Chapter 12
Analysis of Variance
Introduction
Chapters 2-4: techniques to describe data
(敘述性統計 圖表 統計量)
Chapters 5-8: probability, probability distributions
Sampling methods and Central Limit Theorem (CLT)
(機率分配,抽樣方法與中央極限定理)
Introduction
Inferential statistics (推論性統計):
(1) Estimation 估計
(2) Hypothesis testing假說檢定;假設檢定
Chapter 9 Estimation
- point estimation
- interval estimation
Chapter 10 One-sample tests of hypothesis
一個樣本的假設檢定
* population mean
* population proportion
Chapter 11 Two-sample tests of hypothesis
* two population means
* two population proportions
Chapter 12 Analysis of variance (ANOVA, 變異數分析)
* two population variances
* three or more population means
H0: µ1 = µ2 = µ3 = µ4
H1: The means are not all equal
Learning Objectives
LO1 the F distribution
LO2 test the variances of two normal populations
Test statistic F=s²1/s²2
LO3 test three or more population means
- one-way ANOVA (單因子變異數分析)
- two-way ANOVA (雙因子變異數分析)
12-5
LO1: The F Distribution
 It was named to honor
Sir Ronald Fisher
(1890-1962),
father of modern statistics.
12-6
The F Distribution
 It is applied when we want to
 Compare two populations variances
 Compare several population means
simultaneously. It is called “analysis of
variance” (ANOVA).
In both situations, the populations must follow
a normal distribution.
12-7
Characteristics of F-Distribution
1. There is a “family” of F
Distributions.
2. Each F distribution is
determined by two
parameters:
12-8
Characteristics of F-Distribution
1. There is a “family” of F
Distributions.
2. Each F distribution is
determined by two
parameters:
(1) the degrees of freedom
in the numerator
(分子自由度)
(2) the degrees of freedom
in the denominator.
12-9
Characteristics of F-Distribution
3. It is a continuous
distribution.
4. F value is nonnegative.
5. The F distribution is
positively skewed.
6. It is asymptotic. As F 
 the curve approaches
the X-axis but never
touches it.
12-10
Comparing variances of two normal populations
Examples:

A car manufacturer is about to unveil a new, faster
car. However, initial tests indicate there is more
variation in the processing time than the current cars.

