### 6.3 - ANOVA

```CHAPTER 6
Statistical Inference & Hypothesis Testing
• 6.1 - One Sample
 Mean μ, Variance σ 2, Proportion π
• 6.2 - Two Samples
 Means, Variances, Proportions
μ1 vs. μ2 σ12 vs. σ22
π1 vs. π2
• 6.3 - Multiple Samples
 Means, Variances,
μ1, …, μk σ12, …, σk2
Proportions
π1, …, πk
CHAPTER 6
Statistical Inference & Hypothesis Testing
• 6.1 - One Sample
 Mean μ, Variance σ 2, Proportion π
• 6.2 - Two Samples
 Means, Variances, Proportions
μ1 vs. μ2 σ12 vs. σ22
π1 vs. π2
• 6.3 - Multiple Samples
 Means, Variances,
μ1, …, μk σ12, …, σk2
Proportions
π1, …, πk
Example: Y = “\$ Cost of a certain medical service”
Assume Y is known to be normally distributed at each of k = 2 health care facilities (“groups”).
Hospital: Y1 ~ N(μ1, σ1)
Clinic: Y2 ~ N(μ2, σ2)
• Null Hypothesis H0: μ1 = μ2,
i.e., μ1 – μ2 = 0
(“No difference exists.")
2-sided test at significance level α = .05
• Data: Sample 1 = {667, 653, 614, 612, 604}; n1 = 5
• Analysis via T-test (if equivariance holds):
“Group Means” y1 
“Group
2
Variances” s1 
5
2
( 667  630 ) 2 
2
s ppooled

o o led
2
2
 ( 604  630 ) 2
5 1
SS1
s2 = SS/df
Pooled
Variance
667  653  614  612  604
 630
Sample 2 = {593, 525, 520}; n2 = 3
y 
Point estimates
y2 
 788.5 s 2 
2
593  525  520
3
2
( 593  546 ) 2 

y1  y 2  84
 546
2
 ( 520  546 ) 2
3 1
yi / n
 1663
F 
1663
788.5
NOTE:
>0

 2.11  4
SS2
2
2
( 5n11)(
3 s1)(
1663 )
1) 7s18 8.5()n2 ( 1)
2
n1  n52 32 2
 1080
The pooled variance is a weighted average of the group
variances, using the degrees of freedom as the weights.
Example: Y = “\$ Cost of a certain medical service”
Assume Y is known to be normally distributed at each of k = 2 health care facilities (“groups”).
Hospital: Y1 ~ N(μ1, σ1)
Clinic: Y2 ~ N(μ2, σ2)
• Null Hypothesis H0: μ1 = μ2,
i.e., μ1 – μ2 = 0
(“No difference exists.")
2-sided test at significance level α = .05
• Data: Sample 1 = {667, 653, 614, 612, 604}; n1 = 5
Sample 2 = {593, 525, 520}; n2 = 3
• Analysis via T-test (if equivariance holds): Point estimates
“Group Means” y1 
“Group
2
Variances” s1 
5
2
( 667  630 ) 2 
s2 = SS/df
Pooled
Variance
667  653  614  612  604
2
 ( 604  630 ) 2
5 1
 630
y2 
 788.5 s 2 
2
593  525  520
3
2
( 593  546 ) 2 
y 

yi / n
y1  y 2  84
 546
2
 ( 520  5546
46 ) 2
3 1
 1663
F 
1663
788.5
NOTE:
>0
 2.11  4
SSErr = 6480
2
s ppooled

o o led
2
2
2
( 5n11)(
3 s1)(
1663 )
1) 7s18 8. 5()n2 ( 1)
2
n1  n5232 2
dfErr = 6
Standard
s.e.0 
Error
2
pooled
s
1080
11 1 1
  24
5n1 3 n 2
 1080
The pooled variance is a weighted average of the group
variances, using the degrees of freedom as the weights.
