively.
The design structure of the experiment involves the random selection of the machines which are considered as the blocks, these blocks are the larger experimental units wherein the smaller experimental units within these blocks are the three different operators which are chosen at random. However, these operators are not the same from each blocks which can be defined as the treatment structure of the experiment.
\(Y_ijk=μ+α_i+β_{j(i)}+ϵ_{k(ij)}\)
where:
\(Y_{ijk}\) = the observed response for
the \(k^{th}\) unit in the \(j^{th}\) level of B within the \(i^{th}\) level of A, \(i=1,2,⋯,a;j=1,2,⋯,b;k=1,2,⋯,r\) ,
\(μ\) = the overall mean,
\(α_i\) = the effect of the \(i^{th}\) level of A,
\(β_{j(i)}\) = the effect of the \(^{jth}\) level of B within the \(i^{th}\) level of A,
\(ϵ_{k(ij)}\) = random error associated
with the \(k^{th}\) unit in the \(j^{th}\) level of B within the \(i^{th}\) level of A.
\(E[MSA]=br\sigma^2_\alpha+r\sigma^2_{\beta(\alpha)}+\sigma^2_\epsilon\)
\(E[MSB(A)]=r\sigma^2_{\beta(\alpha)}+\sigma^2_\epsilon\) \(E[MSE]=\sigma^2_\epsilon\)
problem1 <- read.csv("problem1.csv")
problem1$Machine <- as.factor(problem1$Machine)
problem1$Operator <- as.factor(problem1$Operator)
answer <- aov(Specimen ~ Machine/Operator, data= problem1)
knitr::kable(anova(answer))| Df | Sum Sq | Mean Sq | F value | Pr(>F) | |
|---|---|---|---|---|---|
| Machine | 3 | 3617.667 | 1205.8889 | 14.270874 | 0.0002910 |
| Machine:Operator | 8 | 2817.667 | 352.2083 | 4.168146 | 0.0134083 |
| Residuals | 12 | 1014.000 | 84.5000 | NA | NA |
The Revised ANOVA.
| Source of Variation | df | SS | MSE | F | p-value |
| Machine | 3 | 3617.667 | 1205.8889 | 3.424 | 0.05253732 |
| Machine:Operator | 8 | 2817.667 | 352.2083 | 4.170 | 0.01338585 |
| Residual | 12 | 1014.000 | 84.5000 | ||
| Total | 23 | 7449.334 |
Effects of A:
\(H_0:\sigma^2_\alpha=0\)
\(H_1:\sigma^2_\alpha>0\)
\(\alpha=0.05\)
Test Statistics: \(F=\frac{MSA}{MSB(A)}=\frac{1205.8889}{352.2083}\approx3.424\)
Decision Rule: We reject \(H_0\) if
p-value is less than \(\alpha\), and
otherwise.
Decision: We cannot reject the null since our
p-value=0.05253732>0.05=\(alpha\).
Conclusion: At the 5% level of significance, there is no significant
difference on the surface finish of metal parts among the machines.
Effects of B:
\(H_0:\sigma^2_{\beta(\alpha)}=0\)
\(H_1:\sigma^2_{\beta(\alpha)}>0\)
\(\alpha=0.05\)
Test Statistic: \(F=\frac{MSB(A)}{MSE}=\frac{352.2083}{84.5}\approx4.170\)
Decision Rule: We reject \(H_0\) if
p-value is less than \(\alpha\), and
otherwise.
Decision: We will reject the null since our
p-value=0.01338585<0.05=\(alpha\).