A sample of 15 technology and 15 utility stocks shows the
same mean rate of return, but there is more variation in the
Internet stocks.
utility stock: 公用事業公司股票
12-11
Equal means but different variances
LO2: Test for Equal Variances
Test statistic F=s²1/s²2
Note: F value is nonnegative.
12-13
Test statistic F=s²1/s²2
Question: When do we reject Hₒ ?
Answer: reject Hₒ if F is not close to 1.
Next Question: How close is close ?
12-14
Rejection region
F=s²1/s²2 follows Fv1,v2 when σ12 = σ22
where v1=n1-1
v2=n2-1
12-16
One-sided and two-sided tests
(a)
H0: σ12 = σ22
H1: σ12 > σ22
(b)
H0: σ12 = σ22
H1: σ12 < σ22
(c)
H0: σ12 = σ22
H1: σ12 ≠ σ22
Test statistic F=s²1/s²2
(a) Reject Hₒ if F is far above 1
(b) Reject Hₒ if F is far below 1
(c) Reject Hₒ if F is not close to 1
One-sided and two-sided tests
(a)
H0: σ12 = σ22
H1: σ12 > σ22
(b)
H0: σ12 = σ22
H1: σ12 < σ22
(c)
H0: σ12 = σ22
H1: σ12 ≠ σ22
Test statistic F=s²1/s²2
(a) Reject Hₒ if F is far above 1 (if F > F,v1,v2)
(b) Reject Hₒ if F is far below 1 (if F < F1- ,v1,v2)
(c) Reject Hₒ if F is not close to 1
(if F > F/2,v1,v2 or if F < F1- /2,v1,v2)
More on two-sided test
H0: σ12 = σ22
H1: σ12 ≠ σ22
Reject Hₒ if F is not close to 1
Reject Hₒ if F > F/2,v1,v2 or if F < F1- /2,v1,v2
Question:
F/2,v1,v2 > 1 or < 1 ?
F1- /2,v1,v2 > 1 or < 1 ?
More on two-sided test
H0: σ12 = σ22
H1: σ12 ≠ σ22
Reject Hₒ if F is not close to 1
Reject Hₒ if F > F/2,v1,v2 or if F < F1- /2,v1,v2
Answer:
F/2,v1,v2 > 1
F1- /2,v1,v2 < 1
Test for Equal Variances - Example
The following are the mean rate of returns of
7 technology and 8 utility stocks.
Technology: 52 67 56 45 70 54 64
Utility
: 59 60 61 51 56 63 57 65
Using the .10 significance level, is there a difference in
the variation in the driving times for the two types of
stocks?
X 1  58.28571; X 2  59.
12-21
Test for Equal Variances - Example
Step 1: The hypotheses are:
H0: σ12 = σ22
H1: σ12 ≠ σ22
Step 2: The significance level is .10.
Step 3: The test statistic is the F distribution.
12-22
Test for Equal Variances - Example
Step 4: State the decision rule.
Reject H0 if F > F/2,v1,v2
F > F.10/2,7-1,8-1
F > F.05,6,7
12-23
Test for Equal Variances - Example
Step 5: Compute the value of F and make a decision
The decision is to reject the null hypothesis, because the
computed F value (4.23) is larger than the critical value (3.87).
We conclude that there is a difference in the variation of the
mean rate of returns of the two types of stocks.
12-24
Wait. Shouldn’t we check two critical values?
H0: σ12 = σ22
H1: σ12 ≠ σ22
Reject Hₒ if F > F/2,v1,v2 or if F < F1- /2,v1,v2
Reject H0 if F > F0.05,6,7
or F < F.95,6,7
Wait. Shouldn’t we check two critical values?
Decision rule:
Reject H0 if F > F0.05,6,7
or F < F.95,6,7
However, the table only provides the upper
critical values !
Approach 1: Place the larger sample variance in the
numerator; hence, we always have F=s²1/s²2 > 1.
Decision rule:
Reject H0 if F > F0.05,6,7
or F < F.95,6,7
Thus, F < F.95,6,7 would never happen;
only the right-tail critical value is required.
Advantage: reduce the size of the
table of critical values
Approach 2: Use the relation : F1 ,v 2 ,v1 
1
F ,v1,v 2
1
1
So, F0.95 , 6 , 7 

