Design of Experiments Questions and Answers

Q: In an ANOVA for an experimental design involving 3 treatments and 10 observations per treatment, if SSE = 399.6, then the estimated variance of error term is :

The estimated error variance is 14.8.The error variance is estimated by the mean square error, SSE divided by its degrees of freedom. With k = 3 treatments and 10 observations each, N = 30, and the error degrees of freedom are \(N - k = 27\):\[ \begin{aligned} \text{MSE} &= \frac{\text{SSE}}{N - k} \\ &= \frac{399.6}{27} = 14.8 \end{aligned} \]133.2 is SSE divided by 3, the number of treatments, and 66.6 is SSE divided by 6; neither is the error degrees of freedom. 87.8 has no basis in the data given.

Q: In a \(2^{5}\) factorial design in block of 8 plots each, total number of interaction confounded with blocks is :

3 interactions are confounded with blocks.A \(2^{5}\) factorial has 32 treatment combinations. In blocks of 8 plots each, one replicate needs \(32/8 = 4\) blocks, and 4 blocks carry \(4 - 1 = 3\) degrees of freedom between them. Each confounded effect in a \(2^{n}\) design uses one degree of freedom, so 3 effects are confounded.In practice two independent interactions are chosen and their generalised interaction is confounded automatically; for example ABC and CDE force \(ABC \times CDE = ABDE\). That is why 2 is too few: the third comes for free. 7 would correspond to 8 blocks of 4 plots, and 4 is not of the form \(2^{p} - 1\), so it is never the number confounded.

Q: Identify the treatments \(x_{1}, x_{2}, x_{3}\) and \(x_{4}\) from blocks 1, 2, 3, 4 respectively so that the design is BIBD :Block 1 : A, B, C, \(x_{1}\)Block 2 : A, \(x_{2}\), C, EBlock 3 : A, B, D, \(x_{3}\)Block 4 : A, \(x_{4}\), D, EBlock 5 : B, C, D, E

The design is a BIBD with \(x_{1}\) = D, \(x_{2}\) = B, \(x_{3}\) = E, \(x_{4}\) = C.With these values the five blocks are {A, B, C, D}, {A, B, C, E}, {A, B, D, E}, {A, C, D, E} and {B, C, D, E}: every 4-element subset of the five treatments, each exactly once. So v = b = 5, k = 4, every treatment appears in r = 4 blocks, and every pair of treatments appears together in \(\lambda = 3\) blocks. Check: \(\lambda(v - 1) = r(k - 1)\) gives \(3 \times 4 = 4 \times 3\).Each of the other choices repeats a treatment within a block, which a BIBD does not allow: the first puts B twice in block 1, the third puts C twice in block 1, and the last puts C twice in block 2 and D twice in block 3.

Q: Consider a two-factor factorial, fixed effect model with main effects A and B used at \(a\) and \(b\) levels respectively. Then the estimate of operating characteristic curve parameter for A is :

The parameter for factor A is \(\Phi^{2} = \dfrac{nb\sum_{i=1}^{a}\tau_{i}^{2}}{a\sigma^{2}}\).In the two-factor fixed effects model with \(n\) replicates, each level of A is observed \(nb\) times. The operating characteristic curves for testing A use\[ \Phi^{2} = \frac{nb\sum_{i=1}^{a}\tau_{i}^{2}}{a\sigma^{2}} \]with \(a - 1\) and \(ab(n - 1)\) degrees of freedom (Montgomery). The multiplier is the number of observations per level of A, and the denominator is the number of levels of A.Option A swaps \(a\) and \(b\). Options C and D sum the \(\beta_{j}^{2}\), which belong to factor B; C is in fact the corresponding parameter for B.

Q: Consider the following statements :(1) If the presence of interaction inflates the error mean square, one should use factorial designs.(2) Two contrasts with coefficient \(\{c_{i}\}\) and \(\{d_{i}\}\) are orthogonal if \(\sum_{i=1}^{a} c_{i}d_{i} = 0\).(3) Confounding is a design technique for arranging a complete factorial experiment in blocks.Which of the above statements are correct ?

