Chapter 11.1, Problem 13E

a.

To determine

Prove that a constant d can be added or subtracted to each $x_{ij}$ without affecting the sum of squares in ANOVA table.

Explanation of Solution

Calculation:

The model used to predict $Y_{ij}$ is given below:

$Y_{ij}=X_{ij}$

To this model a constant d is added,

$Y_{ij}=X_{ij}+d$

The mean response due to the $i^{th}$ level of factor A is,

${\overline{Y}}_{i\cdot }={\overline{X}}_{i\cdot }+d$

Where,

${\overline{X}}_{i\cdot }=\frac{{\displaystyle \sum _j^J{X}_{ij}}}{J}$

The mean response due to the $j^{th}$ level of factor B is,

${\overline{Y}}_{\cdot j}={\overline{X}}_{\cdot j}+d$

Where,

${\overline{X}}_{\cdot j}=\frac{{\displaystyle \sum _i^I{X}_{ij}}}{I}$

The overall mean response of Y is given below:

${\overline{Y}}_{\cdot \cdot }={\overline{X}}_{\cdot \cdot }+d$

Where,

${\overline{X}}_{\cdot \cdot }=\frac{{\displaystyle \sum _i^I{\displaystyle \sum _j^J{X}_{ij}}}}{IJ}$

The total sum of squares is given below:

$\begin{array}{c}\text{SST}={\displaystyle \sum _i^I{{\displaystyle \sum _j^J\left(X_{ij}-{\overline{X}}_{\cdot \cdot }\right)}}^2}\\ ={\displaystyle \sum _i^I{{\displaystyle \sum _j^J\left(Y_{ij}-\cancel{d}-{\overline{Y}}_{\cdot \cdot }+\cancel{d}\right)}}^2}\\ ={\displaystyle \sum _i^I{{\displaystyle \sum _j^J\left(Y_{ij}-{\overline{Y}}_{\cdot \cdot }\right)}}^2}\end{array}$

The error sum of squares is given below:

$\begin{array}{c}\text{SSE}={\displaystyle \sum _i^I{{\displaystyle \sum _j^J\left(X_{ij}-{\overline{X}}_{i\cdot }-{\overline{X}}_{\cdot j}+{\overline{X}}_{\cdot \cdot }\right)}}^2}\\ ={\displaystyle \sum _i^I{{\displaystyle \sum _j^J\left(Y_{ij}-\cancel{d}-{\overline{Y}}_{\cdot \cdot }+\cancel{d}\right)}}^2}\\ ={\displaystyle \sum _i^I{{\displaystyle \sum _j^J\left(Y_{ij}-d-\left({\overline{Y}}_{i\cdot }-d\right)-\left({\overline{Y}}_{\cdot j}-d\right)+{\overline{Y}}_{\cdot \cdot }-d\right)}}^2}\\ ={\displaystyle \sum _i^I{{\displaystyle \sum _j^J\left(Y_{ij}-\cancel{d}-{\overline{Y}}_{i\cdot }+\cancel{d}-{\overline{Y}}_{\cdot j}+\cancel{d}+{\overline{Y}}_{\cdot \cdot }-\cancel{d}\right)}}^2}\end{array}$

$={\displaystyle \sum _i^I{{\displaystyle \sum _j^J\left(Y_{ij}-{\overline{Y}}_{i\cdot }-{\overline{Y}}_{\cdot j}+{\overline{Y}}_{\cdot \cdot }\right)}}^2}$

Using the similar method, the sum of squares due to factor A and B are calculated.

Sum of squares due to factor A:

$\text{SSA}={\displaystyle \sum _i^I{{\displaystyle \sum _j^J\left({\overline{Y}}_{i\cdot }-{\overline{Y}}_{\cdot \cdot }\right)}}^2}$

Sum of squares due to factor B:

$\text{SSB}={\displaystyle \sum _i^I{{\displaystyle \sum _j^J\left({\overline{Y}}_{\cdot j}-{\overline{Y}}_{\cdot \cdot }\right)}}^2}$

From all the sum of squares it can observed that the addition of a constant d has no effect on the sum of squares.

b.

To determine

Identify the change in the sum of squares due to multiplication of constant c.

Identify the changes in the F-statistic for factor A and B.

Find the effect of $y_{ij}=c{x}_{ij}+d$ on the conclusions of ANOVA.

Answer to Problem 13E

The sum of squares remains unchanged and the F-statistic for factor A and B also remains unchanged.

The conclusions for ANOVA remains unchanged by using $y_{ij}=c{x}_{ij}+d$.

