STAT:5201 Design: Exam 3 22S:158/165 Design: Exam 3 April 30, 2008 SolutionsShow your work. Calculator and 2 pages of crib notes are allowed. Some statistical tables are included at the end of the exam.There may be more than one correct answer to certain questions; if so, you may choose the method you want to use, and make itclear what you are doing.

1. An experiment was conducted in a factory to examine the sources of variation in the product. Theexperiment involves 3 work shifts, 3 operators in each shift, and 4 suppliers of raw material. The expectedand observed mean squares from the experiment are shown in the table below.

shift3

|3OPER9

⇥ SUPPL4 Source df Expected MS Observed MS

shift 2 12Q(s) + 4�2O(s) + 3�2

sS + �2 555.00

OPER(shift) 6 4�2O(s) + �2 333.00

SUPPL 3 9�2S + 3�2

sS + �2 777.00

shift*SUPPL 6 3�2sS + �2 222.00

RESIDUAL 18 �2 99.00

10 pts A. Calculate an F statistic for testing the SUPPL effect: H0 : �2S = 0 versus H1 : �2

S > 0, and its degrees offreedom. At a significance level of E = .05, do you reject, or fail to reject, H0?

If H0 is true, EMS(SUPPL) = 3�2sS + �2 = EMS(shift ⇤ SUPPL). Thus, we should use the latter as the

denominator.F =

MS(SUPPL)MS(shift ⇤ SUPPL)

=777222

= 3.5

with (3, 6) df. The critical value is F.05,3,6 = 4.76. The observed F does not exceed this, so we fail to rejectthe null.

10 pts B. Calculate an estimate of the variance component �2sS due to shift*SUPPL. (You do not need to find a

confidence interval for this.)

Note that E{MS(shift ⇤ SUPPL)� MS(RESIDUAL)} = 3�2sS; thus, and unbiased estimate of the required

variance component is

�2sS =

MS(shift ⇤ SUPPL)� MS(RESIDUAL)3

=222� 99

3= 41

10 pts C. We wish to compare the means for shifts 1 and 2 (averaging over operators and suppliers). Give anexpression for Var(y1·· � y2··) in terms of �2

S , �2O(s), �2

sS, and �2. (Note: The RESIDUAL term is the same asSUPPL*OPER(shift).)

Using the dot notation in the handout, the means involve random variations due to SUPPL•,OPER•(shift), shift*SUPPL•, and SUPPL•*OPER•(shift). Accordingly,

Var(yi··) = �2S/4 + �2

O(s)/3 + �2sS/4 + �2/12

Var(y1·· � y2··) = 2

��2

O(s)

3+

�2sS4

+�2

12

⇥

(The �2S term goes away because the SUPPL• terms cancel out.)

10 pts D. Use the results of the preceding question and the observed mean squares to calculate an estimate of thestandard error and the LSD (at E = .05) for this comparison of means.

Using a common denominator and the expected mean squares,

Var(y1·· � y2··) =4�2

O(s) + 3�2sS + �2

6⇤=

MSO(s) + MSsS � MSE

6=

333 + 222� 996

= 76

22S:158/165 Design: Exam 3 April 30, 2008 SolutionsShow your work. Calculator and 2 pages of crib notes are allowed. Some statistical tables are included at the end of the exam.There may be more than one correct answer to certain questions; if so, you may choose the method you want to use, and make itclear what you are doing.

1. An experiment was conducted in a factory to examine the sources of variation in the product. Theexperiment involves 3 work shifts, 3 operators in each shift, and 4 suppliers of raw material. The expectedand observed mean squares from the experiment are shown in the table below.

shift3

|3OPER9

⇥ SUPPL4 Source df Expected MS Observed MS

shift 2 12Q(s) + 4�2O(s) + 3�2

sS + �2 555.00

OPER(shift) 6 4�2O(s) + �2 333.00

SUPPL 3 9�2S + 3�2

sS + �2 777.00

shift*SUPPL 6 3�2sS + �2 222.00

RESIDUAL 18 �2 99.00

10 pts A. Calculate an F statistic for testing the SUPPL effect: H0 : �2S = 0 versus H1 : �2

S > 0, and its degrees offreedom. At a significance level of E = .05, do you reject, or fail to reject, H0?

If H0 is true, EMS(SUPPL) = 3�2sS + �2 = EMS(shift ⇤ SUPPL). Thus, we should use the latter as the

denominator.F =

MS(SUPPL)MS(shift ⇤ SUPPL)

=777222

= 3.5

with (3, 6) df. The critical value is F.05,3,6 = 4.76. The observed F does not exceed this, so we fail to rejectthe null.

