In a \(2^{3}\) factorial design with three treatments A, B, C each at two levels, blocks of 4 plots are available. The design is replicated twice and the allocation of treatment combination is :Replication 1Replication 2Block IBlock IIBlock IBlock II1b1aacbcbcabacababcacabcbcHence the two confounded treatment combinations in replication 1 and 2 respectively are :

The confounded effects are BC in replication 1 and AC in replication 2.A two-factor interaction XY is confounded when one block holds the combinations where X and Y are at the same level and the other block those where they differ.Replication 1, block I = {1, a, bc, abc}: b and c are either both absent or both present, so BC is confounded.Replication 2, block I = {1, b, ac, abc}: a and c are both absent or both present, so AC is confounded.AB would need a block {1, ab, c, abc}, and ABC a block {1, ab, ac, bc}; neither appears, which rules out options A, B and C.

Design of Experiments Questions and Answers

Q: A block design with \(v\) treatments and \(b\) blocks is connected. Hence the rank of \(c\) matrix is :

The rank of the C matrix is \(v - 1\).The C matrix of a block design is \(C = R - NK^{-1}N'\). Its rows add to zero, so its rank is at most \(v - 1\). A design is connected exactly when every elementary treatment contrast is estimable, which happens if and only if \(\mathrm{rank}(C) = v - 1\).Rank \(v\) is impossible because of the zero row sums, and \(b - 1\) or \(b\) refer to the number of blocks, which does not determine the rank of the treatment information matrix.

Q: For a BIBD with parameters \((v, b, r, k, \lambda)\) to exist, the condition \(b \ge v\) is :

Fisher's inequality \(b \ge v\) is necessary but not sufficient.Every BIBD satisfies \(b \ge v\) (Fisher's inequality), so a parameter set with \(b \lt v\) cannot exist: the condition is necessary.It is not sufficient. The parameters \(v = 15\), \(b = 21\), \(r = 7\), \(k = 5\), \(\lambda = 2\) satisfy \(vr = bk\), \(\lambda(v - 1) = r(k - 1)\) and \(b \ge v\), yet no such design exists. Similarly, by the Bruck-Ryser-Chowla theorem there is no symmetric (22, 7, 2) design.

Q: A row-column design with 7 treatments arranged in 3 rows and 7 columns is given below :1 2 3 4 5 6 72 3 4 5 6 7 14 5 6 7 1 2 3Hence the design is :

The design is a Youden square design.Each row contains all seven treatments once, so rows are complete blocks. Each column contains three distinct treatments, and the columns form the blocks {1, 2, 4}, {2, 3, 5}, ..., the Fano plane, a BIBD with \(v = 7\), \(b = 7\), \(k = 3\), \(\lambda = 1\). A rectangle that is complete in one direction and a symmetric BIBD in the other is a Youden square, an 'incomplete Latin square'.It is not a Latin square, which would need 7 rows. A quasi-Latin square is a confounded factorial layout, and a split-plot design has main plots and sub-plots, neither of which fits.

Q: For a given data, the two-way ANOVA table is :SourcedfSsMssFratioTreat56-cBlock34.5--Errora2.0-Total2312.5Hence the values of a, b and c are :

The values are \(a = 15\), \(b = 6.0\) and \(c = 9\).Degrees of freedom add up, so the error df is \(a = 23 - 5 - 3 = 15\). The F ratio for treatments is the treatment mean square over the error mean square:\[ \begin{aligned} \text{MS}_{\text{treat}} &= \frac{6}{5} = 1.2 \\ \text{MS}_{\text{error}} &= \frac{2.0}{15} = 0.1333 \\ c &= \frac{1.2}{0.1333} = 9 \end{aligned} \]The letter \(b\) is not marked anywhere in the printed table; the value 6.0 in the keyed option matches the treatment sum of squares. In any case \(c\) decides the question, since only the last option has \(c = 9\). The values 1, 2.25 and 3 come from dividing the wrong mean squares or using sums of squares in place of mean squares (for instance 6/2 = 3).

Q: Which of the following properties is not true about the incidence matrix M of a BIBD \((r, b, s, k, \lambda)\) ?

