Chapter 14
Sections 1, 2, 3: ANOVA
Recall in Chapter 10 we tested the hypotheses
In this chapter we test the hypothesis
HI: There is at least one difference among the pairs of population means.
Assumptions we make:
1. The populations have approximately normal distributions.
2. The samples are random samples.
3. The samples are independent.
4. The standard deviations from the populations are equal. (it has been shown that as long as
the sample sizes are nearly equal, the variances can differ by amounts that make the largest up
to nine times the smallest, and ANOVA still is reliable.
Now suppose we have three groups of data
Grand Mean
Over All
ni Total scores
#of groups
It turns out that the total variation about the grand mean is embodied in the sum of sum of
squared deviations, ier X-ry called the total sum of squares SST
so SST
Xr- or equivalently, it can be shown algebraically that
SST
The computations ofANOVA involve partitioning the total variation about the grand mean
into two components:
1. The variation that originates from the differences among the group means, i.e., Yu, Yn, Tc,
etc., called (SSG) (the group sum of squares).
2. The random variation within the groups themselves called error, SSE)
(error sum of
squares)
We do this by partitioning SST into the group sum of squares, SSG, and the error sum of
squares, SSE. Thus we have
SST SSG +SSE
Now the formula for SSG is as follows:
SSG
Chapter 14 Sections 1, 2, 3: ANOVA Recall in Chapter 10 we tested the hypotheses In this chapter we test the hypothesis HI: There is at least one difference among the pairs of population means. Assumptions we make: 1. The populations have approximately normal distributions. 2. The samples are random samples. 3. The samples are independent. 4. The standard deviations from the populations are equal. (it has been shown that as long as the sample sizes are nearly equal, the variances can differ by amounts that make the largest up to nine times the smallest, and ANOVA still is reliable. Now suppose we have three groups of data Grand Mean Over All ni Total scores #of groups It turns out that the total variation about the grand mean is embodied in the sum of sum of squared deviations, ier X-ry called the total sum of squares SST so SST Xr- or equivalently, it can be shown algebraically that SST The computations ofANOVA involve partitioning the total variation about the grand mean into two components: 1. The variation that originates from the differences among the group means, i.e., Yu, Yn, Tc, etc., called (SSG) (the group sum of squares). 2. The random variation within the groups themselves called error, SSE) (error sum of squares) We do this by partitioning SST into the group sum of squares, SSG, and the error sum of squares, SSE. Thus we have SST SSG +SSE Now the formula for SSG is as follows: SSGconditions with the following results. The letters, A, M, and L denote automatic, manual, and
lockup torque. At the 0.05 significance level, test the claim that the mean fuel consumption
values are the same
23
23
29
26
20
25
24
23
24
We will do this in clas
Example: A pilot does extensive bad weather flying and decides to buy a battery-powered radio
as an independent backup for his regular radios, which depend on the airplanes electrical
system. Ha has the choice of 3 brands that vary in cost. His obtains the sample data shown. He
cts 5 batt
d test
recharging is necessary. Do the 3 brands have the same mean usable time before recharging is
required? Test at the 0.05 significance level
pud lsts in L
Brand
X
Brand Y
26.0
29.0
30.0
28.5
26.3
27.3
27.6
29.2
d :0.05
25.9
27.1
29.8
28.2
27.0
We will do this in clas
0,24
0,05
ent sampl
SSE
do not
SST SSG SSE
Now the degrees of freedom for SST i
DFT nr.
-1
The degrees of freedom for k groups (SSG) is
DFG k
So the degrees of freedom for in-group variation (SSE) is
DFE DFT DFG which is
DFE
Now we find the variance estimate (mean squares) for groups and error.
S

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