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.
This is an example of a balanced design, since each factor (i.e. Sorority) has the same number of observations.
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)=k1=41=3
df(denom)
=
n

k
=
20

4
=
16
df(denom)=nk=204=16
Calculate the test statistic:
F
=
2.23
F=2.23
Graph:
Probability statement:
pvalue
=
P
(
F
>
2.23
)
=
0.1241
pvalue=P(F>2.23)=0.1241
Compare
α
α and the
pvalue
pvalue:
α
=
0.01
pvalue
=
0.1241
α=0.01pvalue=0.1241
α
<
pvalue
α<pvalue
Make a decision:
Since
α
<
pvalue
α<pvalue, you cannot reject
H
o
H
o
.
Conclusion: There is not sufficient evidence to conclude that there is a
difference among the mean grades for the sororities.
TI83+ 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 pvalue
pvalue
is 0.1241.
df(numerator) = 3df(numerator) = 3 (under "Factor"
) and df(denominator) = 16df(denominator) = 16 (under Error
).
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