In a \(2^{4}\) factorial design with two blocks of 8 plots each the treatment combinations are allotted as below :Block IBlock IIa1babaccbcabcdadabdbdcdacdabcdbcdHence the confounded treatment combination is :

The confounded interaction is ABD.Block II contains the control (1), so it is the principal block: every treatment in it has an even number of letters in common with the confounded interaction. Test ABD against Block II:ab: a, b (2); c: none (0); abc: a, b (2)ad: a, d (2); bd: b, d (2)acd: a, d (2); bcd: b, d (2)All even, while every Block I treatment (a, b, ac, bc, d, abd, cd, abcd) shares an odd number of letters with ABD. ACD fails at once on ab in Block II (one letter, a), BCD fails on ab (one letter, b), and ABCD fails on c (one letter).

Design of Experiments Questions and Answers

Q: The incidence matrix of a block design is given by :\[ N = \begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 1 \end{bmatrix} \]Hence the design is :

The design is not connected and not orthogonal.Read the rows as treatments and the columns as blocks. Treatment 2 (row 2) appears only in block 3, and block 3 contains no other treatment. So treatment 2 is never compared with any other treatment within a block, and no chain of blocks links it to the rest: the design is disconnected, and contrasts involving treatment 2 are not estimable.Orthogonality needs \(n_{ij} = r_{i}k_{j}/n\) for every cell. Here the treatment replications are 2, 1, 2, 2, the block sizes 1, 3, 1, 2, and \(n = 7\); cell (1, 1) has \(n_{11} = 1\) but \(r_{1}k_{1}/n = 2/7\). So it is not orthogonal either. (Reading the rows as blocks leads to the same conclusion, with treatment 3 isolated in block 2.)

Q: Suppose N is the incidence matrix of a BIBD with parameters \((v, b, r, k, \lambda)\), then :

rank \((NN') = v\).For a BIBD, every treatment occurs in \(r\) blocks and every pair together in \(\lambda\) blocks, so\[ NN' = (r - \lambda)I_{v} + \lambda J_{v} \]Its eigenvalues are \(rk\) (once) and \(r - \lambda\) (\(v - 1\) times), and \(r \gt \lambda\) in any proper BIBD, so the determinant \(rk(r - \lambda)^{v-1}\) is non-zero and \(NN'\) is a non-singular \(v \times v\) matrix of rank \(v\).This is the proof of Fisher's inequality \(b \ge v\): rank \(N \ge\) rank \(NN' = v\). So the rank is \(b\) only for symmetric designs, and never \(v - 1\) or \(b - 1\).

Q: 8 treatments are arranged in a row-column design as given below :1234567838167452Hence it is a :

It is a quasi-Latin square design.The layout is \(4 \times 4\), but it holds 8 treatments, each appearing twice, and each row and each column contains only 4 of them. A Latin square would need 4 treatments, each exactly once in every row and column, so this is not one.A quasi-Latin square (Yates) arranges the \(2^{3} = 8\) treatment combinations of a factorial in a square of side 4, with rows and columns acting as incomplete blocks that confound different interactions: here rows 1 and 2 hold {1, 2, 3, 4} and {5, 6, 7, 8}, rows 3 and 4 hold {1, 3, 6, 8} and {2, 4, 5, 7}, and the columns split the treatments a third way.A Youden square is an incomplete Latin square whose rows are complete blocks, which these are not, and 'incomplete block design' describes one-way blocking, not a row-column layout.

Q: Consider a two-way classification model \(y_{ij} = \mu + \tau_{i} + \beta_{j} + \varepsilon_{ij}\) \(i = 1 \ldots v\), \(j = 1, \ldots, b\) with \(E(\varepsilon_{ij}) = 0\), \(\mathrm{Var}(\varepsilon_{ij}) = \sigma^{2}\), \(\varepsilon_{ij}\) uncorrelated where \(\tau_{i}\ \delta = 1 \ldots v\) are fixed effects and \(\beta_{j}\ j = 1, \ldots, b\) are random effects with mean 0 and variance \(\sigma_{\beta}^{2}\). Hence the hypothesis regarding the insignificance of random effects is :

The hypothesis is \(H_{0} : \sigma_{\beta}^{2} = 0\).In a mixed model the block effects \(\beta_{j}\) are random variables drawn from a population with variance \(\sigma_{\beta}^{2}\). Interest is in that population, not in the particular blocks observed, so 'no block effect' means the random effects have no variability: \(\sigma_{\beta}^{2} = 0\).\(\beta_{1} = \cdots = \beta_{b}\) is the hypothesis for fixed effects, \(\tau_{1} = \cdots = \tau_{v}\) concerns the fixed treatment effects, and \(\sigma^{2} = 0\) would mean no experimental error at all.