p-value = 2 P (Y1  Y2  84 )  2 P  T6 
84  0
24
  2 P  T6
 3.5
> 2 * (1 - pt(3.5, 6)) Reject H0 at α = .05
stat signif, Hosp > Clinic
[1] 0.01282634

R code:
> y1 = c(667, 653, 614, 612, 604)
> y2 = c(593, 525, 520)
>
> t.test(y1, y2, var.equal = T)
Formal Conclusion
Two Sample t-test
p-value < α = .05
Reject H0 at this level.
data: y1 and y2
t = 3.5, df = 6, p-value = 0.01283
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
25.27412 142.72588
Interpretation
sample estimates:
mean of x mean of y
The samples provide evidence that the
630
546
difference between mean costs is (moderately)
statistically significant, at the 5% level,
with the hospital being higher than the clinic
(by an average of \$84).
Alternate method ~
 Main Idea: Among several (k  2) independent, equivariant,
normally-distributed “treatment groups”…
“Total Variability” = “Variability between groups” + “Variability within groups”
Y1
Y2
Yk
k
1
1
H0 :
1
k
2
2
=
2
=
HA: “At least one ‘treatment mean’ μi is
significantly different from the others.
= k
Example: Y = “\$ Cost of a certain medical service”
Assume Y is known to be normally distributed at each of k = 2 health care facilities (“groups”).
Hospital: Y1 ~ N(μ1, σ1)
Clinic: Y2 ~ N(μ2, σ2)
• Null Hypothesis H0: μ1 = μ2,
i.e., μ1 – μ2 = 0
(“No difference exists.")
2-sided test at significance level α = .05
• Data: Sample 1 = {667, 653, 614, 612, 604}; n1 = 5
• ANOVA F-test (if equivariance holds): 
“Group Means” y1 
667  653  614  612  604
5
 630
Sample 2 = {593, 525, 520}; n2 = 3
Point estimates
y2 
593  525  520
3
y 
 546
3 (546)
5 (630)
“Grand Mean”
y 
667  653  614  612  604  593  525  520
53
 598.50
The grand mean is a weighted average of the group
means, using the sample sizes as the weights.

yi / n
y1  y 2  84
NOTE:
>0
Alternate method ~
 Main Idea: Among several (k  2) independent, equivariant,
normally-distributed “treatment groups”…
“Total Variability” = “Variability between groups” + “Variability within groups”
Y1
Y2
Yk
k
1
1
H0 :
1
k
2
2
=
2
=
HA: “At least one ‘treatment mean’ μi is
significantly different from the others.
= k
Example: Y = “\$ Cost of a certain medical service”
Assume Y is known to be normally distributed at each of k = 2 health care facilities (“groups”).
Hospital: Y1 ~ N(μ1, σ1)
Clinic: Y2 ~ N(μ2, σ2)
• Null Hypothesis H0: μ1 = μ2,
i.e., μ1 – μ2 = 0
(“No difference exists.")
2-sided test at significance level α = .05
• Data: Sample 1 = {667, 653, 614, 612, 604}; n1 = 5
• ANOVA F-test (if equivariance holds): 
“Group Means” y1 
“Grand Mean”
667  653  614  612  604
5
y 
 630
Point estimates
y2 
5 ( 630 )  3( 546 )
53
Sample 2 = {593, 525, 520}; n2 = 3
593  525  520
3
y 
 546
 598.50
How far is the “total” sample from the grand mean?

yi / n
Example: Y = “\$ Cost of a certain medical service”
Assume Y is known to be normally distributed at each of k = 2 health care facilities (“groups”).
Hospital: Y1 ~ N(μ1, σ1)
Clinic: Y2 ~ N(μ2, σ2)
• Null Hypothesis H0: μ1 = μ2,
i.e., μ1 – μ2 = 0
(“No difference exists.")
