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时间：2021-03-25

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.

Page

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

Page

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.

Page

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

Page

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.

Page

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

Page

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%.

Page

0Question 29. [1 Mark]

The plot of standardized residuals against fitted values is given as follows

l

l

l

l

l l

l

l

l

l

l

l

l

l

l

l

l

l

l

l

550 560 570 580 590 600 610

−

3

−

2

−

1

0

1

2

3

Standardized Residuals vs Fitted Values

St

an

da

rd

ize

d

Re

sid

ua

ls

Fitted Values

St

an

da

rd

ize

d

Re

sid

ua

ls

(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

l

l

l

l

l

l

l

l

l

l

l

l

l

l

l

l

l

l

l

l

−2 −1 0 1 2

−

1.

5

−

1.

0

−

0.

5

0.

0

0.

5

1.

0

1.

5

Normal Q−Q Plot

Sa

m

pl

e

Qu

an

tile

s

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.

Page

0STAT2401 Analysis of Experiments – Test 1

Answer

1 A B C D

2 A B C D

3 A B C D

4 A B C D

5 A B C D

6 A B C D

7 A B C D

8 A B C D

9 A B C D

10 A B C D

11 A B C D

12 A B C D

13 A B C D

14 A B C D

15 A B C D

16 A B C D

17 A B C D

18 A B C D

19 A B C D

20 A B C D

21 A B C D

22 A B C D

23 A B C D

24 A B C D

25 A B C D

26 A B C D

27 A B C D

28 A B C D

29 A B C D

30 A B C D

Page

学霸联盟

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.

Page

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

Page

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.

Page

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

Page

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.

Page

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

Page

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%.

Page

0Question 29. [1 Mark]

The plot of standardized residuals against fitted values is given as follows

l

l

l

l

l l

l

l

l

l

l

l

l

l

l

l

l

l

l

l

550 560 570 580 590 600 610

−

3

−

2

−

1

0

1

2

3

Standardized Residuals vs Fitted Values

St

an

da

rd

ize

d

Re

sid

ua

ls

Fitted Values

St

an

da

rd

ize

d

Re

sid

ua

ls

(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

l

l

l

l

l

l

l

l

l

l

l

l

l

l

l

l

l

l

l

l

−2 −1 0 1 2

−

1.

5

−

1.

0

−

0.

5

0.

0

0.

5

1.

0

1.

5

Normal Q−Q Plot

Sa

m

pl

e

Qu

an

tile

s

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.

Page

0STAT2401 Analysis of Experiments – Test 1

Answer

1 A B C D

2 A B C D

3 A B C D

4 A B C D

5 A B C D

6 A B C D

7 A B C D

8 A B C D

9 A B C D

10 A B C D

11 A B C D

12 A B C D

13 A B C D

14 A B C D

15 A B C D

16 A B C D

17 A B C D

18 A B C D

19 A B C D

20 A B C D

21 A B C D

22 A B C D

23 A B C D

24 A B C D

25 A B C D

26 A B C D

27 A B C D

28 A B C D

29 A B C D

30 A B C D

Page

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