This time, we will perform the calculations that lead to the F' statistic. Notice that
each group has the same number of plants so we will use the formula
F'
=
n
⋅
(
s
x¯
)
2
(
s
pooled
)
2
F'
=
n
⋅
(
s
x
)
2
(
s
pooled
)
2
size 12{F= { {n cdot \( s rSub { size 8{  x} } \) rSup { size 8{2} } } over { \( s rSub { size 8{ ital "pooled"} } \) rSup { size 8{2} } } } } {}
.
First, calculate the sample mean and sample variance of each group.
Table 3

Tommy's Plants 
Tara's Plants 
Nick's Plants 
Sample Mean 
24.2 
25.4 
24.4 
Sample Variance 
11.7 
18.3 
16.3 
Next, calculate the variance of the three group means (Calculate the variance of 24.2,
25.4, and 24.4). Variance of the group means = 0.413 =
(s
x¯
)2(s
x
)2
Then
MS
between
=
n
(s
x¯
)2
=
(
5
)
(
0.413
)
MS
between
=n(s
x
)2=(5)(0.413) where n=5n=5 is the sample size (number of plants
each child grew).
Calculate the average of the three sample variances (Calculate the average of 11.7,
18.3, and 16.3). Average of the sample variances = 15.433
=
(spooled)2
=(spooled)2 size 12{ \( s rSub { size 8{ ital "pooled"} } \) rSup { size 8{2} } ={}} {}
Then
MS
within
=
(spooled)2
=
15.433
MS
within
=(spooled)2 size 12{ \( s rSub { size 8{ ital "pooled"} } \) rSup { size 8{2} } ={}} {}=15.433.
The
F
F statistic (or
F
F ratio) is
F
=
MS
between
MS
within
=
n
⋅
(
s
x¯
)
2
(
s
pooled
)
2
=
(
5
)
⋅
(
0.413
)
15.433
=
0.134
F=
MS
between
MS
within
=
n
⋅
(
s
x
)
2
(
s
pooled
)
2
=
(
5
)
⋅
(
0.413
)
15.433
=0.134
The dfs for the numerator =
the number of groups

1
=
3

1
=
2
the number of groups1=31=2
The dfs for the denominator =
the
total number of samples

the number
of groups
=
15

3
=
12
the
total number of samplesthe number
of groups=153=12
The distribution for the test is
F
2
,
12
F
2
,
12
and the F statistic is
F
=
0.134
F=0.134
The pvalue is
P
(
F
>
0.134
)
=
0.8759
P(F>0.134)=0.8759.
Decision: Since
α
=
0.03
α=0.03
and the
pvalue
=
0.8759
pvalue=0.8759, do not reject
H
o
H
o
. (Why?)
Conclusion: With a 3% the level of significance, from the sample data, the evidence is
not sufficient to conclude that the average heights of the bean plants are not different.
Of the three media tested, it appears that it does not matter which one the bean plants
are grown in.
(This experiment was actually done by three classmates of the son of one of the
authors.)