2-sided test at significance level α = .05
• Data: Sample 1 = {667, 653, 614, 612, 604}; n1 = 5
• ANOVA F-test (if equivariance holds): 
“Group Means” y1 
667  653  614  612  604
5
y 
“Grand Mean”
SSTot = (667  598.5 )
 630
2
Point estimates
y2 
5 ( 630 )  3( 546 )
53
593  525  520
3
y 

yi / n
 546
 598.50
 (653  598.5 )  (614  598.5 )  (612  598.5 )  (604  598.5 )
2
 (593  598.5 )  (525  598.5 )  (520  598.5 )
2
Sample 2 = {593, 525, 520}; n2 = 3
2
2
2
2
= 19710
2
dfTot = (5+3) –1 = 7
Alternate method ~
 Main Idea: Among several (k  2) independent, equivariant,
normally-distributed “treatment groups”…
“Total Variability” = “Variability between groups” + “Variability within groups”
Y1
Y2
Yk
k
1
1
H0 :
1
k
2
2
=
How can we measure this?
2
=
= k
Imagine zero variability within groups…
Alternate method ~
 Main Idea: Among several (k  2) independent, equivariant,
normally-distributed “treatment groups”…
“Total Variability” = “Variability between groups” + “Variability within groups”
Y1
Y2
Yk
k
1
1
H0 :
1
k
2
2
=
How can we measure this?
2
=
= k
Imagine zero variability within groups…
Example: Y = “\$ Cost of a certain medical service”
Assume Y is known to be normally distributed at each of k = 2 health care facilities (“groups”).
Hospital: Y1 ~ N(μ1, σ1)
Clinic: Y2 ~ N(μ2, σ2)
• Null Hypothesis H0: μ1 = μ2,
i.e., μ1 – μ2 = 0
(“No difference exists.")
2-sided test at significance level α = .05
• Data: Sample 1 = {667,
{630, 653,
630, 614,
630, 612,
630, 630
604};
} n1 = 5
• ANOVA F-test (if equivariance holds): 
“Group Means” y1 
667  653  614  612  604
5
y 
“Grand Mean”
SSTot = (667  598.5 )
 630
2
Point estimates
y2 
5 ( 630 )  3( 546 )
53
593  525  520
3
y 

yi / n
 546
 598.50
 (653  598.5 )  (614  598.5 )  (612  598.5 )  (604  598.5 )
2
2
 (593  598.5 )  (525  598.5 )  (520  598.5 )
2
SSTrt =
Sample 2 = {593,
{546, 525,
546, 520};
546} n2 = 3
2
5 ( 630  598.5 )  3 ( 546  598.5 )
2
“The Clonemaster”
2
= 13230
2
2
= 19710
2
dfTot = (5+3) –1 = 7
dfTrt = (2) –1
=1
Alternate method ~
 Main Idea: Among several (k  2) independent, equivariant,
normally-distributed “treatment groups”…
“Total Variability” = “Variability between groups” + “Variability within groups”
Y1
Y2
Yk
k
1
1
H0 :
1
k
2
2
=
2
=
= k
Example: Y = “\$ Cost of a certain medical service”
Assume Y is known to be normally distributed at each of k = 2 health care facilities (“groups”).
Hospital: Y1 ~ N(μ1, σ1)
Clinic: Y2 ~ N(μ2, σ2)
• Null Hypothesis H0: μ1 = μ2,
i.e., μ1 – μ2 = 0
(“No difference exists.")
2-sided test at significance level α = .05
• Data: Sample 1 = {667, 653, 614, 612, 604}; n1 = 5
• ANOVA F-test (if equivariance holds): 
“Group Means” y1 
667  653  614  612  604
5
y 
“Grand Mean”
SSTot = (667  598.5 )
 630
2
Point estimates
y2 
5 ( 630 )  3( 546 )
53
593  525  520
3
y 

yi / n
 546
 598.50
 (653  598.5 )  (614  598.5 )  (612  598.5 )  (604  598.5 )
2
2
 (593  598.5 )  (525  598.5 )  (520  598.5 )
2
SSTrt =
Sample 2 = {593, 525, 520}; n2 = 3
2
5 ( 630  598.5 )  3 ( 546  598.5 )
2
2
2
2
= 19710
= 13230
How far is each sample from its own group mean?
