Let
μ
1
μ
1
,
μ
2
μ
2
,
μ
3
μ
3
,
μ
4
μ
4
be the population means of the sororities. Remember that the null
hypothesis claims that the sorority groups are from the same normal distribution.
The alternate hypothesis says that at least two of the sorority groups come from
populations with different normal distributions. Notice that the four sample sizes are
each size 5.
H
o
:
μ
1
=
μ
2
=
μ
3
=
μ
4
H
o
:
μ
1
=
μ
2
=
μ
3
=
μ
4
H
a
H
a
: Not all of the means
μ
1
,
μ
2
,
μ
3
,
μ
4
μ
1
,
μ
2
,
μ
3
,
μ
4
are equal.
Distribution for the test:
F
3
,
16
F
3
,
16
where
k
=
4 groupsk=4 groups
and
N
=
20 samples in total
N=20 samples in total
df(num)
=
k
-
1
=
4
-
1
=
3
df(num)=k-1=4-1=3
df(denom)
=
N
-
k
=
20
-
4
=
16
df(denom)=N-k=20-4=16
Calculate the test statistic:
F
=
2.23
F=2.23
Graph:
Probability statement:
p-value
=
P
(
F
>
2.23
)
=
0.1241
p-value=P(F>2.23)=0.1241
Compare
α
α and the
p-value
p-value:
α
=
0.01
p-value
=
0.1242
α
<
p-value
α=0.01p-value=0.1242α<p-value.
Make a decision:
Since
α
<
p-value
α<p-value, you cannot reject
H
o
H
o
.
This means that the population averages appear to be the same.
Conclusion: There is not sufficient evidence to conclude that there is a
difference among the grade averages for the sororities.
TI-83+ or TI 84: Put the data into lists L1, L2, L3, and L4. Press STAT and
arrow over to TESTS. Arrow down to F:ANOVA. Press ENTER and Enter
(L1,L2,L3,L4). The F statistic is 2.2303 and the p-value
p-value
is 0.1241.
df(numerator) = 3df(numerator) = 3 (under "Factor") and df(denominator) = 16df(denominator) = 16 (under Error).