The property that does not hold is M is a symmetric matrix.The paper lists the parameters as \((r, b, s, k, \lambda)\) in the position of the usual \((v, b, r, k, \lambda)\), so \(s\) plays the role of the replication number. The incidence matrix M has a row for each treatment and a column for each block, with entry 1 when the treatment occurs in the block. Then:each column (block) has exactly \(k\) ones, the block size;each row (treatment) has exactly \(s\) ones, the number of replications;two distinct rows share 1s in exactly \(\lambda\) columns, since every pair of treatments occurs together in \(\lambda\) blocks, which is the defining balance property.M need not be square (there are usually more blocks than treatments), and even for a symmetric BIBD, where it is square, the matrix itself is generally not symmetric. So symmetry is the property that fails.

Q: Suppose we wish to have a confounding arrangement in \(2^{r}\) blocks of a \(2^{n}\) factorial experiment. Then, how many interactions have to be confounded ?

The number of confounded interactions is \(2^{r} - 1\).Dividing the \(2^{n}\) treatment combinations into \(2^{r}\) blocks uses up \(2^{r} - 1\) degrees of freedom between blocks. In a \(2^{n}\) factorial every effect carries exactly one degree of freedom, so \(2^{r} - 1\) effects are confounded with blocks. Of these, \(r\) are chosen independently and the remaining \(2^{r} - r - 1\) are their generalised interactions.For example, a \(2^{3}\) experiment in \(2^{2} = 4\) blocks confounds 3 effects, such as AB, BC and their generalised interaction AC.\(r - 1\) and \(2^{r-1}\) undercount, and \(n - r + 1\) does not count block degrees of freedom at all.

Q: In a completely randomized design with unequal group numbers, that is, \(n_{1} = 5\), \(n_{2} = 7\), \(n_{3} = 6\), what is the degrees of freedom for the error term ?

The error degrees of freedom are 15.In a completely randomised design with \(k\) groups and \(N\) observations in all, the degrees of freedom split as total \(N - 1\), treatments \(k - 1\), error \(N - k\). Here\[ \begin{aligned} N &= 5 + 7 + 6 = 18,\quad k = 3 \\ \text{error df} &= 18 - 3 = 15 \end{aligned} \]Equivalently, each group contributes \(n_{i} - 1\): \(4 + 6 + 5 = 15\). Unequal group sizes do not change the formula.17 is the total df \(N - 1\), 18 is \(N\), and 120 is not related to the design.

Q: When the conditional effect (also known as sample effect) for one of the factors in a \(2^{2}\) design differ only in sign, then the corresponding main effect estimate will be :

The main effect estimate is zero.In a \(2^{2}\) design, the main effect of A is the average of its two simple (conditional) effects, one at each level of B:\[ A = \frac{1}{2}\left[(a - (1)) + (ab - b)\right] \]If the two conditional effects are equal in size and opposite in sign, say \(+d\) and \(-d\), their average is 0. The interaction, which is half their difference, is then \(\pm d\): the whole effect of A shows up as interaction, and the main effect hides it.'One' has no basis, and \((a + b)/2\) and \((a - b)/2\) treat the treatment-combination symbols as if they were the two simple effects.

Q: In a completely randomized design with unequal group numbers, that is, n1 = 5, n2 = 7, n3 = 6, what is the degree of freedom for the error term ?

The error degrees of freedom are 15.In a completely randomized design with k groups and N observations in total, the error degrees of freedom are N − k. Here\[ N - k = (5 + 7 + 6) - 3 = 18 - 3 = 15 \]17 is the total degrees of freedom N − 1, 18 is N itself, and 120 is the product of the group sizes, which has no role in the analysis of variance.

21.

A block design with \(v\) treatments and \(b\) blocks is connected. Hence the rank of \(c\) matrix is :

A.\(v - 1\)
B.\(v\)
C.\(b - 1\)
D.\(b\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

The rank of the C matrix is \(v - 1\).

The C matrix of a block design is \(C = R - NK^{-1}N'\). Its rows add to zero, so its rank is at most \(v - 1\). A design is connected exactly when every elementary treatment contrast is estimable, which happens if and only if \(\mathrm{rank}(C) = v - 1\).

Rank \(v\) is impossible because of the zero row sums, and \(b - 1\) or \(b\) refer to the number of blocks, which does not determine the rank of the treatment information matrix.

Discuss in Forum

22.

