3.28. A semiconductor manufacturer has developed three different methods for red
ID: 2921341 • Letter: 3
Question
3.28. A semiconductor manufacturer has developed three different methods for reducing particle counts on wafers. All three methods are tested on six different wafers and the after treatment particle count obtained. The data are shown below: Method Count 4 1 11 24 30 35 43 27 120 97 6880 10 31 62 4 53 21 2 (a) Do all methods have the same effect on mean particle count? (b) Plot the residuals versus the predicted response. Construct a normal probability plot of the residuals. re there potential concerns about the validity of the e vaidity of the assumptions? ased on your answer to part (b) conduct another analysis of the particle count data and draw appropri- ate conclusions.Explanation / Answer
Answer:
a).
MINITAB used
One-way ANOVA: Method1, Method2, Method3
Method
Null hypothesis
All means are equal
Alternative hypothesis
Not all means are equal
Significance level
= 0.05
Equal variances were assumed for the analysis.
Factor Information
Factor
Levels
Values
Factor
3
Method1, Method2, Method3
Analysis of Variance
Source
DF
Adj SS
Adj MS
F-Value
P-Value
Factor
2
11308
5654.1
12.36
0.001
Error
15
6861
457.4
Total
17
18169
Model Summary
S
R-sq
R-sq(adj)
R-sq(pred)
21.3867
62.24%
57.20%
45.62%
Means
Factor
N
Mean
StDev
95% CI
Method1
6
13.00
11.19
(-5.61, 31.61)
Method2
6
39.00
13.18
(20.39, 57.61)
Method3
6
74.2
32.8
(55.6, 92.8)
Pooled StDev = 21.3867
Calculated F=12.36, P=0.001 which is < 0.05 level. Ho is rejected. We conclude that the three methods are not equal.
b).
The plots shows that assumptions for ANOVA is not met.
c).
Nonparametric method, Kruskal-Wallis Test used.
Kruskal-Wallis Test: data versus method
Descriptive Statistics
method
N
Median
Mean Rank
Z-Value
Method1
6
10.5
4.0
-3.09
Method2
6
37.5
10.2
0.37
Method3
6
74.0
14.3
2.72
Overall
18
9.5
Test
Null hypothesis
H: All medians are equal
Alternative hypothesis
H: At least one median is different
DF
H-Value
P-Value
2
11.38
0.003
Calculated H=11.38, P=0.003 which is < 0.05 level. Ho is rejected. We conclude that the three methods are not equal.
Null hypothesis
All means are equal
Alternative hypothesis
Not all means are equal
Significance level
= 0.05
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