0STAT2401 Analysis of Experiments – Test 1
Solutions
In many integrated circuit manufacturing steps, wafers are completely coated with a layer of ma-
terial such as silicon dioxide or a metal. The unwanted material is then selectively removed by
etching through a mask, thereby creating circuit patterns, electrical interconnects, and areas in
which diffusions or metal depositions are to be made. A plasma etching process is widely used for
this operation, particularly in small geometry applications. Energy is supplied by a radio frequency
(RF) generator causing plasma to be generated in the gap between the electrodes. The chemical
species in the plasma are determined by the particular gases used. Fluorocarbons, such as CF4
(tetrafluoromethane) or C2F6 (hexafluoroethane), are often used in plasma etching, but other gases
and mixtures of gases are relatively common, depending on the application.
An engineer is interested in investigating the relationship between the RF power setting and the
etch rate for this tool. The objective of an experiment like this is to model the relationship between
etch rate and RF power, and to specify the power setting that will give a desired target etch rate.
She is interested in a particular gas (C2F6) and gap (0.80 cm) and wants to test four levels of RF
power: 160, 180, 200, and 220 W. She decided to test five wafers at each level of RF power. She
has run a completely randomized experiment with four levels of RF power and five replicates. For
convenience, we give the summary of the data:
RF Power (W) Etch rate (A˚/min) sample mean sample variance size
160 y11 y12 y13 y14 y15 y¯1· = 549.4 s21 = 80.3 n1 = 5
180 y21 y22 y23 y24 y25 y¯2· = 588.0 s22 = 100.5 n2 = 5
200 y31 y32 y33 y34 y35 y¯3· = 602.0 s23 = 50.5 n3 = 5
220 y41 y42 y43 y44 y45 y¯4· = 613.0 s24 = 77.5 n4 = 5
We will use the Analysis of Variance (ANOVA) to test
H0 : µ1 = µ2 = µ3 = µ4
against the alternative H1: some means are different. Here we assume that
Etch rate (A˚/min)
RF Power (W) treatment mean variance
160 µ1 σ
2
180 µ2 σ
2
200 µ3 σ
2
220 µ4 σ
2
The ANOVA table for the test is summarized as follows:
Df Sum Sq Mean Sq F value Pr(>F)
RF Power 3 B C 49.89 2.42e-08
Residuals 16 1235.20 A
Total D 12789.80
0Question 1. [1 Mark]
The F -statistic or F -ratio for the test is
(a) Fobs = 1235.20,
(b) Fobs = 49.89, X
(c) Fobs = 12789.80, or
(d) Fobs = 2.42e-08.
There are two usual approaches to make the decision, p-value approach and critical value approach.
Let’s take the significance level be α = 5%.
Question 2. [1 Mark]
The critical value for the test
(a) F3,16,α = qf(1-0.05,3,16) = 3.24, X
(b) F16,3,α/2 = qf(1-0.05/2,16,3) = 14.23,
(c) F16,3,α = qf(1-0.05,16,3) = 8.69, or
(d) F3,16,α/2 = qf(1-0.05/2,3,16) = 4.08.
Question 3. [1 Mark]
So we reject H0 at the (α =)5% significance level because
(a) Fobs < Fv1,v2,α,
(b) Fobs > Fv1,v2,α, X
(c) Fobs > Fv1,v2,α/2, or
(d) Fobs < Fv1,v2,α/2,
where (v1, v2) are the degrees of freedom determined in Question 2, and the values Fobs and
Fv1,v2,α are determined in Question 1 and Question 2 respectively.
Question 4. [1 Mark]
or we reject H0 at the (α =)5% significance level because
(a) p-value = 12789.80 > 0.05 = α,
(b) p-value = 2.42e-08 < 0.05 = α, X
(c) p-value = 1235.20 > 0.05 = α, or
(d) p-value = 49.89 > 0.05 = α.
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0Question 5. [1 Mark]
The conclusion is that the small p-value is strong evidence that over the the RF power setting,
(a) the 4 treatment means of the etch rates are not all different,
(b) the 4 treatment means of the etch rates are not all the same, X
(c) the 4 treatment means of the etch rates are all the same, or
(d) the 4 treatment means of the etch rates are all different.