10 pts B. Calculate an estimate of the variance component �2sS due to shift*SUPPL. (You do not need to find a

confidence interval for this.)

Note that E{MS(shift ⇤ SUPPL)� MS(RESIDUAL)} = 3�2sS; thus, and unbiased estimate of the required

variance component is

�2sS =

MS(shift ⇤ SUPPL)� MS(RESIDUAL)3

=222� 99

3= 41

10 pts C. We wish to compare the means for shifts 1 and 2 (averaging over operators and suppliers). Give anexpression for Var(y1·· � y2··) in terms of �2

S , �2O(s), �2

sS, and �2. (Note: The RESIDUAL term is the same asSUPPL*OPER(shift).)

Using the dot notation in the handout, the means involve random variations due to SUPPL•,OPER•(shift), shift*SUPPL•, and SUPPL•*OPER•(shift). Accordingly,

Var(yi··) = �2S/4 + �2

O(s)/3 + �2sS/4 + �2/12

Var(y1·· � y2··) = 2

��2

O(s)

3+

�2sS4

+�2

12

⇥

(The �2S term goes away because the SUPPL• terms cancel out.)

10 pts D. Use the results of the preceding question and the observed mean squares to calculate an estimate of thestandard error and the LSD (at E = .05) for this comparison of means.

Using a common denominator and the expected mean squares,

Var(y1·· � y2··) =4�2

O(s) + 3�2sS + �2

6⇤=

MSO(s) + MSsS � MSE

6=

333 + 222� 996

= 76

22S:158/165 Design: Exam 3 April 30, 2008 SolutionsShow your work. Calculator and 2 pages of crib notes are allowed. Some statistical tables are included at the end of the exam.There may be more than one correct answer to certain questions; if so, you may choose the method you want to use, and make itclear what you are doing.

1. An experiment was conducted in a factory to examine the sources of variation in the product. Theexperiment involves 3 work shifts, 3 operators in each shift, and 4 suppliers of raw material. The expectedand observed mean squares from the experiment are shown in the table below.

shift3

|3OPER9

⇥ SUPPL4 Source df Expected MS Observed MS

shift 2 12Q(s) + 4�2O(s) + 3�2

sS + �2 555.00

OPER(shift) 6 4�2O(s) + �2 333.00

SUPPL 3 9�2S + 3�2

sS + �2 777.00

shift*SUPPL 6 3�2sS + �2 222.00

RESIDUAL 18 �2 99.00

10 pts A. Calculate an F statistic for testing the SUPPL effect: H0 : �2S = 0 versus H1 : �2

S > 0, and its degrees offreedom. At a significance level of E = .05, do you reject, or fail to reject, H0?

If H0 is true, EMS(SUPPL) = 3�2sS + �2 = EMS(shift ⇤ SUPPL). Thus, we should use the latter as the

denominator.F =

MS(SUPPL)MS(shift ⇤ SUPPL)

=777222

= 3.5

with (3, 6) df. The critical value is F.05,3,6 = 4.76. The observed F does not exceed this, so we fail to rejectthe null.

10 pts B. Calculate an estimate of the variance component �2sS due to shift*SUPPL. (You do not need to find a

confidence interval for this.)

Note that E{MS(shift ⇤ SUPPL)� MS(RESIDUAL)} = 3�2sS; thus, and unbiased estimate of the required

variance component is

�2sS =

MS(shift ⇤ SUPPL)� MS(RESIDUAL)3

=222� 99

3= 41

10 pts C. We wish to compare the means for shifts 1 and 2 (averaging over operators and suppliers). Give anexpression for Var(y1·· � y2··) in terms of �2

S , �2O(s), �2

sS, and �2. (Note: The RESIDUAL term is the same asSUPPL*OPER(shift).)

Using the dot notation in the handout, the means involve random variations due to SUPPL•,OPER•(shift), shift*SUPPL•, and SUPPL•*OPER•(shift). Accordingly,

Var(yi··) = �2S/4 + �2

O(s)/3 + �2sS/4 + �2/12

Var(y1·· � y2··) = 2

��2

O(s)

3+

�2sS4

+�2

12

⇥

(The �2S term goes away because the SUPPL• terms cancel out.)

10 pts D. Use the results of the preceding question and the observed mean squares to calculate an estimate of thestandard error and the LSD (at E = .05) for this comparison of means.

