Chapter 12, Problem 17P

A researcher used ANOVA and computed and F-ration for the following data.

	Treatments
I	II	III
$n=10$	$n=10$	$n=10$
$M=20$	$M=28$	$M=35$
$SS=105$	$SS=191$	$SS=180$

If the mean for treatment III were changed to $M=25$

1. , what would happen to the size of the F-ratio (increase or decrease)? Explain your answer.

If the SS for treatment I were changed to $SS=1400$

, what would happen to the size of the F-ratio (increase or decrease)? Explain your answer.

To determine

1. If the mean for treatment III were changed to $M=25$, what would happen to the size of the F-ratio (increase or decrease)? Explain your answer.
2. If the $SS$ for treatment I were changed to $SS=1400$, what would happen to the size of the F-ratio (increase or decrease)? Explain your answer.

Answer to Problem 17P

Solution:

1. If the mean for the treatment III were changed to $M=25$ from $M=35,$ the size of the F-ratio will decrease. This change will affect the numerator of the F-ratio. Also we can see after changing the sample mean for treatment III, the sample means will come closer to each other and therefore the treatments have less between effects.
2. If the $SS$ for treatment I were changed to $SS=1400$ from $SS=105$, the size of the F-ratio will decrease. This change will affect the denominator of the F-ratio. Actually it will increase the Mean sum of squares within treatments, thereby reducing the F-ratio.

Explanation of Solution

1. If the mean for the treatment III were changed to $M=25$ from $M=35,$ the size of the F-ratio will decrease. This change will affect the numerator of the F-ratio. Also we can see after changing the sample mean for treatment III, the sample means will come closer to each other and therefore the treatments have less between effects.

   Let's take a look how it decreases the F-ratio:

   We are given:

   	Treatments
   	I	II	III
   n	10	10	10
   M	20	28	35
   SS	105	191	180
   T	200	280	350

   S S within and $S{S}_{Between}$ :

   $S{S}_{within}={\displaystyle \sum _{}^{}S{S}_{\text{inside each treatment}}=S{S}_I+S{S}_{II}+S{S}_{III}=105+191+180=476}$

   $S{S}_{Between}={\displaystyle \sum _{}^{}\frac{T^2}{n}}-\frac{G^2}{N}$

   $=\left(\frac{{200}^2}{10}+\frac{{280}^2}{10}+\frac{{350}^2}{10}\right)-\left(\frac{{(200+280+350)}^2}{30}\right)$

   $=\left(4000+7840+12250\right)-22963.33$

   $=24090-22963.33=1126.7$

   $d{f}_{Between}=k-1=3-1=2$

   $d{f}_{within}=N-k=30-3=27$

   $M{S}_{Between}=\frac{S{S}_{Between}}{d{f}_{Between}}=\frac{1126.7}{2}=563.35$

   $M{S}_{within}=\frac{S{S}_{within}}{d{f}_{within}}=\frac{476}{27}=17.63$

   Therefore the F-ratio $=\frac{M{S}_{Between}}{M{S}_{within}}=\frac{563.35}{17.63}=31.95$

   Now if the mean for treatment III were changed to $M=25$, there will be impact on the F-ratio.

   	Treatments
   	I	II	III
   n	10	10	10
   M	20	28	25
   SS	105	191	180
   T	200	280	250

   $S{S}_{within}={\displaystyle \sum _{}^{}S{S}_{\text{inside each treatment}}=S{S}_I+S{S}_{II}+S{S}_{III}=105+191+180=476}$

   $S{S}_{Between}={\displaystyle \sum _{}^{}\frac{T^2}{n}}-\frac{G^2}{N}$

   $=\left(\frac{{200}^2}{10}+\frac{{280}^2}{10}+\frac{{250}^2}{10}\right)-\left(\frac{{(200+280+250)}^2}{30}\right)$

   $=\left(4000+7840+6250\right)-17763.33$

   $=18090-17763.33=326.7$

   $d{f}_{Between}=k-1=3-1=2$

   $d{f}_{within}=N-k=30-3=27$

   $M{S}_{Between}=\frac{S{S}_{Between}}{d{f}_{Between}}=\frac{326.7}{2}=163.35$

   $M{S}_{within}=\frac{S{S}_{within}}{d{f}_{within}}=\frac{476}{27}=17.63$

   Therefore the F-ratio $=\frac{M{S}_{Between}}{M{S}_{within}}=\frac{163.35}{17.63}=9.27$

   Thus we clearly see if we change the mean for treatment III to $M=25,$ the F-ratio got decreased.