2
dfTot = (5+3) –1 = 7
dfTrt = (2) –1
=1
Example: Y = “\$ Cost of a certain medical service”
Assume Y is known to be normally distributed at each of k = 2 health care facilities (“groups”).
Hospital: Y1 ~ N(μ1, σ1)
Clinic: Y2 ~ N(μ2, σ2)
• Null Hypothesis H0: μ1 = μ2,
i.e., μ1 – μ2 = 0
(“No difference exists.")
2-sided test at significance level α = .05
• Data: Sample 1 = {667, 653, 614, 612, 604}; n1 = 5
• ANOVA F-test (if equivariance holds): 
“Group Means” y1 
667  653  614  612  604
 630
5
y 
“Grand Mean”
SSTot = (667  598.5 )
2
Point estimates
y2 
5 ( 630 )  3( 546 )
53
593  525  520
2
2
2
2
2
2
dfTrt = (2) –1
 (653  630 )  (614  630 )  (612  630 )  (604  630 )
2
2
2
2
2
dfTot = (5+3) –1 = 7
= 19710
= 13230
 (593  546 )  (525  546 )  (520  546 )
2
yi / n
 598.50
2
5 ( 630  598.5 )  3 ( 546  598.5 )
2

 (653  598.5 )  (614  598.5 )  (612  598.5 )  (604  598.5 )
2
SSErr = (667  630 )
y 
 546
3
 (593  598.5 )  (525  598.5 )  (520  598.5 )
SSTrt =
Sample 2 = {593, 525, 520}; n2 = 3
2
2
=1
Example: Y = “\$ Cost of a certain medical service”
Assume Y is known to be normally distributed at each of k = 2 health care facilities (“groups”).
Hospital: Y1 ~ N(μ1, σ1)
Clinic: Y2 ~ N(μ2, σ2)
• Null Hypothesis H0: μ1 = μ2,
i.e., μ1 – μ2 = 0
(“No difference exists.")
2-sided test at significance level α = .05
• Data: Sample 1 = {667, 653, 614, 612, 604}; n1 = 5
Sample 2 = {593, 525, 520}; n2 = 3
• Analysis via T-test (if equivariance holds): Point estimates
“Group Means” y1 
“Group
2
Variances” s1 
5
2
( 667  630 ) 
2
s ppooled

o o led
2
 (6
04  630 )
604
2
5 1
SS1
s2 = SS/df
Pooled
Variance
667  653  614  612  604
 630
y2 
 788.5 s 2 
2
593  525  520
3
2
( 593  546 ) 2 
y 

y1  y 2  84
 546
2
 ( 520  5546
46 ) 2
3 1
yi / n
 1663
F 
1663
788.5
NOTE:
>0
 2.11  4
SS2
2
2
( 5n11)(
3 s1)(
1663 )
1) 7s18 8.5()n2 ( 1)
2
n1  n52 32 2
 1080
The pooled variance is a weighted average of the group
variances, using the degrees of freedom as the weights.
Example: Y = “\$ Cost of a certain medical service”
Assume Y is known to be normally distributed at each of k = 2 health care facilities (“groups”).
Hospital: Y1 ~ N(μ1, σ1)
Clinic: Y2 ~ N(μ2, σ2)
• Null Hypothesis H0: μ1 = μ2,
i.e., μ1 – μ2 = 0
(“No difference exists.")