Conclusion: At the 5% level of significance, there is a significant
difference on the surface finish of metal parts among operators within
the selected machines.
emmeans::emmeans(answer, pairwise ~ Operator | Machine, adjust = "bonferroni")## NOTE: A nesting structure was detected in the fitted model:
## Operator %in% Machine## $emmeans
## Machine = M1:
## Operator emmean SE df lower.CL upper.CL
## O1 70.5 6.5 12 56.3 84.7
## O2 84.0 6.5 12 69.8 98.2
## O3 51.5 6.5 12 37.3 65.7
##
## Machine = M2:
## Operator emmean SE df lower.CL upper.CL
## O4 95.5 6.5 12 81.3 109.7
## O5 82.0 6.5 12 67.8 96.2
## O6 72.0 6.5 12 57.8 86.2
##
## Machine = M3:
## Operator emmean SE df lower.CL upper.CL
## O7 81.5 6.5 12 67.3 95.7
## O8 54.5 6.5 12 40.3 68.7
## O9 51.5 6.5 12 37.3 65.7
##
## Machine = M4:
## Operator emmean SE df lower.CL upper.CL
## O10 44.5 6.5 12 30.3 58.7
## O11 48.0 6.5 12 33.8 62.2
## O12 54.5 6.5 12 40.3 68.7
##
## Confidence level used: 0.95
##
## $contrasts
## Machine = M1:
## contrast estimate SE df t.ratio p.value
## O1 - O2 -13.5 9.19 12 -1.469 0.5030
## O1 - O3 19.0 9.19 12 2.067 0.1831
## O2 - O3 32.5 9.19 12 3.536 0.0123
##
## Machine = M2:
## contrast estimate SE df t.ratio p.value
## O4 - O5 13.5 9.19 12 1.469 0.5030
## O4 - O6 23.5 9.19 12 2.556 0.0755
## O5 - O6 10.0 9.19 12 1.088 0.8941
##
## Machine = M3:
## contrast estimate SE df t.ratio p.value
## O7 - O8 27.0 9.19 12 2.937 0.0373
## O7 - O9 30.0 9.19 12 3.264 0.0204
## O8 - O9 3.0 9.19 12 0.326 1.0000
##
## Machine = M4:
## contrast estimate SE df t.ratio p.value
## O10 - O11 -3.5 9.19 12 -0.381 1.0000
## O10 - O12 -10.0 9.19 12 -1.088 0.8941
## O11 - O12 -6.5 9.19 12 -0.707 1.0000
##
## P value adjustment: bonferroni method for 3 tests| Machine | Operator | Means |
| Machine 1 | Operator1 | \(70.5^{ab}\) |
| Operator2 | \(84.0^a\) | |
| Operator3 | \(51.5^b\) | |
| Machine 2 | Operator4 | \(95.5^a\) |
| Operator5 | \(82.0^{ab}\) | |
| Operator6 | \(72.0^b\) | |
| Machine 3 | Operator7 | \(81.5^a\) |
| Operator8 | \(54.5^b\) | |
| Operator9 | \(51.5^b\) | |
| Machine 4 | Operator10 | \(44.5^a\) |
| Operator11 | \(48.0^a\) | |
| Operator12 | \(54.5^a\) |
\(MSE=84.50\)
\(MSA=brσ^2_α+rσ^2_{β(α)}+σ^2_ϵ\)
\(MSB(A)=rσ^2_{β(α)}+σ^2_ϵ\)
\(a=4,b=9,r=2\)
\(MSB(A)=rσ^2_{β(α)}+σ^2_ϵ\)
\(352.21=(2)σ^2_{β(α)}+84.50\)
\(σ^2_{β(α)}=352.21−84.502\)
\(σ^2_{β(α)}=133.855\)
\(MSA=brσ^2_α+rσ^2_{β(α)}+σ^2_ϵ\)
\(MSA=brσ^2_α+MSB(A)\)
\(1205.89=9(2)σ^2_α+352.21\)
\(σ^2_α=1205.89−352.2118\)
\(σ^2_α=47.42667\)
hence we have an estimated variance component of:
\(V(Y_{ijk})=σ^2_{β(α)}+σ^2_α+σ^2_ϵ\)
\(V(Y_{ijk})=133.855+47.42667+84.50\)
\(V(Y_{ijk})=265.7817\)
The estimated variance component is 265.7817.