 0.2375.
F0.05 , 7 , 6
4.21
The reason is below :
s12
  P( 2
s2
s12
 F ,v1,v 2 ) where 2 ~ F(v1, v2)
s2
s22
1
s22
 P( 2 
) where 2 ~ F(v2, v1)
s1
F ,v1,v 2
s1
s22
 P ( 2  F1 ,v 2 ,v1 )
s1
So, F1 ,v 2 ,v1 
1
F ,v1,v 2
.
Approach 2 (continued)
For the derivation in the previous slide,
the 1st equality: definition
the 2nd equality: taking reciprocal on both sides
the 3rd equality: definition
The last line follows by comparing the 2nd and 3rd
equalities.
Approach 2 (continued)
Decision rule:
Reject H0 if F > F0.05,6,7 =3.87
or F < F.95,6,7 =0.2375
So, we have obtained the two critical values.
And we reject H0 .
Now, use approach 2 with smaller
variance in the numerator.
The original F statistic is
If the larger sample variance is not placed in the
numerator, then the computed F statistic = ?
F statistic: F=s²2/s²1 =1/(4.23)=0.2364.
So, we reject the null hypothesis because 0.2364 < 0.2583.
Decision rule:
Reject H0 if F > F0.05,7,6 =4.21
or F < F.95,7,6 =1/(3.87) =0.2583
LO3: Comparing Means of Two or More
Populations
- One-way analysis of variance
One-factor analysis of variance
(單因子變異數分析)
- Two-way analysis of variance
Two-factor analysis of variance
(雙因子變異數分析)
12-33
Comparing Means of Two or More Populations
The F distribution is also used for testing whether
three or more population means are equal.
H0: µ1 = µ2 =…= µk
H1: The means are not all equal
Assumptions:
– The populations follow the normal distribution.
– The populations have equal variance.
– The random samples are independent.
12-34
Comparing Means of Two or More Populations –
Example
A manager of a regional financial center
wishes to compare the productivity, as
measured by the number of customers
served, among three employees.
Four days are randomly selected and the
number of customers served by each
employee is recorded.
12-35
Comparing Means of Two or More Populations
Are the figures below fulfill the three assumptions ?
– The populations follow the normal distribution? (Yes)
– The populations have equal variance? (Yes)
– The random samples are independent? (Not sure)
12-36
ANOVA (變異數分析)
Goal:
Test whether there is a significant difference
between the treatment effect.
Treatment: 處方
*ANOVA was first developed for agriculture applications.
Idea:
decomposition of total variation (總變異) into
several parts
Wolfe
Whtie
Korosa
X G  58, grand mean
55
66
47
X 1  56, X 2  70, X 3  48 : group means
54
76
51
group 1 : X11 , X12 , X13 , X14
59
67
46
group 2 : X 21 , X 22 , X 23 , X 24
56
71
48
group 3 : X 31 , X 32 , X 33 , X 34
total variation :  ( X  X G ) 2
4
variation within group i :  ( X ij  X i ) 2
j 1
3
variation between group :  ( X i  X G ) 2
i 1
Idea: decomposition of total variation (總變異)
total variat ion
( X  X G )
2
 ( X  X c  X c  X G )
2
 ( X  X c )  ( X c  X G )  2( X  X c )( X c  X G )
2
2
 ( X  X c )  ( X c  X G )
2
2
組內變異
(within groups)
vs
組間變異
(between groups)
Idea: decomposition of total variation (總變異)
SSTotal (sum of squares total)
 ( X  X G )2
 ( X  X c )2  ( X c  X G )2
 SSE  SSTrt
 (SS for error)  (SS for treatm ent)
When H 0 is true, SST will be smaller;
when H 1 is true, SST will be larger.
SST
So, we will reject H 0 if
is too (large, small) ?
SSE
Analysis of variance (變異數分析)
SSTotal (sum of squares total)
 SSE  SSTrt
 (SS error)  (SS trtment)
SSTrt
is too large.
SSE
SSTrt/(k - 1)
test statistic : F 
SSE/(n - k)
So, we will reject H 0 if
When H 0 is true, the test statistic follows F distri.
with degrees of freedom k - 1 and n - k.
In this case, k  3, n  12.
Why (k-1) and (n-k) in
SSTrt /( k  1)
F 
SSE /( n  k )
?
Compare with F in equation (9-1):
s12
( X  X ) 2 /( n1  1)
F  2 
s2
 (Y  X ) 2 /( n2  1)
Recall: Each F distribution is determined by two
parameters:
(1) the degrees of freedom in the numerator
(分子自由度)
(2) the degrees of freedom in the denominator.
Analysis of Variance – F statistic
 k: number of populations
 n: number of observations
 The test statistic is computed by:
F
SST k  1
SSE n  k 
12-43
Anova table
SST/(k - 1) MST
F

SSE/(n - k) MSE
12-44
MSE: mean squared error (均方差)
Have you seen this MSE before?
What does the MSE estimate ?
MSE: mean squared error (均方差)
SSE   ( X  X c ) 2
SSE
MSE 
, n  n1  n 2  ...  n k
n-k
The values of residentia l homes :
k  2, n1  n 2  10,
10
10
 ( X 1i  X 1 )   ( X 2 i  X 2 ) 2
So, MSE  i 1
2
i 1
10  10  2
It is the pooled variance !
It estimates the common population variances.
Procedure of hypothesis testing
H0: µ1 = µ2 =…= µk
H1: The means are not all equal

Null: the population means are all the same.
Alternative: at least one of the means is different.