2-sided test at significance level α = .05
• Data: Sample 1 = {667, 653, 614, 612, 604}; n1 = 5
Sample 2 = {593, 525, 520}; n2 = 3
• Analysis via T-test (if equivariance holds): Point estimates
“Group Means” y1 
“Group
2
Variances” s1 
5
2
( 667  630 ) 
s2 = SS/df
Pooled
Variance
667  653  614  612  604
 (6
04  630 )
604
2
5 1
 630
y2 
 788.5 s 2 
2
593  525  520
3
2
( 593  546 ) 2 
y 

y1  y 2  84
 546
2
 ( 520  5546
46 ) 2
3 1
yi / n
 1663
F 
1663
788.5
NOTE:
>0
 2.11  4
SSErr = 6480
2
s ppooled

o o led
2
2
2
( 5n11)(
3 s1)(
1663 )
1) 7s18 8.5()n2 ( 1)
2
n1  n52 32 2
dfErr = 6
 1080
The pooled variance is a weighted average of the group
variances, using the degrees of freedom as the weights.
Example: Y = “\$ Cost of a certain medical service”
Assume Y is known to be normally distributed at each of k = 2 health care facilities (“groups”).
Hospital: Y1 ~ N(μ1, σ1)
Clinic: Y2 ~ N(μ2, σ2)
• Null Hypothesis H0: μ1 = μ2,
i.e., μ1 – μ2 = 0
(“No difference exists.")
2-sided test at significance level α = .05
• Data: Sample 1 = {667, 653, 614, 612, 604}; n1 = 5
• ANOVA F-test (if equivariance holds): 
“Group Means” y1 
667  653  614  612  604
 630
5
y 
“Grand Mean”
SSTot = (667  598.5 )
2
Point estimates
y2 
5 ( 630 )  3( 546 )
53
593  525  520
2
2
2
2
2
2
dfTrt = (2) –1
 (653  630 )  (614  630 )  (612  630 )  (604  630 )
2
2
2
2
2
dfTot = (5+3) –1 = 7
= 19710
= 13230
 (593  546 )  (525  546 )  (520  546 )
2
yi / n
 598.50
2
5 ( 630  598.5 )  3 ( 546  598.5 )
2

 (653  598.5 )  (614  598.5 )  (612  598.5 )  (604  598.5 )
2
SSErr = (667  630 )
y 
 546
3
 (593  598.5 )  (525  598.5 )  (520  598.5 )
SSTrt =
Sample 2 = {593, 525, 520}; n2 = 3
2
2
=1
Example: Y = “\$ Cost of a certain medical service”
Assume Y is known to be normally distributed at each of k = 2 health care facilities (“groups”).
Hospital: Y1 ~ N(μ1, σ1)
Clinic: Y2 ~ N(μ2, σ2)
• Null Hypothesis H0: μ1 = μ2,
i.e., μ1 – μ2 = 0
(“No difference exists.")
2-sided test at significance level α = .05
• Data: Sample 1 = {667, 653, 614, 612, 604}; n1 = 5
• ANOVA F-test (if equivariance holds): 
“Group Means” y1 
667  653  614  612  604
5
y 
“Grand Mean”
SSTot = (667  598.5 )
 630
2
Point estimates
y2 
5 ( 630 )  3( 546 )
53
593  525  520
3
y 

yi / n
 546
 598.50
 (653  598.5 )  (614  598.5 )  (612  598.5 )  (604  598.5 )
2
2
 (593  598.5 )  (525  598.5 )  (520  598.5 )
2
SSTrt =
Sample 2 = {593, 525, 520}; n2 = 3
2
5 ( 630  598.5 )  3 ( 546  598.5 )
2
2
= 13230
SSErr = 4 ( 788.5 )  2 (1663 ) = 6480
SSTot = SSTrt + SSErr
2
2
= 19710
2
dfTot = (5+3) –1 = 7
dfTrt = (2) –1
=1
dfErr = (5+3) –2 = 6
dfTot = dfTrt + dfErr
SSTot = SSTrt + SSErr
Tot
dfTot = dfTrt + dfErr
Err
Trt
MS 
ANOVA Table
SS
F 
M S T rt
F1,6
M S E rr
df
12.25
Source
Treatment
df
SS
MS
1
13230
13230
  s b etw een 
F-ratio
p-value
12.25
.01282634
2
Error
Total
6
7
6480
19710
SSTot = SSTrt + SSErr
1080
  s w2 ith in 
–
on F1, 6 1–pf(12.25, 1, 6)
F-table: comp w/ α
Note:
2
This is also s p o o led .