We are now going to identify the missing values in the ANOVA table
Question 6. [1 Mark]
(a) A = 12789.80/16 = 799.36,
(b) A = 1235.20× 16 = 19763.20,
(c) A = 12789.80× 16 = 204636.80, or
(d) A = 1235.20/16 = 77.20. X
Question 7. [1 Mark]
(a) B = 1235.20/16 = 77.20,
(b) B = 12789.80/16 = 799.36,
(c) B = 12789.80 + 1235.20 = 14025.00, or
(d) B = 12789.80− 1235.20 = 11554.60. X
Question 8. [1 Mark]
(a) C = (12789.80 + 1235.20)/3 = 4675.00,
(b) C = (12789.80 + 1235.20)/16 = 876.56,
(c) C = (12789.80− 1235.20)/16 = 722.16, or
(d) C = (12789.80− 1235.20)/3 = 3851.53. X
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0Question 9. [1 Mark]
(a) D = 16 + 3 = 19, X
(b) D = 16− 3− 1 = 12,
(c) D = 16 + 3− 1 = 18, or
(d) D = 16− 3 = 13.
Another point of view for this testing problem is to compare the following two models. (1) Under
H0, we have the model Yij = µ+ ij where ij are independent and identically normal distributed
random variables with mean 0 and variance σ2. (2) Under H1, we have the model Yij = µi + ij
where ij are independent and identically normal distributed random variables with mean 0 and
variance σ2.
Question 10. [1 Mark]
The model under H0 is called
(a) Random Model,
(b) Reduced Model, X
(c) Full Model, or
(d) Normal Model.
Question 11. [1 Mark]
The model under H1 is called
(a) Full Model, X
(b) Reduced Model,
(c) Random Model, or
(d) Normal Model.
Question 12. [1 Mark]
Consider the model under H0, the estimate of the parameter µ is given by
(a) 5×549.4+5×588.0+5×602.0+5×613.0
5+5+5+5 = 588.10, X
(b) 80.3×549.4+100.5×588.0+50.5×602.0+77.5×613.0
549.4+588.0+602.0+613.0 = 76.99,
(c) 80.3×549.4+100.5×588.0+50.5×602.0+77.5×613.0
80.3+100.5+50.5+77.5 = 586.53, or
(d) 80.3×5+100.5×5+50.5×5+77.5×5
5+5+5+5 = 77.20.
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0Question 13. [1 Mark]
Consider the model under H0, the estimate of the parameter σ
2 is equal to
(a) C,
(b) B,
(c) A, or
(d) 12789.80/D, X
where A, B, C, D are the values from the ANOVA table.
Question 14. [1 Mark]
Consider the model under H0, the Residual Sum of Squares is
(a) 3× 80.3 + 3× 100.5 + 3× 50.5 + 3× 77.5 = 926.40,
(b) 4× 80.3 + 4× 100.5 + 4× 50.5 + 4× 77.5 = 1235.20,
(c) 12789.80 (from the ANOVA table), or X
(d) 5× 80.3 + 5× 100.5 + 5× 50.5 + 5× 77.5 = 1544.00.
Question 15. [1 Mark]
Consider the model under H1, the estimate of the parameter µ1 is given by
(a) 549.4, X
(b) 5×549.4+5×588.0+5×602.0+5×613.0
5+5+5+5 = 588.10,
(c) 588.0, or
(d) 4×549.4+4×588.0+4×602.0+4×613.0
5+5+5+5 = 470.48.
Question 16. [1 Mark]
Consider the model under H1, the estimate of the parameter µ2 is given by
(a) 4×549.4+4×588.0+4×602.0+4×613.0
5+5+5+5 = 470.48,
(b) 5×549.4+5×588.0+5×602.0+5×613.0
5+5+5+5 = 588.10,
(c) 549.4, or
(d) 588.0. X
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0Question 17. [1 Mark]
Consider the model under H1, the estimate of the parameter σ
2 is given by
(a) 80.3×5+100.5×5+50.5×5+77.5×5
6+6+6+6 = 64.33,
(b) 80.3×5+100.5×5+50.5×5+77.5×5
4+4+4+4 = 96.50,
(c) 80.3×4+100.5×4+50.5×4+77.5×4
5+5+5+5 = 61.76, or
(d) 80.3×4+100.5×4+50.5×4+77.5×4
4+4+4+4 = 77.20. X
Question 18. [1 Mark]
Consider the model under H1, the estimate of the parameter σ
2 is also equal to
(a) 12789.80/D,
(b) B,
(c) A, or X
(d) C,
where A, B, C, D are the values from the ANOVA table.