All are correct.(1): if factors interact and a one-factor-at-a-time or main-effects analysis is used, the interaction is left in the error term, inflating the error mean square and hiding real effects. A factorial design estimates the interaction separately, so it should be used.(2): with equal replication, contrasts \(\sum c_{i}\mu_{i}\) and \(\sum d_{i}\mu_{i}\) are orthogonal when \(\sum c_{i}d_{i} = 0\); their sums of squares are then independent components of the treatment sum of squares.(3): confounding is the technique of arranging a complete factorial in blocks smaller than a full replicate, by deliberately making some (usually high-order) interactions indistinguishable from block effects. Since all three hold, the answer is D.

Q: Consider the following statements :(1) In Resolution-III designs main effects are aliased with two-factor interactions.(2) In Resolution-IV designs, two-factor interactions are aliased with each other.(3) In Resolution-V designs two-factor interactions cannot be aliased with three-factor interactions.Which of the above statements are correct ?

Statements 1 and 2 are correct.The resolution of a fractional factorial is the length of the shortest word in its defining relation. In a Resolution III design, main effects are not aliased with one another but are aliased with two-factor interactions (1). In Resolution IV, main effects are clear of two-factor interactions, but two-factor interactions are aliased with each other (2).Statement 3 is false. In a Resolution V design no main effect or two-factor interaction is aliased with another of these, but two-factor interactions are aliased with three-factor interactions. A word of length 5, such as ABCDE, links AB with CDE. So the codes including 3 are wrong.

Q: A \(2^{3}\) design with 4 replicates are under consideration. Two of the replicates are shown below :Replicate 1Replicate 2(1)cababcabacbc(1)abcabcbcabacIdentify which of the treatment combinations are partially confounded in each replicate ?

The confounded interactions are AB and BC.In each replicate, the confounded effect is the one whose plus and minus treatment combinations form the two blocks. The sign of a treatment in AB is + when it contains both or neither of \(a\) and \(b\).Replicate 1: the block {(1), \(c\), \(ab\), \(abc\)} contains exactly the treatments with an even number of letters from \(\{a, b\}\), the + set of AB, and the other block is the - set. So AB is confounded.Replicate 2: the block {(1), \(a\), \(bc\), \(abc\)} has an even number of letters from \(\{b, c\}\) in each treatment, the + set of BC. So BC is confounded. Each is lost only in its own replicate, which is partial confounding.

Q: A completely randomized design has \(n\) plots and 10 treatments. While calculating the F-ratio for assessing equality of treatment effects, the experimenter forgot to divide the numerator and denominator sums of squares by corresponding degrees of freedom. But the statistician said that the calculated F-ratio is correct. Hence \(n\) is equal to :

\(n\) is 19.In a completely randomised design with \(t = 10\) treatments and \(n\) plots, the F-ratio is\[ \begin{aligned} F &= \frac{SS_{T}/(t - 1)}{SS_{E}/(n - t)} \\ &= \frac{n - t}{t - 1}\cdot\frac{SS_{T}}{SS_{E}} \end{aligned} \]Forgetting both divisions gives \(SS_{T}/SS_{E}\), which equals the correct F exactly when \(n - t = t - 1\), that is\[ n - 10 = 9 \Rightarrow n = 19 \]With 10, 20 or 30 plots the error degrees of freedom would be 0, 10 or 20, not 9.

Q: Under completely randomized design with model :\(Ey_{ij} = \mu + \alpha_{i} + \in_{ij}\) \(i = 1, ......, p\)\(j = 1, ......, n_{i}\)\(\in_{ij} \sim N(0, \sigma^{2})\) and independently distributed, a parametric function \(c\mu + \sum_{i=1}^{p} d_{i}\alpha_{i}\) where \(c\) and \(d_{1}, ......., d_{p}\) are known constants is estimable if and only if :

The function is estimable if and only if \(\sum d_{i} = c\).The estimable functions are linear combinations of the expected observations \(E(y_{ij}) = \mu + \alpha_{i}\). A combination \(\sum_{i} l_{i}(\mu + \alpha_{i})\) equals \(\left(\sum l_{i}\right)\mu + \sum l_{i}\alpha_{i}\), so it has the form \(c\mu + \sum d_{i}\alpha_{i}\) with \(d_{i} = l_{i}\) and \(c = \sum d_{i}\).Conversely any function with \(c = \sum d_{i}\) is such a combination. Contrasts (\(c = 0\), \(\sum d_{i} = 0\)) are the special case; \(c = 0\) alone, or \(\sum d_{i} = 0\) with \(c \gt 0\), break the required equality, and \(\mu\) itself (\(c = 1\), all \(d_{i} = 0\)) is not estimable.