Q: The incidence matrix of a general block design is :\[ N = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \end{bmatrix} \]Hence the number of estimable orthogonal block contrasts is :

The number of estimable orthogonal block contrasts is 2.N has 4 rows (treatments) and 3 columns (blocks): block 1 contains treatments 1 and 2, block 2 contains 1, 3 and 4, and block 3 contains 1 and 4. Treatment 1 appears in every block, linking them all, so the design is connected.In a connected design all block contrasts are estimable, and with \(b = 3\) blocks there are \(b - 1 = 2\) independent (orthogonal) block contrasts. 3 would be the number of blocks, and 1 or 4 do not match \(b - 1\).

Q: Consider a BIBD with parameter \((v, b, r, k, \lambda)\). If \(v = b = 11\) and \(r = 5\), then :

The parameters are \(k = 5, \lambda = 2\).For a BIBD, \(bk = vr\) and \(\lambda(v - 1) = r(k - 1)\). With \(v = b = 11\) and \(r = 5\):\[ \begin{aligned} 11k &= 11 \cdot 5 \Rightarrow k = 5 \\ 10\lambda &= 5 \cdot 4 \Rightarrow \lambda = 2 \end{aligned} \](A symmetric design has \(k = r\).) \(\lambda = 1\) would need \(10 = 20\), and \(k = 2\) or \(k = 11\) contradicts \(bk = vr\).

Q: To examine whether five different skin creams A, B, C, D and E have different effect on the human body, \(n\) randomly chosen persons were enrolled in an experiment. The five creams were applied on each of one randomly chosen fingres of the same person; and the procedure is repeated for each person. The resulting design is :

The design is a randomized block design with \(n\) blocks and five treatments.Each person is a block: all five creams are applied to that person, on randomly chosen fingers, so differences between people are removed from the comparison of creams. With \(n\) persons and 5 creams, there are \(n\) blocks of 5 plots and 5 treatments.A completely randomized design would assign creams to persons without blocking. Option B reverses the roles of blocks and treatments, and the creams are applied simultaneously to different fingers rather than in sequence over time, so this is treated as an RBD rather than a repeated measures design.

Q: For one-way classification model the F-test is valid under the assumption that :

The F-test is valid when the errors are independently normally distributed with mean zero and common positive variance.In the model \(y_{ij} = \mu + \tau_{i} + e_{ij}\), the ratio of the between-group to within-group mean squares has an exact F distribution under \(H_{0}\) only if the \(e_{ij}\) are independent \(N(0, \sigma^{2})\). Independence and normality make the two sums of squares independent chi-square variables.Uncorrelated errors (B) or observations (A) with common variance give unbiased estimates, but not an exact F distribution. Option C states normality but omits independence, which the F distribution also needs.

Q: Under a \(2^{3}\) factorial design with factors A, B, C arranged in a single replicate of two blocks, the containts of block-I (in Yate's notation) are {1, a, ab, b}. Which of the following effects is confounded with blocks ?

The effect confounded with blocks is C.Block I contains 1, a, b, ab, exactly the treatment combinations in which factor C is at its low level, and block II therefore contains c, ac, bc, abc. The main effect of C is the contrast (c + ac + bc + abc) - (1 + a + b + ab), which is the same as the block difference, so C cannot be separated from blocks.For AB the block would have to be {1, ab, c, abc}, for AC {1, ac, b, abc}, and for ABC {1, ab, ac, bc}. Confounding a main effect is unusual in practice, but it is what this allocation does.

11.

The incidence matrix of a block design is given by :

\[ N = \begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 1 \end{bmatrix} \]

Hence the design is :

A.connected and not orthogonal
B.not connected and not orthogonal
C.not connected and orthogonal
D.connected and orthogonal

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

The design is not connected and not orthogonal.

Read the rows as treatments and the columns as blocks. Treatment 2 (row 2) appears only in block 3, and block 3 contains no other treatment. So treatment 2 is never compared with any other treatment within a block, and no chain of blocks links it to the rest: the design is disconnected, and contrasts involving treatment 2 are not estimable.

Orthogonality needs \(n_{ij} = r_{i}k_{j}/n\) for every cell. Here the treatment replications are 2, 1, 2, 2, the block sizes 1, 3, 1, 2, and \(n = 7\); cell (1, 1) has \(n_{11} = 1\) but \(r_{1}k_{1}/n = 2/7\). So it is not orthogonal either. (Reading the rows as blocks leads to the same conclusion, with treatment 3 isolated in block 2.)

Discuss in Forum

12.