Question 19. [1 Mark]
Consider the model under H1, the Residual Sum of Squares is
(a) 6× 80.3 + 6× 100.5 + 6× 50.5 + 6× 77.5 = 1852.80,
(b) 4× 80.3 + 4× 100.5 + 4× 50.5 + 4× 77.5 = 1235.20, X
(c) 5× 80.3 + 5× 100.5 + 5× 50.5 + 5× 77.5 = 1544.00, or
(d) 12789.80 (from the ANOVA table).
Question 20. [1 Mark]
The Extra Sum of Squares between the models under H1 and H0 is
(a) 1235.20,
(b) 12789.80− 1235.20 = 11554.60, X
(c) 12789.80, or
(d) 1235.20− 12789.80 = -11554.60.
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0Question 21. [1 Mark]
We prefer the model under H1 if the Extra Sum of Squares is greater than
(a) 16× qf(1-0.05,3,16)× 1235.20
3
= 21336.82,
(b) 3× qf(1-0.05,16,3)× 1235.20
16
= 2013.13,
(c) 3× qf(1-0.05,3,16)× 1235.20
16
= 750.12, or X
(d) 16× qf(1-0.05,16,3)× 1235.20
3
= 57262.46.
We are going to perform diagnostic checking for the ANOVA model
Question 22. [1 Mark]
The residual for an observation is given by
(a) fitted value - observation,
(b) fitted value - standardized residual,
(c) observation - standardized residual, or
(d) observation - fitted value. X
Question 23. [1 Mark]
An overall measure of how well the model fits the data could be
(a) The Total Sum of Squares,
(b) The Factor Sum of Squares,
(c) The Between Group Sum of Squares, or
(d) The Residual Sum of Squares. X
Question 24. [1 Mark]
To judge whether the normality assumption is satisfied, we examine
(a) the historgam and QQ-plot of the fitted values,
(b) the historgam and QQ-plot of the raw data,
(c) the historgam and QQ-plot of the estimates, or
(d) the historgam and QQ-plot of the standardized residuals. X
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0Question 25. [1 Mark]
To judge whether the constant variance assumption is satisftied, we could examine
(a) the plot of the observations against the fitted values,
(b) the plot of the standardized residuals against the fitted values, X
(c) the plot of the fitted values against the observations, or
(d) the plot of the standardized residuals against the residuals.
Question 26. [1 Mark]
The standardized residuals are
(a) independent,
(b) linearly independent,
(c) correlated, or X
(d) uncorrelated.
Question 27. [1 Mark]
If a standardized residual is less than −2 or greater than 2, then the corresponding observation
is called
(a) satisfy the normality assumption,
(b) satisfy the constant variance assumption,
(c) satisfy the independent assumption, or
(d) an unusual observation. X
Question 28. [1 Mark]
Apporiximately, a standard normal random variable is less than −2 or greater than 2 with
probability
(a) 97.5%,
(b) 95%, X
(c) 99.7%, or
(d) 68%.
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0Question 29. [1 Mark]
The plot of standardized residuals against fitted values is given as follows
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(a) There are two unusual observations,
(b) There are three unusual observations,
(c) There is one unusual observation, or
(d) There is no unusual observation. X
Question 30. [1 Mark]
The QQ-plot of the standardized residuals is given as follows
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Normal Q−Q Plot
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We could determine whether
(a) the constant variance assumption is satisfied,
(b) the normality assumption is satisfied, X
(c) the independence assumption is satisfied, or
(d) the constant variance assumption is NOT satisfied.
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0STAT2401 Analysis of Experiments – Test 1
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