Using a common denominator and the expected mean squares,

Var(y1·· � y2··) =4�2

O(s) + 3�2sS + �2

6⇤=

MSO(s) + MSsS � MSE

6=

333 + 222� 996

= 76

22S:158/165 Design: Exam 3 April 30, 2008 SolutionsShow your work. Calculator and 2 pages of crib notes are allowed. Some statistical tables are included at the end of the exam.There may be more than one correct answer to certain questions; if so, you may choose the method you want to use, and make itclear what you are doing.

1. An experiment was conducted in a factory to examine the sources of variation in the product. Theexperiment involves 3 work shifts, 3 operators in each shift, and 4 suppliers of raw material. The expectedand observed mean squares from the experiment are shown in the table below.

shift3

|3OPER9

⇥ SUPPL4 Source df Expected MS Observed MS

shift 2 12Q(s) + 4�2O(s) + 3�2

sS + �2 555.00

OPER(shift) 6 4�2O(s) + �2 333.00

SUPPL 3 9�2S + 3�2

sS + �2 777.00

shift*SUPPL 6 3�2sS + �2 222.00

RESIDUAL 18 �2 99.00

10 pts A. Calculate an F statistic for testing the SUPPL effect: H0 : �2S = 0 versus H1 : �2

S > 0, and its degrees offreedom. At a significance level of E = .05, do you reject, or fail to reject, H0?

If H0 is true, EMS(SUPPL) = 3�2sS + �2 = EMS(shift ⇤ SUPPL). Thus, we should use the latter as the

denominator.F =

MS(SUPPL)MS(shift ⇤ SUPPL)

=777222

= 3.5

with (3, 6) df. The critical value is F.05,3,6 = 4.76. The observed F does not exceed this, so we fail to rejectthe null.

10 pts B. Calculate an estimate of the variance component �2sS due to shift*SUPPL. (You do not need to find a

confidence interval for this.)

Note that E{MS(shift ⇤ SUPPL)� MS(RESIDUAL)} = 3�2sS; thus, and unbiased estimate of the required

variance component is

�2sS =

MS(shift ⇤ SUPPL)� MS(RESIDUAL)3

=222� 99

3= 41

10 pts C. We wish to compare the means for shifts 1 and 2 (averaging over operators and suppliers). Give anexpression for Var(y1·· � y2··) in terms of �2

S , �2O(s), �2

sS, and �2. (Note: The RESIDUAL term is the same asSUPPL*OPER(shift).)

Using the dot notation in the handout, the means involve random variations due to SUPPL•,OPER•(shift), shift*SUPPL•, and SUPPL•*OPER•(shift). Accordingly,

Var(yi··) = �2S/4 + �2

O(s)/3 + �2sS/4 + �2/12

Var(y1·· � y2··) = 2

��2

O(s)

3+

�2sS4

+�2

12

⇥

(The �2S term goes away because the SUPPL• terms cancel out.)

10 pts D. Use the results of the preceding question and the observed mean squares to calculate an estimate of thestandard error and the LSD (at E = .05) for this comparison of means.

Using a common denominator and the expected mean squares,

Var(y1·· � y2··) =4�2

O(s) + 3�2sS + �2

6⇤=

MSO(s) + MSsS � MSE

6=

333 + 222� 996

= 76

22S:158/165 Design: Exam 3 2

The degrees of freedom are

d f =(333 + 222� 99)2

3332/6 + 2222/6 + 992/18=

207, 93627, 240

= 7.63

Interpolating in the t tables, the critical value is t.025,7.6 ⌅ 2.33, so

LSD = 2.33⇧

76 = 20.31

An answer based on the nearest d.f. in the table, t.025,8 = 2.31, is acceptable.

2. A consumer organization conducts an experiment where subjects are asked to rate chicken pot pies in a blindtasting. The factors in the experiment are SUBJ (there are 13 subjects), brand (2 brands of pot pie), and lite (2levels, regular and low-calorie). Each subject tastes all four combinations of brand and lite, in random order.

10 pts A. Draw a diagram of this experiment.

These factors are all crossed:SUBJ13 ⇥ brand2 ⇥ lite2

15 pts B. Give a table showing the model terms, degrees of freedom, and expected mean squares. You must beconsistent with the diagram. Include all interactions except the one that corresponds to individualmeasurements.