Q: Under a Latin-Square design with V treatments, which of the following statements is not correct ?

The incorrect statement is it requires exactly \(V^{3}\) experimental units.A Latin square for V treatments is a \(V \times V\) arrangement, one plot per cell, so it uses \(V^{2}\) units, each treatment once in every row and every column.The other statements are the usual properties. The design classifies the variation three ways, by rows, columns and treatments; row, column and treatment contrasts are all estimated with the same variance \(2\sigma^{2}/V\) because the design is balanced; and the error degrees of freedom are \((V^{2} - 1) - 3(V - 1) = (V - 1)(V - 2)\).

In an ANOVA for an experimental design involving 3 treatments and 10 observations per treatment, if SSE = 399.6, then the estimated variance of error term is :

A.133.2
B.87.8
C.66.6
D.14.8

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The estimated error variance is 14.8.

The error variance is estimated by the mean square error, SSE divided by its degrees of freedom. With k = 3 treatments and 10 observations each, N = 30, and the error degrees of freedom are \(N - k = 27\):

\[ \begin{aligned} \text{MSE} &= \frac{\text{SSE}}{N - k} \\ &= \frac{399.6}{27} = 14.8 \end{aligned} \]

133.2 is SSE divided by 3, the number of treatments, and 66.6 is SSE divided by 6; neither is the error degrees of freedom. 87.8 has no basis in the data given.

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In a \(2^{5}\) factorial design in block of 8 plots each, total number of interaction confounded with blocks is :

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

3 interactions are confounded with blocks.

A \(2^{5}\) factorial has 32 treatment combinations. In blocks of 8 plots each, one replicate needs \(32/8 = 4\) blocks, and 4 blocks carry \(4 - 1 = 3\) degrees of freedom between them. Each confounded effect in a \(2^{n}\) design uses one degree of freedom, so 3 effects are confounded.

In practice two independent interactions are chosen and their generalised interaction is confounded automatically; for example ABC and CDE force \(ABC \times CDE = ABDE\). That is why 2 is too few: the third comes for free. 7 would correspond to 8 blocks of 4 plots, and 4 is not of the form \(2^{p} - 1\), so it is never the number confounded.

Discuss in Forum

Identify the treatments \(x_{1}, x_{2}, x_{3}\) and \(x_{4}\) from blocks 1, 2, 3, 4 respectively so that the design is BIBD :

Block 1 : A, B, C, \(x_{1}\)
Block 2 : A, \(x_{2}\), C, E
Block 3 : A, B, D, \(x_{3}\)
Block 4 : A, \(x_{4}\), D, E
Block 5 : B, C, D, E

A.\(x_{1}\) = B, \(x_{2}\) = E, \(x_{3}\) = C, \(x_{4}\) = D
B.\(x_{1}\) = D, \(x_{2}\) = B, \(x_{3}\) = E, \(x_{4}\) = C
C.\(x_{1}\) = C, \(x_{2}\) = D, \(x_{3}\) = B, \(x_{4}\) = E
D.\(x_{1}\) = E, \(x_{2}\) = C, \(x_{3}\) = D, \(x_{4}\) = B

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The design is a BIBD with \(x_{1}\) = D, \(x_{2}\) = B, \(x_{3}\) = E, \(x_{4}\) = C.

With these values the five blocks are {A, B, C, D}, {A, B, C, E}, {A, B, D, E}, {A, C, D, E} and {B, C, D, E}: every 4-element subset of the five treatments, each exactly once. So v = b = 5, k = 4, every treatment appears in r = 4 blocks, and every pair of treatments appears together in \(\lambda = 3\) blocks. Check: \(\lambda(v - 1) = r(k - 1)\) gives \(3 \times 4 = 4 \times 3\).

Each of the other choices repeats a treatment within a block, which a BIBD does not allow: the first puts B twice in block 1, the third puts C twice in block 1, and the last puts C twice in block 2 and D twice in block 3.

