STAT 495, Fall 2010
Homework Assignment #8
1. A nested design was used to obtain information on possible sources of variability
in the hardness of a metal alloy. There are two alloy chemistries. There are three
heats of the metal within both chemistries. Within each heat, two ingots are
selected. For each ingot the hardness is measured twice. Note that this nested
design has 4 levels. Below are the data. The next page has formulas that extend
what was done in class.
Chemistry
Heats
Ingots
1
2
1
1
40
63
2
27
30
1
95
67
3
2
69
47
1
65
54
2
2
1
2
76
45
1
30
18
2
13
17
1
81
50
3
2
43
32
1
49
66
2
43
50
a) Use JMP to construct the following graphs.



A histogram of all of the data. Use a scale from 0 to 100 with increments
of 20.
A graph that shows the variability at each level.
Construct an s chart.
Interpret what the graphs are telling you. Where does there appear to be large
variation?
b) Go through the hand calculations to construct an analysis of variance table
with sources of variation: Chemistry, Heat, Ingot and Measurement.
c) Give the formulas for the Expected Mean Squares for the components:
Chemistry, Heat, Ingot and Measurement.
d) Compute variance components and express them as a percentage of the total.
e) Use JMP to verify your results in parts b), c) and d).
f) Interpret your results in terms of where improvement efforts should be
focused.
1
Formulas for Nested Designs
Sums of Squares
SSA = dcb(a – 1)(sample variance of the A means)
SSB(A) = dc(b – 1)(sum of the a sample variances of the B(A) means)
SSC(B(A)) = d(c – 1)(sum of the ab sample variances of the C(B(A)) means)
SSD(C(B(A))) = (d – 1)(sum of the abc sample variances of the D(C(B(A))) means)
SSTotal = (abcd – 1)(grand sample variance of all abcd data values)
Analysis of Variance
Source
A
Sum
Squares
SSA
B(A)
SSB(A)
a(b–1)
SSB(A)/a(b–1)
 2  d C2  cd B2
C(B(A))
SSC(B(A))
ab(c–1)
SSC(B(A))/ab(c–1)
 2  d C2
D(C(B(A))) SSD(C(B(A)))
abc(d–1)
SSD(C(B(A)))/abc(d–1)
2
Total
abcd – 1
SSTotal
of d.f.
Mean Square
Expected Mean Square
a–1
SSA/(a–1)
2
 2  d C2  cd B2  bcd A
2