dfTot = dfTrt + dfErr
SSTot = SSTrt + SSErr
dfTot = dfTrt + dfErr
Err
MS 
ANOVA Table
SS
df
F 
M S T rt
F1,6
M S E rr
12.25
Source
Treatment
df
SS
1
MS
F-ratio
p-value
12.25
.01282634
13230
  s b etw een 
2
Error
Total
6
6480
on F1, 6 1–pf(12.25, 1, 6)
F-table: comp w/ α
–
7
Thus, the treatment accounts for
1080
  s w2 ith in 
13230
19710
= 67.1% of the total variability in the response Y.
R code:
# ANOVA FOR UNBALANCED DESIGN
> y1 = c(667, 653, 614, 612, 604)
> y2 = c(593, 525, 520)
>
> Data = data.frame(
+
Y = c(y1, y2),
+
X = factor(rep(c("y1", "y2"), times = c(length(y1),
length(y2))))
+
)
>
> var.test(Y ~ X, data = Data)
# EQUIVARIANCE?
F test to compare two variances
data: Y by X
F = 0.4741, num df = 4, denom df = 2,
p-value = 0.4738
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
0.01208057 5.04920249
sample estimates:
ratio of variances
0.4741431

R code:
# ANOVA FOR UNBALANCED DESIGN
> out = aov(Y ~ X, data = Data)
> anova(out)
Analysis of Variance Table
Response: Y
Df Sum Sq Mean Sq F value Pr(>F)
X
1 13230
13230
12.25 0.01283 *
Residuals 6
6480
1080
--Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Note: Vis-à-vis T-test vs. F-test,
• p-value is the same using either method (.01283), since the sample is unchanged!
• The square of the Tdf -score (3.5) is equal to the F1, df -score (12.25).
(Recall that the square of the Z-score is equal to the  1 -score.)
2
Alternate method ~
 Main Idea: Among several (k  2) independent, equivariant,
normally-distributed “treatment groups”…
Y1
Y2
Yk
k
1
1
H0:
1
k
2
2
=
2
=
= k
Alternate method ~
 Main Idea: Among several (k  2) independent, equivariant,
normally-distributed “treatment groups”…
• Equivariance can be tested via very similar “two variances” F-test in
6.2.2 (but this is very sensitive to normality assumption), or others.
If violated, can extend Welch Test for two means.
Y1
Y2
Yk
k
1
1
H0:
1
k
2
2
=
2
=
= k
Alternate method ~
 Main Idea: Among several (k  2) independent, equivariant,
normally-distributed “treatment groups”…
• Normality can be tested via usual methods.
If violated, use nonparametric Kruskal-Wallis Test.
Y1
Y2
Yk
k
1
1
H0:
1
k
2
2
=
2
=
= k
Alternate method ~
 Main Idea: Among several (k  2) independent, equivariant,
normally-distributed “treatment groups”…
• Extensions of ANOVA for data in matched “blocks” designs,
repeated measures, multiple factor levels within groups, etc.
Y1
Y2
Yk
k
1
1
H0:
1
k
2
2
=
2
=
= k
Alternate method ~
 Main Idea: Among several (k  2) independent, equivariant,
normally-distributed “treatment groups”…
• How to identify significant group(s)? Pairwise testing, with correction
(e.g., Bonferroni) for spurious significance.
• Example: k = 5 groups result in 10 such tests, so let each α* = α / 10.
Y1
Y2
Yk
k
1
1
H0:
1
k
2
2
=
2
=
= k
```