Design of Experiments Questions and Answers

Answer: D

It is equivalent to a t-test with alternative hypothesis 'not equal to'.

With a single factor at two levels, the one-way ANOVA F statistic equals the square of the pooled two-sample t statistic, \(F = t^{2}\), with 1 and N − 2 degrees of freedom. Because squaring removes the sign, large F corresponds to large \(|t|\) in either direction, so the ANOVA F-test is the two-sided t-test.

A one-sided t-test ('greater than') is not equivalent, because F cannot tell the direction. An F-test with a 'less than' alternative and a chi-square test are not the matching procedures.

Answer: B

The C-matrix is positive semidefinite.

The C-matrix of a block design, \(C = R - NK^{-1}N'\), is symmetric, and \(\tau'C\tau\) is the adjusted treatment sum of squares for a vector of effects, which is never negative. Its rows sum to zero, so \(C\mathbf{1} = 0\) and C is singular, with rank at most r − 1 for r treatments (exactly r − 1 for a connected design).

So C is not positive definite and not of rank r. Being symmetric, it is not asymmetric.

Answer: B

The variance is σ²(N − n_i)/(Nn_i).

\(\bar{y}_{..}\) is the overall mean of all N observations and includes the \(n_{i}\) observations of group i, so

\[ \begin{aligned} Cov(\bar{y}_{i.}, \bar{y}_{..}) &= \frac{n_{i}}{N} \cdot \frac{\sigma^{2}}{n_{i}} \\ &= \frac{\sigma^{2}}{N} \end{aligned} \]

Then

\[ \begin{aligned} Var(\bar{y}_{i.} - \bar{y}_{..}) &= \frac{\sigma^{2}}{n_{i}} + \frac{\sigma^{2}}{N} - \frac{2\sigma^{2}}{N} \\ &= \sigma^{2}\frac{N - n_{i}}{Nn_{i}} \end{aligned} \]

The other options do not follow from this calculation: \(\sigma^{2}/Nn_{i}\) is far too small, and the others put \(n_{i}\) in the numerator where it does not belong.

Answer: B

The values are a = 7.5 and b = 19.

The error degrees of freedom are \(11 - 3 - 2 = 6\), so the error mean square is \(\frac{9.0}{6} = 1.5\). The block F ratio is the block mean square over the error mean square:

\[ \begin{aligned} \frac{a/2}{1.5} &= 2.5 \Rightarrow a = 7.5 \\ b &= 2.5 + 7.5 + 9.0 = 19 \end{aligned} \]

The total sum of squares is the sum of the component sums of squares. \(a = 5.0\) uses the wrong divisor, and \(b = 10\) or 7.5 leave out one of the components.

Answer: D

Neither (I) nor (II) is untrue: both are properties of the C matrix.

The C matrix of a block design is \(C = R - N K^{-1} N'\), where R and K are diagonal matrices of replications and block sizes and N is the incidence matrix. Each term is symmetric, so C is symmetric, which is (I).

Every row of C sums to 0, because

\[ \begin{aligned} C\mathbf{1} &= \mathbf{r} - NK^{-1}\mathbf{k} \\ &= \mathbf{r} - N\mathbf{1} = \mathbf{0}. \end{aligned} \]

By symmetry every column also sums to 0, so any row sum equals any column sum, which is (II).

Since the question asks which is NOT true, the answer is neither.

Answer: C

The total sum of squares is 150.

In a 5 × 5 Latin square, the treatment sum of squares has 4 degrees of freedom and the error has \((5 - 1)(5 - 2) = 12\). Use the F statistic to find the error sum of squares:

\[ \begin{aligned} MS_{T} &= \frac{60}{4} = 15 \\ F &= \frac{15}{MS_{E}} = 30 \Rightarrow MS_{E} = 0.5 \\ SS_{E} &= 0.5 \times 12 = 6 \end{aligned} \]

Then the total is \(80 + 4 + 60 + 6 = 150\).

The other totals do not fit the given F. A total of 146 would leave \(SS_{E} = 2\) and F = 90; 158 would leave 14 and F of about 12.9; 175 would leave 31 and F of about 5.8.