An experimenter is designing an experiment in which she plans to compare nine different formulations of a meat product. One factor, F, is percent fat (10%, 15%, 20%) in the meat. The other factor, C, is cooking method (broil, bake, fry). She will prepare samples of each of the nine combinations and present them to tasters who will score the samples based on various criteria. Four tasters are available for the study. Each taster will taste nine samples. There are taster-to-taster differences, but the order in which the samples are tasted will not influence the taste scores. The samples will be prepared in the following manner so that the meat samples can be prepared and kept warm for the tasters. A portion of meat containing 10% fat will be divided into three equal portions. Each of the three methods of cooking will then be randomly assigned to one of the three portions. This procedure will be repeated for meat samples having 15% and 20% fat. The nine meat samples will then be tasted and scored by the taster. The whole process is repeated for the other three tasters. The taste scores (0 to 100) are given in the following table.
The design structure of the experiment uses a split-plot design. It includes two factors wherein factor F (percent of fat in the meat) is the main plot factor and factor C (cooking method) is the subplot factor. However, the treatment structure comes at which the levels of factor F were randomized over the main plots and after which, the levels of factor C were also randomized over the subplots within each main plot.
problem2 <- read.csv("problem2.csv")
library(ExpDes)
with(problem2,split2.rbd(fat,Cooking,Taster,Score,
quali= c(TRUE,TRUE),
mcomp = "tukey",
fac.names = c("Fat Percentage","Cooking Method"),
sigT = 0.01,
sigF = 0.01,
unfold = NULL))## ------------------------------------------------------------------------
## Legend:
## FACTOR 1 (plot): Fat Percentage
## FACTOR 2 (split-plot): Cooking Method
## ------------------------------------------------------------------------
##
## ------------------------------------------------------------------------
## Analysis of Variance Table
## ------------------------------------------------------------------------
## DF SS MS Fc Pr(>Fc)
## Fat Percentage 2 804.39 402.19 31.822 0.000639 ***
## Block 3 539.00 179.67 14.215 0.003910 **
## Error a 6 75.83 12.64
## Cooking Method 2 369.06 184.53 27.077 4e-06 ***
## Fat Percentage*Cooking Method 4 4.94 1.24 0.181 0.945068
## Error b 18 122.67 6.81
## Total 35 1915.89
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## ------------------------------------------------------------------------
## CV 1 = 4.44699 %
## CV 2 = 3.265418 %
##
## No significant interaction: analyzing the simple effects
## ------------------------------------------------------------------------
## Fat Percentage
## Tukey's test
## ------------------------------------------------------------------------
## Groups Treatments Means
## a 20 86.08333
## b 15 79.16667
## b 10 74.58333
## ------------------------------------------------------------------------
##
## Cooking Method
## Tukey's test
## ------------------------------------------------------------------------
## Groups Treatments Means
## a 3 83.75
## b 2 80.16667
## c 1 75.91667
## ------------------------------------------------------------------------
##
##
##
##
## Significant interaction: analyzing the interaction
## ------------------------------------------------------------------------
##
## Analyzing Fat Percentage inside of each level of Cooking Method
## ------------------------------------------------------------------------
## DF SS MS Fc
## Fat Percentage : Cooking Method 1 2.00000 265.1667 132.583333 15.14170
## Fat Percentage : Cooking Method 2 2.00000 290.6667 145.333333 16.59781
## Fat Percentage : Cooking Method 3 2.00000 253.5000 126.750000 14.47550
## Pooled Error 18.67788 163.5468 8.756173 NA
## p.value
## Fat Percentage : Cooking Method 1 0.000123
## Fat Percentage : Cooking Method 2 0.000072
## Fat Percentage : Cooking Method 3 0.000160
## Pooled Error NA
## ------------------------------------------------------------------------
##
##
## Fat Percentage inside of Cooking Method 