Choose significance level

Test Statistic: F distribution.

The Decision rule:

Reject H0 if F > F,k-1,n-k
Compute F and make decision
Question: Is the rejection
region on one or two tails?
Why ?
12-47
Commonly seen mistakes
H0: µ1 = µ2 =…= µk
H1: µ1 ≠ µ2 ≠…≠ µk

The Decision rule:
(not quite right)
Reject H0 if F > F/2,k-1,n-k
or F < F1-/2,k-1,n-k
(It is wrong!)
12-48
Example
A marketing researcher randomly selected and
surveyed customers from four stores regarding
their level of satisfaction with a recent purchase.
Twenty-five questions offered a range of possible
answers: excellent, good, fair, or poor; with a
score of 4, 3, 2, and1, respectively. These
responses were then totaled.
Is there a difference in the mean satisfaction level among
the four stores? Use the .01 significance level.
12-49
Eaton
Tony
Aden
Oz
94
75
70
68
90
68
73
70
85
77
76
72
80
83
78
65
88
80
74
68
65
total
65
Total
349
391
510
414
1,664
mean
87.25
78.20
72.86
69.00
75.64
Grand mean : 75.64
Group means: 87.25, 78.20, 72.86, 69.00
Comparing Means of Two or More Populations –
Example
Step 1: State the null and alternate hypotheses.
H0: µ1 = µ2 = µ3 = µk
H1: The means are not all equal
Reject H0 if F > F,k-1,n-k
Step 2: State the level of significance.
The .01 significance level is stated in the problem.
Step 3: Find the appropriate test statistic.
Because we are comparing means of more than two
groups, use the F statistic
12-51
Comparing Means of Two or More Populations –
Example
Step 4: State the decision rule.
Reject H0 if
F > F,k-1,n-k
F > F.01,4-1,22-4
F > F.01,3,18
F > 5.09
12-52
Step 5: Compute the value of F and make a decision
12-53
Eaton
Tony
Aden
Oz
94
75
70
68
90
68
73
70
85
77
76
72
80
83
78
65
88
80
74
68
65
total
65
mean
87.25
78.20
72.86
69.00
75.64
Need to calculate the following:
total variation :  ( X  X G ) 2
4
variation within group i :  ( X ij  X i ) 2
j 1
3
variation between group :  ( X i  X G ) 2
i 1
Eaton
Tony
Aden
Oz
18.36
-0.64
- 5.64
-7.64
14.36
-7.64
- 2.64
-5.64
9.36
1.36
0.36
-3.64
4.36
7.36
2.36
-10.64
12.36
4.36
-1.64
- 7.64
-10.64
Eaton
Tony
Aden
Oz
337.09
0.41
31.81
58.37
206.21
58.37
6.97
31.81
87.61
1.85
0.13
13.25
19.0
54.17
5.57
113.21
152.77 19.01
2.69
58.37
113.21
113.21
-10.64
total
649.91
267.57 235.07
332.54
So, SS total = 649.91+267.57+235.07+332.54=1485.09
Eaton
Tony
Aden
Oz
Eaton
Tony
Aden
Oz
6.75
-3.2
-2.86
-1
45.5625
10.24
8.18
1
2.75
-10.2
0.14
1
7.5625
104.04
0.02
1
-2.25
-1.2
3.14
3
5.0625
1.44
9.86
9
-7.25
4.8
5.14
-4
52.5625
23.04
26.42
16
7.14
5
96.04
50.98
25
-4.86
-4
23.62
16
61.78
-7.86
total
110.75
234.80
So, SSE = 110.75+234.80+180.86+68=594.41
180.86
68
Computing SST
4
(SS due to Treatment)
nc
SSTrt    ( X c  X G ) 2
c 1 j 1
 n1 ( X 1  X G ) 2  n2 ( X 2  X G ) 2  n3 ( X 3  X G ) 2  n4 ( X 4  X G ) 2
Why ?
SSTotal (sum of squares total)
4
nc
   ( X cj  X G ) 2
c 1 j 1
4
nc
4
nc
   ( X cj  X c )    ( X c  X G ) 2
c 1 j 1
2
c 1 j 1
 SSE  SSTrt
Another way to obtain SST:
12-57
What is the estimated value of the common variance ?
What is the computed value of F ?
12-58
The computed value of F is 8.99, which is
greater than the critical value of 5.09, so the null
hypothesis is rejected.
Conclusion: The mean scores are not the same
for the four stores.
Note: At this point we can only conclude there
is a difference in the treatment means. We
cannot determine which treatment groups differ.
12-59
Further question: Which treatment means differ ?
One procedure:
Use confidence intervals for the difference between two
means to test H0: μ1 = μ2
Confidence interval for μ1 - μ2 :
X  X t
1
2
 1
1 