For a BIBD with parameters \((v, b, r, k, \lambda)\) to exist, the condition \(b \ge v\) is :

A.neither necessary nor sufficient
B.not necessary but sufficient
C.necessary but not sufficient
D.necessary and sufficient

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

Fisher's inequality \(b \ge v\) is necessary but not sufficient.

Every BIBD satisfies \(b \ge v\) (Fisher's inequality), so a parameter set with \(b \lt v\) cannot exist: the condition is necessary.

It is not sufficient. The parameters \(v = 15\), \(b = 21\), \(r = 7\), \(k = 5\), \(\lambda = 2\) satisfy \(vr = bk\), \(\lambda(v - 1) = r(k - 1)\) and \(b \ge v\), yet no such design exists. Similarly, by the Bruck-Ryser-Chowla theorem there is no symmetric (22, 7, 2) design.

Discuss in Forum

23.

A row-column design with 7 treatments arranged in 3 rows and 7 columns is given below :

1 2 3 4 5 6 7
2 3 4 5 6 7 1
4 5 6 7 1 2 3

Hence the design is :

A.Latin Square Design
B.Youden Square Design
C.Quasi-Latin Square Design
D.Split-Plot design

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The design is a Youden square design.

Each row contains all seven treatments once, so rows are complete blocks. Each column contains three distinct treatments, and the columns form the blocks {1, 2, 4}, {2, 3, 5}, ..., the Fano plane, a BIBD with \(v = 7\), \(b = 7\), \(k = 3\), \(\lambda = 1\). A rectangle that is complete in one direction and a symmetric BIBD in the other is a Youden square, an 'incomplete Latin square'.

It is not a Latin square, which would need 7 rows. A quasi-Latin square is a confounded factorial layout, and a split-plot design has main plots and sub-plots, neither of which fits.

Discuss in Forum

24.

In a \(2^{3}\) factorial design with three treatments A, B, C each at two levels, blocks of 4 plots are available. The design is replicated twice and the allocation of treatment combination is :

Replication 1	Replication 2
Block I	Block II	Block I	Block II
1	b	1	a
a	c	b	c
bc	ab	ac	ab
abc	ac	abc	bc

Hence the two confounded treatment combinations in replication 1 and 2 respectively are :

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

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

The confounded effects are BC in replication 1 and AC in replication 2.

A two-factor interaction XY is confounded when one block holds the combinations where X and Y are at the same level and the other block those where they differ.

Replication 1, block I = {1, a, bc, abc}: b and c are either both absent or both present, so BC is confounded.
Replication 2, block I = {1, b, ac, abc}: a and c are both absent or both present, so AC is confounded.

AB would need a block {1, ab, c, abc}, and ABC a block {1, ab, ac, bc}; neither appears, which rules out options A, B and C.

Discuss in Forum

25.

For a given data, the two-way ANOVA table is :

Source	df	Ss	Mss	F_ratio
Treat	5	6	-	c
Block	3	4.5	-	-
Error	a	2.0	-
Total	23	12.5

Hence the values of a, b and c are :

A.\(a = 15\ \ b = 6.5\ \ c = 1\)
B.\(a = 15\ \ b = 4.5\ \ c = 2.25\)
C.\(a = 15\ \ b = 6.0\ \ c = 3\)
D.\(a = 15\ \ b = 6.0\ \ c = 9\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: D

The values are \(a = 15\), \(b = 6.0\) and \(c = 9\).

Degrees of freedom add up, so the error df is \(a = 23 - 5 - 3 = 15\). The F ratio for treatments is the treatment mean square over the error mean square:

\[ \begin{aligned} \text{MS}_{\text{treat}} &= \frac{6}{5} = 1.2 \\ \text{MS}_{\text{error}} &= \frac{2.0}{15} = 0.1333 \\ c &= \frac{1.2}{0.1333} = 9 \end{aligned} \]

The letter \(b\) is not marked anywhere in the printed table; the value 6.0 in the keyed option matches the treatment sum of squares. In any case \(c\) decides the question, since only the last option has \(c = 9\). The values 1, 2.25 and 3 come from dividing the wrong mean squares or using sums of squares in place of mean squares (for instance 6/2 = 3).

Discuss in Forum

26.

Which of the following properties is not true about the incidence matrix M of a BIBD \((r, b, s, k, \lambda)\) ?

