Download AP Statistics: ANOVA Section 2

AP Statistics: ANOVA Section 2
In the previous section, we saw how to use
ANOVA to test for a difference in means among
several groups. However, that test only tells us
when differences exist, not which specific
groups differ. The goal of this section is to adapt
the inference procedures of section 13.1 to use
the results of the ANOVA analysis. This will allow
us to find a confidence interval for the mean of
any group, find a confidence interval for a
difference in means between two groups and
test when that difference is significant.
In earlier inference work, we used formulas such
as those below for doing inference about a
single mean or a difference in means.
If we have found an ANOVA table based on samples
from several groups, we make a couple of small
adjustments to these formulas.
* Estimate any standard deviation with
from the
ANOVA table.
Since one of the conditions for the ANOVA is that
the standard deviation is the same for each
group, using
gives an estimate that is based
on all of the samples and not just one.
* Use the error degrees of freedom, n – k, for any tdistributions.
We often call
the pooled standard deviation
Inference for Means After ANOVA
After doing an ANOVA for a difference in means
among k groups based on sample of size
Confidence interval for :
xi  t
*
MSE
ni
Confidence interval for :
( xi  x j )  t *
1 1
MSE   
n n 
j 
 i
If the ANOVA indicates that there are differences
among the means:
Pairwise test of i vs  j :
t
xi  x j
1 1
MSE   
n n 
j 
 i
where MSE is the mean square error from the
ANOVA table and the t-distribution use n – k
degrees of freedom
Example: Let’s go back to the ants and
sandwich fillings example from section 1.
Find and interpret a 95% confidence interval for the
mean number of ants attracted to a peanut butter
sandwich.
138.7
34  2.080
8
(25.34,42.66)
We are 95% confident that the mean number of ants attracted
to a peanut butter sandwich is between 25.34 and 42.66
Find and interpret a 95% confidence interval for the
difference in average ant counts between vegemite
and ham & pickles sandwiches.
1 1
30.75  49.25  2.080 138.7  
8 8
 30.75,6.25
We are 95% confident the difference between the mean number of
ants attacted to vegemite vs ham & pickles is between - 30.75 and - 6.25.
Note: Since our confidence interval
contains only negative numbers, this
implies that our two population means
differ – this is not surprising since
ANOVA done earlier indicated that at
least two of the groups have different
means.
Test at the 5% level for a difference in mean number
of ants between vegemite and peanut butter
sandwiches.
H 0 : V   PB
H a : V   PB
where each  is the mean number of
ants attracted to that sandwich filling
conditions were checked in ANOVA section 1
t
30.75  34
1 1
138.7  
8 8
 0.55
tcdf (1000,0.55,21)  .294
p - value  2(.294)  .588
Since our p - value is larger than any common
significance level we do not reject the H 0 . We
cannot conclude that there is a difference in the
mean number of attracted to vegemite vs peanut
butter sandwiches.