2. If the $SS$ for treatment I were changed to $SS=1400$ from $SS=105$, the size of the F-ratio will decrease. This change will affect the denominator of the F-ratio. Actually it will increase the Mean sum of squares within treatments, thereby reducing the F-ratio.
Let's take a look how it decreases the F-ratio:

We are given:

	Treatments
	I	II	III
n	10	10	10
M	20	28	35
SS	105	191	180
T	200	280	350

S S within and $S{S}_{Between}$ :

$S{S}_{within}={\displaystyle \sum _{}^{}S{S}_{\text{inside each treatment}}=S{S}_I+S{S}_{II}+S{S}_{III}=105+191+180=476}$

$S{S}_{Between}={\displaystyle \sum _{}^{}\frac{T^2}{n}}-\frac{G^2}{N}$

$=\left(\frac{{200}^2}{10}+\frac{{280}^2}{10}+\frac{{350}^2}{10}\right)-\left(\frac{{(200+280+350)}^2}{30}\right)$

$=\left(4000+7840+12250\right)-22963.33$

$=24090-22963.33=1126.7$

$d{f}_{Between}=k-1=3-1=2$

$d{f}_{within}=N-k=30-3=27$

$M{S}_{Between}=\frac{S{S}_{Between}}{d{f}_{Between}}=\frac{1126.7}{2}=563.35$

$M{S}_{within}=\frac{S{S}_{within}}{d{f}_{within}}=\frac{476}{27}=17.63$

Therefore the F-ratio $=\frac{M{S}_{Between}}{M{S}_{within}}=\frac{563.35}{17.63}=31.95$ Now if the $SS$ for treatment I were changed to $SS=1400$, there will be impact on the F-ratio.

Now let's see how the F-ratio will be impacted by changing the $SS$ for treatment I.

	Treatments
	I	II	III
n	10	10	10
M	20	28	35
SS	1400	191	180
T	200	280	350

$S{S}_{within}={\displaystyle \sum _{}^{}S{S}_{\text{inside each treatment}}=S{S}_I+S{S}_{II}+S{S}_{III}=1400+191+180=1771}$

$S{S}_{Between}={\displaystyle \sum _{}^{}\frac{T^2}{n}}-\frac{G^2}{N}$

$=\left(\frac{{200}^2}{10}+\frac{{280}^2}{10}+\frac{{350}^2}{10}\right)-\left(\frac{{(200+280+350)}^2}{30}\right)$ $=\left(4000+7840+12250\right)-22963.33$

$=24090-22963.33=1126.7$

$d{f}_{Between}=k-1=3-1=2$

$d{f}_{within}=N-k=30-3=27$

$M{S}_{Between}=\frac{S{S}_{Between}}{d{f}_{Between}}=\frac{1126.7}{2}=563.35$

$M{S}_{within}=\frac{S{S}_{within}}{d{f}_{within}}=\frac{1771}{27}=65.6$

Therefore the F-ratio $=\frac{M{S}_{Between}}{M{S}_{within}}=\frac{163.35}{65.6}=8.59$ Thus we clearly see if we change the $SS$ for treatment I to $SS=1400,$ the F-ratio got decreased.

Conclusion:

1. If the mean for the treatment III were changed to $M=25$ from $M=35,$ the size of the F-ratio will decrease. This change will affect the numerator of the F-ratio. Also we can see after changing the sample mean for treatment III, the sample means will come closer to each other and therefore the treatments have less between effects.
2. If the $SS$ for treatment I were changed to $SS=1400$ from $SS=105$, the size of the F-ratio will decrease.