A.Every column of M contains exactly \(k\) '1's
B.Every row of M contains exactly \(s\) '1's
C.M is a symmetric matrix
D.Two distinct rows of M contains both 1's in exactly \(\lambda\) columns

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The property that does not hold is M is a symmetric matrix.

The paper lists the parameters as \((r, b, s, k, \lambda)\) in the position of the usual \((v, b, r, k, \lambda)\), so \(s\) plays the role of the replication number. The incidence matrix M has a row for each treatment and a column for each block, with entry 1 when the treatment occurs in the block. Then:

each column (block) has exactly \(k\) ones, the block size;
each row (treatment) has exactly \(s\) ones, the number of replications;
two distinct rows share 1s in exactly \(\lambda\) columns, since every pair of treatments occurs together in \(\lambda\) blocks, which is the defining balance property.

M need not be square (there are usually more blocks than treatments), and even for a symmetric BIBD, where it is square, the matrix itself is generally not symmetric. So symmetry is the property that fails.

Discuss in Forum

27.

Suppose we wish to have a confounding arrangement in \(2^{r}\) blocks of a \(2^{n}\) factorial experiment. Then, how many interactions have to be confounded ?

A.\(2^{r-1}\)
B.\(r - 1\)
C.\(n - r + 1\)
D.\(2^{r} - 1\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: D

The number of confounded interactions is \(2^{r} - 1\).

Dividing the \(2^{n}\) treatment combinations into \(2^{r}\) blocks uses up \(2^{r} - 1\) degrees of freedom between blocks. In a \(2^{n}\) factorial every effect carries exactly one degree of freedom, so \(2^{r} - 1\) effects are confounded with blocks. Of these, \(r\) are chosen independently and the remaining \(2^{r} - r - 1\) are their generalised interactions.

For example, a \(2^{3}\) experiment in \(2^{2} = 4\) blocks confounds 3 effects, such as AB, BC and their generalised interaction AC.

\(r - 1\) and \(2^{r-1}\) undercount, and \(n - r + 1\) does not count block degrees of freedom at all.

Discuss in Forum

28.

In a completely randomized design with unequal group numbers, that is, \(n_{1} = 5\), \(n_{2} = 7\), \(n_{3} = 6\), what is the degrees of freedom for the error term ?

A.17
B.18
C.15
D.120

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The error degrees of freedom are 15.

In a completely randomised design with \(k\) groups and \(N\) observations in all, the degrees of freedom split as total \(N - 1\), treatments \(k - 1\), error \(N - k\). Here

\[ \begin{aligned} N &= 5 + 7 + 6 = 18,\quad k = 3 \\ \text{error df} &= 18 - 3 = 15 \end{aligned} \]

Equivalently, each group contributes \(n_{i} - 1\): \(4 + 6 + 5 = 15\). Unequal group sizes do not change the formula.

17 is the total df \(N - 1\), 18 is \(N\), and 120 is not related to the design.

Discuss in Forum

29.

When the conditional effect (also known as sample effect) for one of the factors in a \(2^{2}\) design differ only in sign, then the corresponding main effect estimate will be :

A.zero
B.one
C.\((a + b)/2\)
D.\((a - b)/2\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

The main effect estimate is zero.

In a \(2^{2}\) design, the main effect of A is the average of its two simple (conditional) effects, one at each level of B:

\[ A = \frac{1}{2}\left[(a - (1)) + (ab - b)\right] \]

If the two conditional effects are equal in size and opposite in sign, say \(+d\) and \(-d\), their average is 0. The interaction, which is half their difference, is then \(\pm d\): the whole effect of A shows up as interaction, and the main effect hides it.

'One' has no basis, and \((a + b)/2\) and \((a - b)/2\) treat the treatment-combination symbols as if they were the two simple effects.

Discuss in Forum

30.

In a completely randomized design with unequal group numbers, that is, n₁ = 5, n₂ = 7, n₃ = 6, what is the degree of freedom for the error term ?

A.17
B.15
C.120
D.18

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

The error degrees of freedom are 15.

In a completely randomized design with k groups and N observations in total, the error degrees of freedom are N − k. Here

\[ N - k = (5 + 7 + 6) - 3 = 18 - 3 = 15 \]

17 is the total degrees of freedom N − 1, 18 is N itself, and 120 is the product of the group sizes, which has no role in the analysis of variance.

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