Suppose N is the incidence matrix of a BIBD with parameters \((v, b, r, k, \lambda)\), then :

A.rank \((NN') = v\)
B.rank \((NN') = b\)
C.rank \((NN') = v - 1\)
D.rank \((NN') = b - 1\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: A

rank \((NN') = v\).

For a BIBD, every treatment occurs in \(r\) blocks and every pair together in \(\lambda\) blocks, so

\[ NN' = (r - \lambda)I_{v} + \lambda J_{v} \]

Its eigenvalues are \(rk\) (once) and \(r - \lambda\) (\(v - 1\) times), and \(r \gt \lambda\) in any proper BIBD, so the determinant \(rk(r - \lambda)^{v-1}\) is non-zero and \(NN'\) is a non-singular \(v \times v\) matrix of rank \(v\).

This is the proof of Fisher's inequality \(b \ge v\): rank \(N \ge\) rank \(NN' = v\). So the rank is \(b\) only for symmetric designs, and never \(v - 1\) or \(b - 1\).

Discuss in Forum

13.

8 treatments are arranged in a row-column design as given below :

1	2	3	4
5	6	7	8
3	8	1	6
7	4	5	2

Hence it is a :

A.Latin Square Design
B.Youden Square Design
C.Quasi-Latin Square Design
D.Incomplete Block Design

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

It is a quasi-Latin square design.

The layout is \(4 \times 4\), but it holds 8 treatments, each appearing twice, and each row and each column contains only 4 of them. A Latin square would need 4 treatments, each exactly once in every row and column, so this is not one.

A quasi-Latin square (Yates) arranges the \(2^{3} = 8\) treatment combinations of a factorial in a square of side 4, with rows and columns acting as incomplete blocks that confound different interactions: here rows 1 and 2 hold {1, 2, 3, 4} and {5, 6, 7, 8}, rows 3 and 4 hold {1, 3, 6, 8} and {2, 4, 5, 7}, and the columns split the treatments a third way.

A Youden square is an incomplete Latin square whose rows are complete blocks, which these are not, and 'incomplete block design' describes one-way blocking, not a row-column layout.

Discuss in Forum

14.

Consider a two-way classification model \(y_{ij} = \mu + \tau_{i} + \beta_{j} + \varepsilon_{ij}\) \(i = 1 \ldots v\), \(j = 1, \ldots, b\) with \(E(\varepsilon_{ij}) = 0\), \(\mathrm{Var}(\varepsilon_{ij}) = \sigma^{2}\), \(\varepsilon_{ij}\) uncorrelated where \(\tau_{i}\ \delta = 1 \ldots v\) are fixed effects and \(\beta_{j}\ j = 1, \ldots, b\) are random effects with mean 0 and variance \(\sigma_{\beta}^{2}\). Hence the hypothesis regarding the insignificance of random effects is :

A.\(H_{0} : \beta_{1} = \beta_{2} = \cdots = \beta_{b}\)
B.\(H_{0} : \sigma_{\beta}^{2} = 0\)
C.\(H_{0} : \tau_{1} = \tau_{2} = \cdots = \tau_{v}\)
D.\(H_{0} : \sigma^{2} = 0\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The hypothesis is \(H_{0} : \sigma_{\beta}^{2} = 0\).

In a mixed model the block effects \(\beta_{j}\) are random variables drawn from a population with variance \(\sigma_{\beta}^{2}\). Interest is in that population, not in the particular blocks observed, so 'no block effect' means the random effects have no variability: \(\sigma_{\beta}^{2} = 0\).

\(\beta_{1} = \cdots = \beta_{b}\) is the hypothesis for fixed effects, \(\tau_{1} = \cdots = \tau_{v}\) concerns the fixed treatment effects, and \(\sigma^{2} = 0\) would mean no experimental error at all.

Discuss in Forum

15.

The incidence matrix of a general block design is :

\[ N = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \end{bmatrix} \]

Hence the number of estimable orthogonal block contrasts is :

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The number of estimable orthogonal block contrasts is 2.

N has 4 rows (treatments) and 3 columns (blocks): block 1 contains treatments 1 and 2, block 2 contains 1, 3 and 4, and block 3 contains 1 and 4. Treatment 1 appears in every block, linking them all, so the design is connected.

In a connected design all block contrasts are estimable, and with \(b = 3\) blocks there are \(b - 1 = 2\) independent (orthogonal) block contrasts. 3 would be the number of blocks, and 1 or 4 do not match \(b - 1\).

Discuss in Forum

16.