Explanation of Solution

Calculation:

The model used to predict $Y_{ij}$ is given below:

$Y_{ij}=X_{ij}$

To this model a constant c is multiplied,

$Y_{ij}=c{X}_{ij}$

Then the sum of squares would have the square of the constant c and it is shown below:

Total sum of squares:

$\text{SST}=c^2{\displaystyle \sum _i^I{{\displaystyle \sum _j^J\left(Y_{ij}-{\overline{Y}}_{\cdot \cdot }\right)}}^2}$

Sum of squares due to factor A:

$\text{SSA}=c^2{\displaystyle \sum _i^I{{\displaystyle \sum _j^J\left({\overline{Y}}_{i\cdot }-{\overline{Y}}_{\cdot \cdot }\right)}}^2}$

Sum of squares due to factor B:

$\text{SSB}=c^2{\displaystyle \sum _i^I{{\displaystyle \sum _j^J\left({\overline{Y}}_{\cdot j}-{\overline{Y}}_{\cdot \cdot }\right)}}^2}$

Sum of squares due to error:

$\text{SSE}=c^2{\displaystyle \sum _i^I{{\displaystyle \sum _j^J\left(Y_{ij}-{\overline{Y}}_{i\cdot }-{\overline{Y}}_{\cdot j}+{\overline{Y}}_{\cdot \cdot }\right)}}^2}$

Mean sum of squares for factor A:

$\begin{array}{c}\text{MSA = }\frac{\text{SSA}}{I-1}\\ =\frac{c^2{\displaystyle \sum _i^I{{\displaystyle \sum _j^J\left({\overline{Y}}_{i\cdot }-{\overline{Y}}_{\cdot \cdot }\right)}}^2}}{I-1}\end{array}$

Mean sum of squares for factor B:

$\begin{array}{c}\text{MSB = }\frac{\text{SSB}}{J-1}\\ =\frac{c^2{\displaystyle \sum _i^I{{\displaystyle \sum _j^J\left({\overline{Y}}_{\cdot j}-{\overline{Y}}_{\cdot \cdot }\right)}}^2}}{J-1}\end{array}$

Mean sum of squares for error:

$\begin{array}{c}\text{MSE = }\frac{\text{SSB}}{\left(I-1\right)\left(J-1\right)}\\ =\frac{c^2{\displaystyle \sum _i^I{{\displaystyle \sum _j^J\left(Y_{ij}-{\overline{Y}}_{i\cdot }-{\overline{Y}}_{\cdot j}+{\overline{Y}}_{\cdot \cdot }\right)}}^2}}{\left(I-1\right)\left(J-1\right)}\end{array}$

Thus, the F-statistic for factor A would be,

$\begin{array}{c}{f}_A=\frac{\text{MSA}}{\text{MSE}}\\ =\frac{\frac{\cancel{c^2}{\displaystyle \sum _i^I{{\displaystyle \sum _j^J\left({\overline{Y}}_{i\cdot }-{\overline{Y}}_{\cdot \cdot }\right)}}^2}}{I-1}}{\frac{\cancel{c^2}{\displaystyle \sum _i^I{{\displaystyle \sum _j^J\left(Y_{ij}-{\overline{Y}}_{i\cdot }-{\overline{Y}}_{\cdot j}+{\overline{Y}}_{\cdot \cdot }\right)}}^2}}{\left(I-1\right)\left(J-1\right)}}\\ =\frac{\frac{{\displaystyle \sum _i^I{{\displaystyle \sum _j^J\left({\overline{Y}}_{i\cdot }-{\overline{Y}}_{\cdot \cdot }\right)}}^2}}{I-1}}{\frac{{\displaystyle \sum _i^I{{\displaystyle \sum _j^J\left(Y_{ij}-{\overline{Y}}_{i\cdot }-{\overline{Y}}_{\cdot j}+{\overline{Y}}_{\cdot \cdot }\right)}}^2}}{\left(I-1\right)\left(J-1\right)}}\end{array}$

For factor B:

$\begin{array}{c}{f}_A=\frac{\text{MSA}}{\text{MSE}}\\ =\frac{\frac{\cancel{c^2}{\displaystyle \sum _i^I{{\displaystyle \sum _j^J\left({\overline{Y}}_{\cdot j}-{\overline{Y}}_{\cdot \cdot }\right)}}^2}}{J-1}}{\frac{\cancel{c^2}{\displaystyle \sum _i^I{{\displaystyle \sum _j^J\left(Y_{ij}-{\overline{Y}}_{i\cdot }-{\overline{Y}}_{\cdot j}+{\overline{Y}}_{\cdot \cdot }\right)}}^2}}{\left(I-1\right)\left(J-1\right)}}\\ =\frac{\frac{{\displaystyle \sum _i^I{{\displaystyle \sum _j^J\left({\overline{Y}}_{\cdot j}-{\overline{Y}}_{\cdot \cdot }\right)}}^2}}{J-1}}{\frac{{\displaystyle \sum _i^I{{\displaystyle \sum _j^J\left(Y_{ij}-{\overline{Y}}_{i\cdot }-{\overline{Y}}_{\cdot j}+{\overline{Y}}_{\cdot \cdot }\right)}}^2}}{\left(I-1\right)\left(J-1\right)}}\end{array}$

From part (a) the addition of constant d has no effect on ANOVA and also the multiplication of constant c has no effect on ANOVA.

Hence, coding the model as $y_{ij}=c{x}_{ij}+d$ has no effect on the conclusions of ANOVA.