Source df Expected MS

SUBJ 12 4�2S + 2�2

Sb + 2�2Sl + �2

brand 1 26Q(b) + 2�2Sb + �2

lite 1 26Q(l) + 2�2Sl + �2

brand ⇤ lite 1 13Q(bl) + �2

SUBJ ⇤ brand 12 2�2Sb + �2

SUBJ ⇤ lite 12 2�2Sl + �2

RESID = S ⇤ b ⇤ l 12 �2

10 pts C. Explain why this is a block experiment, and state which factor(s) are blocks. How would it berandomized differently if it were not a block experiment?

It is a block experiment (with SUBJ as the blocking factor) because a separate randomization of theorder of the four brand*lite combinations is done on each subject. If it were not a block experiment, wewould done one randomization of all 13⇥ 2⇥ 2 = 52 combinations of SUBJ, brand, and lite instead of 13separate randomizations.

22S:158/165 Design: Exam 3 2

The degrees of freedom are

d f =(333 + 222� 99)2

3332/6 + 2222/6 + 992/18=

207, 93627, 240

= 7.63

Interpolating in the t tables, the critical value is t.025,7.6 ⌅ 2.33, so

LSD = 2.33⇧

76 = 20.31

An answer based on the nearest d.f. in the table, t.025,8 = 2.31, is acceptable.

2. A consumer organization conducts an experiment where subjects are asked to rate chicken pot pies in a blindtasting. The factors in the experiment are SUBJ (there are 13 subjects), brand (2 brands of pot pie), and lite (2levels, regular and low-calorie). Each subject tastes all four combinations of brand and lite, in random order.

10 pts A. Draw a diagram of this experiment.

These factors are all crossed:SUBJ13 ⇥ brand2 ⇥ lite2

15 pts B. Give a table showing the model terms, degrees of freedom, and expected mean squares. You must beconsistent with the diagram. Include all interactions except the one that corresponds to individualmeasurements.

Source df Expected MS

SUBJ 12 4�2S + 2�2

Sb + 2�2Sl + �2

brand 1 26Q(b) + 2�2Sb + �2

lite 1 26Q(l) + 2�2Sl + �2

brand ⇤ lite 1 13Q(bl) + �2

SUBJ ⇤ brand 12 2�2Sb + �2

SUBJ ⇤ lite 12 2�2Sl + �2

RESID = S ⇤ b ⇤ l 12 �2

10 pts C. Explain why this is a block experiment, and state which factor(s) are blocks. How would it berandomized differently if it were not a block experiment?

It is a block experiment (with SUBJ as the blocking factor) because a separate randomization of theorder of the four brand*lite combinations is done on each subject. If it were not a block experiment, wewould done one randomization of all 13⇥ 2⇥ 2 = 52 combinations of SUBJ, brand, and lite instead of 13separate randomizations.

22S:158/165 Design: Exam 3 2

The degrees of freedom are

d f =(333 + 222� 99)2

3332/6 + 2222/6 + 992/18=

207, 93627, 240

= 7.63

Interpolating in the t tables, the critical value is t.025,7.6 ⌅ 2.33, so

LSD = 2.33⇧

76 = 20.31

An answer based on the nearest d.f. in the table, t.025,8 = 2.31, is acceptable.

2. A consumer organization conducts an experiment where subjects are asked to rate chicken pot pies in a blindtasting. The factors in the experiment are SUBJ (there are 13 subjects), brand (2 brands of pot pie), and lite (2levels, regular and low-calorie). Each subject tastes all four combinations of brand and lite, in random order.

10 pts A. Draw a diagram of this experiment.

These factors are all crossed:SUBJ13 ⇥ brand2 ⇥ lite2

15 pts B. Give a table showing the model terms, degrees of freedom, and expected mean squares. You must beconsistent with the diagram. Include all interactions except the one that corresponds to individualmeasurements.

Source df Expected MS

SUBJ 12 4�2S + 2�2

Sb + 2�2Sl + �2

brand 1 26Q(b) + 2�2Sb + �2

lite 1 26Q(l) + 2�2Sl + �2

brand ⇤ lite 1 13Q(bl) + �2

SUBJ ⇤ brand 12 2�2Sb + �2

SUBJ ⇤ lite 12 2�2Sl + �2

RESID = S ⇤ b ⇤ l 12 �2

10 pts C. Explain why this is a block experiment, and state which factor(s) are blocks. How would it berandomized differently if it were not a block experiment?

It is a block experiment (with SUBJ as the blocking factor) because a separate randomization of theorder of the four brand*lite combinations is done on each subject. If it were not a block experiment, wewould done one randomization of all 13⇥ 2⇥ 2 = 52 combinations of SUBJ, brand, and lite instead of 13separate randomizations.