Consider a BIBD with parameter \((v, b, r, k, \lambda)\). If \(v = b = 11\) and \(r = 5\), then :

A.\(k = 5, \lambda = 2\)
B.\(k = 5, \lambda = 1\)
C.\(k = 2, \lambda = 5\)
D.\(k = 11, \lambda = 1\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The parameters are \(k = 5, \lambda = 2\).

For a BIBD, \(bk = vr\) and \(\lambda(v - 1) = r(k - 1)\). With \(v = b = 11\) and \(r = 5\):

\[ \begin{aligned} 11k &= 11 \cdot 5 \Rightarrow k = 5 \\ 10\lambda &= 5 \cdot 4 \Rightarrow \lambda = 2 \end{aligned} \]

(A symmetric design has \(k = r\).) \(\lambda = 1\) would need \(10 = 20\), and \(k = 2\) or \(k = 11\) contradicts \(bk = vr\).

Discuss in Forum

17.

In a \(2^{4}\) factorial design with two blocks of 8 plots each the treatment combinations are allotted as below :

Block I	Block II
a	1
b	ab
ac	c
bc	abc
d	ad
abd	bd
cd	acd
abcd	bcd

Hence the confounded treatment combination is :

A.ABD
B.ACD
C.BCD
D.ABCD

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The confounded interaction is ABD.

Block II contains the control (1), so it is the principal block: every treatment in it has an even number of letters in common with the confounded interaction. Test ABD against Block II:

ab: a, b (2); c: none (0); abc: a, b (2)
ad: a, d (2); bd: b, d (2)
acd: a, d (2); bcd: b, d (2)

All even, while every Block I treatment (a, b, ac, bc, d, abd, cd, abcd) shares an odd number of letters with ABD. ACD fails at once on ab in Block II (one letter, a), BCD fails on ab (one letter, b), and ABCD fails on c (one letter).

Discuss in Forum

18.

To examine whether five different skin creams A, B, C, D and E have different effect on the human body, \(n\) randomly chosen persons were enrolled in an experiment. The five creams were applied on each of one randomly chosen fingres of the same person; and the procedure is repeated for each person. The resulting design is :

A.a completely randomized design with 5 treatments
B.a randomized block design with 5 blocks and \(n\) treatments
C.a randomized block design with \(n\) blocks and five treatments
D.a repeated measures design

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

The design is a randomized block design with \(n\) blocks and five treatments.

Each person is a block: all five creams are applied to that person, on randomly chosen fingers, so differences between people are removed from the comparison of creams. With \(n\) persons and 5 creams, there are \(n\) blocks of 5 plots and 5 treatments.

A completely randomized design would assign creams to persons without blocking. Option B reverses the roles of blocks and treatments, and the creams are applied simultaneously to different fingers rather than in sequence over time, so this is treated as an RBD rather than a repeated measures design.

Discuss in Forum

19.

For one-way classification model the F-test is valid under the assumption that :

A.The observations on response variable are uncorrelated with common mean and variance
B.The errors are uncorrelated with common mean and variance
C.The errors are normally distributed with common mean zero and common positive variance
D.The errors are independently normally distributed with common mean zero and common +ve variance

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

The F-test is valid when the errors are independently normally distributed with mean zero and common positive variance.

In the model \(y_{ij} = \mu + \tau_{i} + e_{ij}\), the ratio of the between-group to within-group mean squares has an exact F distribution under \(H_{0}\) only if the \(e_{ij}\) are independent \(N(0, \sigma^{2})\). Independence and normality make the two sums of squares independent chi-square variables.

Uncorrelated errors (B) or observations (A) with common variance give unbiased estimates, but not an exact F distribution. Option C states normality but omits independence, which the F distribution also needs.

Discuss in Forum

20.

Under a \(2^{3}\) factorial design with factors A, B, C arranged in a single replicate of two blocks, the containts of block-I (in Yate's notation) are {1, a, ab, b}. Which of the following effects is confounded with blocks ?

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

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

The effect confounded with blocks is C.

Block I contains 1, a, b, ab, exactly the treatment combinations in which factor C is at its low level, and block II therefore contains c, ac, bc, abc. The main effect of C is the contrast (c + ac + bc + abc) - (1 + a + b + ab), which is the same as the block difference, so C cannot be separated from blocks.

For AB the block would have to be {1, ab, c, abc}, for AC {1, ac, b, abc}, and for ABC {1, ab, ac, bc}. Confounding a main effect is unusual in practice, but it is what this allocation does.

Design of Experiments Questions and Answers

Topics

Other Subjects

Tags: Design of Experiments Questions and Answers

About Us

Design of Experiments Questions and Answers

Topics

Other Subjects

Tags: Design of Experiments Questions and Answers

About Us

Stay informed about all important Dates & Notifications