Discuss in Forum

Consider a two-factor factorial, fixed effect model with main effects A and B used at \(a\) and \(b\) levels respectively. Then the estimate of operating characteristic curve parameter for A is :

A.\(\dfrac{na\sum_{i=1}^{a}\tau_{i}^{2}}{b\sigma^{2}}\)
B.\(\dfrac{nb\sum_{i=1}^{a}\tau_{i}^{2}}{a\sigma^{2}}\)
C.\(\dfrac{na\sum_{i=1}^{b}\beta_{j}^{2}}{b\sigma^{2}}\)
D.\(\dfrac{nb\sum_{i=1}^{b}\beta_{j}^{2}}{a\sigma^{2}}\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The parameter for factor A is \(\Phi^{2} = \dfrac{nb\sum_{i=1}^{a}\tau_{i}^{2}}{a\sigma^{2}}\).

In the two-factor fixed effects model with \(n\) replicates, each level of A is observed \(nb\) times. The operating characteristic curves for testing A use

\[ \Phi^{2} = \frac{nb\sum_{i=1}^{a}\tau_{i}^{2}}{a\sigma^{2}} \]

with \(a - 1\) and \(ab(n - 1)\) degrees of freedom (Montgomery). The multiplier is the number of observations per level of A, and the denominator is the number of levels of A.

Option A swaps \(a\) and \(b\). Options C and D sum the \(\beta_{j}^{2}\), which belong to factor B; C is in fact the corresponding parameter for B.

Discuss in Forum

Consider the following statements :

(1) If the presence of interaction inflates the error mean square, one should use factorial designs.
(2) Two contrasts with coefficient \(\{c_{i}\}\) and \(\{d_{i}\}\) are orthogonal if \(\sum_{i=1}^{a} c_{i}d_{i} = 0\).
(3) Confounding is a design technique for arranging a complete factorial experiment in blocks.

Which of the above statements are correct ?

A.Only 1 and 2 are correct
B.Only 1 and 3 are correct
C.Only 2 and 3 are correct
D.All are correct

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

All are correct.

(1): if factors interact and a one-factor-at-a-time or main-effects analysis is used, the interaction is left in the error term, inflating the error mean square and hiding real effects. A factorial design estimates the interaction separately, so it should be used.

(2): with equal replication, contrasts \(\sum c_{i}\mu_{i}\) and \(\sum d_{i}\mu_{i}\) are orthogonal when \(\sum c_{i}d_{i} = 0\); their sums of squares are then independent components of the treatment sum of squares.

(3): confounding is the technique of arranging a complete factorial in blocks smaller than a full replicate, by deliberately making some (usually high-order) interactions indistinguishable from block effects. Since all three hold, the answer is D.

Discuss in Forum

Consider the following statements :

(1) In Resolution-III designs main effects are aliased with two-factor interactions.
(2) In Resolution-IV designs, two-factor interactions are aliased with each other.
(3) In Resolution-V designs two-factor interactions cannot be aliased with three-factor interactions.

Which of the above statements are correct ?

A.1 and 2 are correct
B.1 and 3 are correct
C.2 and 3 are correct
D.All are correct

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

Statements 1 and 2 are correct.

The resolution of a fractional factorial is the length of the shortest word in its defining relation. In a Resolution III design, main effects are not aliased with one another but are aliased with two-factor interactions (1). In Resolution IV, main effects are clear of two-factor interactions, but two-factor interactions are aliased with each other (2).

Statement 3 is false. In a Resolution V design no main effect or two-factor interaction is aliased with another of these, but two-factor interactions are aliased with three-factor interactions. A word of length 5, such as ABCDE, links AB with CDE. So the codes including 3 are wrong.

Discuss in Forum

A \(2^{3}\) design with 4 replicates are under consideration. Two of the replicates are shown below :

Replicate 1		Replicate 2
(1) c ab abc	a b ac bc	(1) a bc abc	b c ab ac

Identify which of the treatment combinations are partially confounded in each replicate ?

A.AB and AC
B.AB and BC
C.AC and BC
D.A and BC

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The confounded interactions are AB and BC.