1
## ------------------------------------------------------------------------
## Tukey's test
## ------------------------------------------------------------------------
## Groups Treatments Means
## a 20 82.25
## b 15 74.5
## b 10 71
## ------------------------------------------------------------------------
##
## Fat Percentage inside of Cooking Method 2
## ------------------------------------------------------------------------
## Tukey's test
## ------------------------------------------------------------------------
## Groups Treatments Means
## a 20 86.5
## b 15 79.5
## b 10 74.5
## ------------------------------------------------------------------------
##
## Fat Percentage inside of Cooking Method 3
## ------------------------------------------------------------------------
## Tukey's test
## ------------------------------------------------------------------------
## Groups Treatments Means
## a 20 89.5
## ab 15 83.5
## b 10 78.25
## ------------------------------------------------------------------------
##
##
## Analyzing Cooking Method inside of each level of Fat Percentage
## ------------------------------------------------------------------------
## DF SS MS Fc p.value
## Cooking Method : Fat Percentage 10 2 105.1667 52.583333 7.716032 0.003802
## Cooking Method : Fat Percentage 15 2 162.6667 81.333333 11.934782 0.000502
## Cooking Method : Fat Percentage 20 2 106.1667 53.083333 7.789402 0.003655
## Error b 18 122.6667 6.814815 NA NA
## ------------------------------------------------------------------------
##
##
## Cooking Method inside of Fat Percentage 10
## ------------------------------------------------------------------------
## Tukey's test
## ------------------------------------------------------------------------
## Groups Treatments Means
## a 3 78.25
## ab 2 74.5
## b 1 71
## ------------------------------------------------------------------------
## ------------------------------------------------------------------------
##
##
## Cooking Method inside of Fat Percentage 15
## ------------------------------------------------------------------------
## Tukey's test
## ------------------------------------------------------------------------
## Groups Treatments Means
## a 3 83.5
## ab 2 79.5
## b 1 74.5
## ------------------------------------------------------------------------
## ------------------------------------------------------------------------
##
##
## Cooking Method inside of Fat Percentage 20
## ------------------------------------------------------------------------
## Tukey's test
## ------------------------------------------------------------------------
## Groups Treatments Means
## a 3 89.5
## ab 2 86.5
## b 1 82.25
## ------------------------------------------------------------------------
## ------------------------------------------------------------------------Block effects:
\(H_o:ρ_j=0\)
\(H_a:ρ_j>0\)
\(\alpha=0.01\)
Test Statistics:
\(F−value=31.822\)
\(p−value=0.000639\)
Decision on \(H_o\):
We reject the null since, pvalue=0.003910<0.01=\(\alpha\).
Conclusion:
There is significant block effect (p−value<0.01). Blocking strategy
is effective.
Interaction Effects:
\(H_o:(αβ)_{ik}=0\)
\(H_a:(αβ)_{ik}>0\)
\(\alpha=0.01\)
Test Statistics:
\(F−value=0.181\)
\(p−value=0945068\)
Decision on \(H_o\):
We fail to reject the null, since pvalue=0.945068>0.01=\(\alpha\).
Conclusion:
There is no interaction between the Fat Percentage and Cooking
Method.
Fat Percentage Effects:
\(H_o:α_i=0\)
\(H_a:α_i>0\)
\(\alpha=0.01\)
Test Statistics:
\(F−value=31.822\)
\(p−value=0.000639\)
Decision on \(H_o\):
We reject the null, since pval<significance level.
Conclusion:
There is a significant effect on Fat percentage.
Post hoc on Fat Percentage means:
| Fat Percentage | Mean |
| 20% | \(86.083^a\) |
| 15% | \(79.167^b\) |
| 10% | \(74.583^b\) |
Cooking Method Effects:
\(H_o:β_k=0\)
\(H_a:β_k>0\)
\(\alpha=0.01\)
Test Statistics:
\(F−value=27.077\)
\(p−value<0.001\)
Decision on Ho:
We reject the null, since pvalue<significance level.
Conclusion:
There is a significant difference on the cooking method.
post hoc on Cooking Method Effect:
| Cooking Method | Mean |
| Fry | \(83.75^a\) |
| Bake | \(80.167^b\) |
| Boil | \(75.917^c\) |