Chapter 13, Problem 9P

For the data in problem 8,

a. Compute $S{S}_{\text{total}}$ and $S{S}_{\text{between treatments}}$ .

b. Eliminate the mean difference between treatments by adding 2 points to each score in treatment I, adding 1 point to each score in treatment II, and subtracting 3 points from each score in treatment II. (All three treatments should end up with $M=\text{3}$ and $T=15$ .)

c. Calculate $S{S}_{\text{total}}$ for the modified scores. (Caution: You first must find the new value for ${\displaystyle \sum X^2}$ .)

d. Because the treatment effects were eliminated in part b, you should find that $S{S}_{\text{total}}$ for the modified scores is smaller than $S{S}_{\text{total}}$ for the original scores. The difference between the two SS values should be exactly equal to the value of $S{S}_{\text{between treatments}}$ for the original scores.

To determine

For the data in problem 8,

1. Compute $S{S}_{total}\text{ and }S{S}_{\text{between treatments}}$ .
2. Eliminate the mean differences between treatments by adding 2 points to each score in treatment I, adding 1 point to each score in treatment II and subtracting 3 points from each score in treatment III (All three treatments should end up with $M=3$ and $T=15.)$
3. Calculate $S{S}_{total}$ for the modified scores. (Caution: You first must find the new value for ${\displaystyle \sum _{}^{}{X}^2}$ )
4. Because the treatment effects were eliminated in part b, you should find that $S{S}_{total}$ for the modified scores is smaller than $S{S}_{total}$ for the original scores. The difference between the two $SS$ values should be exactly equal to the value of $S{S}_{\text{between treatments}}$ for the original scores.

Answer to Problem 9P

Solution:

1. $S{S}_{total}=96$ and $S{S}_{\text{between treatments}}=70$.
2. To eliminate the mean differences between treatments we will add 2 points to each score in treatment I add 1 point to each score in treatment II and subtract 3 points from each score in treatment III. The resulting table is given below:

	Treatments
Person	I	II	III	Person Totals
A	3	2	1	P=6
B	5	5	5	P=15	N=15
C	2	3	4	P=9	G=45
D	2	1	3	P=6	${\displaystyle \sum _{}^{}{X}^2=}$ 161
E	3	4	2	P=9
	M=3	M=3	M=3
	T=15	T=15	T=15

We clearly see that the Mean of all the three treatments = 3 and all the three treatments have $T=15$.

3. The modified $S{S}_{total}$ =26
4. The original $S{S}_{total}$ =96 and modified $S{S}_{total}$ =26, therefore the modified $S{S}_{total}$ is smaller than original $S{S}_{total}$.

The difference between the original $S{S}_{total}$ and modified $S{S}_{total}$ is

$96-26=70$

Also we know the original $S{S}_{\text{between treatments}}=70$. Therefore the difference the two $SS$ values is exactly equal to the value of $S{S}_{\text{between treatments}}$ for the original scores..

Explanation of Solution

1. We are given the below data in problem 8:

	Treatments
Person	I	II	III	Person Totals
A	1	1	4	P=6
B	3	4	8	P=15	N=15
C	0	2	7	P=9	G=45
D	0	0	6	P=6	${\displaystyle \sum _{}^{}{X}^2=}$ 231
E	1	3	5	P=9
	M=1	M=2	M=6
	T=5	T=10	T=30
	SS=6	SS=10	SS=10

$S{S}_{total}={\displaystyle \sum _{}^{}{X}^2-\frac{G^2}{N}=231-\frac{{45}^2}{15}=231-135=96}$

$S{S}_{\text{Between Treatments}}={\displaystyle \sum _{}^{}\frac{T^2}{n}-\frac{G^2}{N}=\frac{5^2}{5}+\frac{{10}^2}{5}+\frac{{30}^2}{5}-\frac{{45}^2}{15}}=5+20+180-135=205-135=70$

2. To eliminate the mean differences between treatments we will add 2 points to each score in treatment I add 1 point to each score in treatment II and subtract 3 points from each score in treatment III. The resulting table is given below:

	Treatments
Person	I	II	III	Person Totals
A	3	2	1	P=6
B	5	5	5	P=15	N=15
C	2	3	4	P=9	G=45
D	2	1	3	P=6	${\displaystyle \sum _{}^{}{X}^2=}$ 161
E	3	4	2	P=9
	M=3	M=3	M=3
	T=15	T=15	T=15

We clearly see that the Mean of all the three treatments = 3 and all the three treatments have T=15 .

3. Calculate $S{S}_{total}$ for the modified scores.

To calculate the $S{S}_{total}$ for the modified scores, we have

$S{S}_{total}={\displaystyle \sum _{}^{}{X}^2-\frac{G^2}{N}=161-\frac{{45}^2}{15}=161-135=26}$

Therefore the modified $S{S}_{total}$ =26

4. The original $S{S}_{total}$ =96 and modified $S{S}_{total}$ =26, therefore the modified $S{S}_{total}$ is smaller than original $S{S}_{total}$.

The difference between the original $S{S}_{total}$ and modified $S{S}_{total}$ is

$96-26=70$

Also we know the original $S{S}_{\text{between treatments}}=70$. Therefore the difference the two $SS$ values is exactly equal to the value of $S{S}_{\text{between treatments}}$ for the original scores.

Conclusion:

1. $S{S}_{total}=96$ and $S{S}_{\text{between treatments}}=70$.
2. To eliminate the mean differences between treatments we will add 2 points to each score in treatment I add 1 point to each score in treatment II and subtract 3 points from each score in treatment III. The resulting table is given below:

	Treatments
Person	I	II	III	Person Totals
A	3	2	1	P=6
B	5	5	5	P=15	N=15
C	2	3	4	P=9	G=45
D	2	1	3	P=6	${\displaystyle \sum _{}^{}{X}^2=}$ 161
E	3	4	2	P=9
	M=3	M=3	M=3
	T=15	T=15	T=15

We clearly see that the Mean of all the three treatments = 3 and all the three treatments have T=15 .

3. The modified $S{S}_{total}$ =26
4. The original $S{S}_{total}$ =96 and modified $S{S}_{total}$ =26, therefore the modified $S{S}_{total}$ is smaller than original $S{S}_{total}$.
The difference between the original $S{S}_{total}$ and modified $S{S}_{total}$ is

$96-26=70$

Also we know the original $S{S}_{\text{between treatments}}=70$. Therefore the difference the two $SS$ values is exactly equal to the value of $S{S}_{\text{between treatments}}$ for the original scores..