In each replicate, the confounded effect is the one whose plus and minus treatment combinations form the two blocks. The sign of a treatment in AB is + when it contains both or neither of \(a\) and \(b\).

Replicate 1: the block {(1), \(c\), \(ab\), \(abc\)} contains exactly the treatments with an even number of letters from \(\{a, b\}\), the + set of AB, and the other block is the - set. So AB is confounded.

Replicate 2: the block {(1), \(a\), \(bc\), \(abc\)} has an even number of letters from \(\{b, c\}\) in each treatment, the + set of BC. So BC is confounded. Each is lost only in its own replicate, which is partial confounding.

Discuss in Forum

A completely randomized design has \(n\) plots and 10 treatments. While calculating the F-ratio for assessing equality of treatment effects, the experimenter forgot to divide the numerator and denominator sums of squares by corresponding degrees of freedom. But the statistician said that the calculated F-ratio is correct. Hence \(n\) is equal to :

A.10
B.20
C.30
D.19

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

\(n\) is 19.

In a completely randomised design with \(t = 10\) treatments and \(n\) plots, the F-ratio is

\[ \begin{aligned} F &= \frac{SS_{T}/(t - 1)}{SS_{E}/(n - t)} \\ &= \frac{n - t}{t - 1}\cdot\frac{SS_{T}}{SS_{E}} \end{aligned} \]

Forgetting both divisions gives \(SS_{T}/SS_{E}\), which equals the correct F exactly when \(n - t = t - 1\), that is

\[ n - 10 = 9 \Rightarrow n = 19 \]

With 10, 20 or 30 plots the error degrees of freedom would be 0, 10 or 20, not 9.

Discuss in Forum

Under completely randomized design with model :

\(Ey_{ij} = \mu + \alpha_{i} + \in_{ij}\) \(i = 1, ......, p\)
\(j = 1, ......, n_{i}\)
\(\in_{ij} \sim N(0, \sigma^{2})\) and independently distributed, a parametric function \(c\mu + \sum_{i=1}^{p} d_{i}\alpha_{i}\) where \(c\) and \(d_{1}, ......., d_{p}\) are known constants is estimable if and only if :

A.\(c = 0\)
B.\(\sum_{i=1}^{p} d_{i} = 0,\ c \gt 0\)
C.\(c = 0,\ \sum_{i=1}^{p} d_{i} \gt 0\)
D.\(\sum_{i=1}^{p} d_{i} = c\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

The function is estimable if and only if \(\sum d_{i} = c\).

The estimable functions are linear combinations of the expected observations \(E(y_{ij}) = \mu + \alpha_{i}\). A combination \(\sum_{i} l_{i}(\mu + \alpha_{i})\) equals \(\left(\sum l_{i}\right)\mu + \sum l_{i}\alpha_{i}\), so it has the form \(c\mu + \sum d_{i}\alpha_{i}\) with \(d_{i} = l_{i}\) and \(c = \sum d_{i}\).

Conversely any function with \(c = \sum d_{i}\) is such a combination. Contrasts (\(c = 0\), \(\sum d_{i} = 0\)) are the special case; \(c = 0\) alone, or \(\sum d_{i} = 0\) with \(c \gt 0\), break the required equality, and \(\mu\) itself (\(c = 1\), all \(d_{i} = 0\)) is not estimable.

Discuss in Forum

10.

Under a Latin-Square design with V treatments, which of the following statements is not correct ?

A.It can accommodate three sources of heterogeneity
B.It requires exactly \(V^{3}\) experimental units
C.The elementary (pairwise) contrasts among column effects, among row effects and among treatment effects are all estimated with common variance
D.The degrees of freedom associated with error are equal to (V - 2) (V - 1)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

The incorrect statement is it requires exactly \(V^{3}\) experimental units.

A Latin square for V treatments is a \(V \times V\) arrangement, one plot per cell, so it uses \(V^{2}\) units, each treatment once in every row and every column.

The other statements are the usual properties. The design classifies the variation three ways, by rows, columns and treatments; row, column and treatment contrasts are all estimated with the same variance \(2\sigma^{2}/V\) because the design is balanced; and the error degrees of freedom are \((V^{2} - 1) - 3(V - 1) = (V - 1)(V - 2)\).

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