MSE 

n2 
 n1
Confidence interval for one population mean μ :
s
X z
n
X  t / 2,n 1
s
n
12-60
Which treatment means differ ? (cont’d)
One procedure:
Use confidence intervals for the difference between two
means to test H0: μ1 = μ2
Confidence interval for μ1 - μ2 :
X  X t
1
2
 1
1 

MSE 

n2 
 n1
where
1. t is t/2,n-k
(obtained from t table with n-k degrees of freedom)
2. MSE=SSE/(n-k): an estimate of the population variance
(mean squared error, 均方差)
12-61
Confidence Interval for the
Difference Between Two Means - Example
Develop a 95% confidence interval for the difference in
the mean between Eaton and Oz.
t/2,n-k =2.101
The 95% confidence interval is (10.46, 26.04).
Can we conclude that there is a difference between
the two stores?
12-62
We conclude these treatment means differ
significantly.
The 95% confidence interval for μ1 - μ2 is (10.46, 26.04).
Both endpoints are positive; hence, we can conclude these
treatment means differ significantly.
That is, customers on Eaton rated service significantly
different from those on Oz.
12-63
Two-Way Analysis of Variance (雙因子變異數分析)
12-64
Two-Way Analysis of Variance (雙因子變異數分析)

For the two-factor ANOVA we test whether there is a
significant difference between the treatment effect
and whether there is a difference in the blocking
effect. Let Br be the block totals (r for rows)
Treatment: 處方
 Blocking variable (集區變數): A second treatment
variable.

*ANOVA was first developed for
agriculture applications.
12-65
Example: The value of residential homes
Home
Taylor
Watson
1
235
228
2
210
205
3
231
219
4
242
240
5
205
198
6
230
223
7
231
227
8
210
215
9
225
222
10
249
245
blocking variable: home
Two-Way ANOVA Table
SST  b( x c  x G ) 2
SSB  k( x b  x G )
2
12-67
Example: Travel times at 4 different times by 5 drivers
The Transportation Department would like to study
whether the travel times of Bus S10 at the following
4 different times are different:
(1) 10:00-12:00, (2) 12:00-14:00, (3) 14:00-16:00,
(4) 16:00-18:00.
Since there are many different drivers, the test was
set up so that each driver drove at each of the 4
different times.
12-68
Example: Travel time (minutes)
10:0012:00
12:0014:00
14:0016:00
16:0018:00
Andy
18
17
21
22
Ben
16
23
23
22
David
21
21
26
22
George
23
22
29
25
Jack
25
24
28
28
At 0.05 significance level, is there a difference in the mean
travel time at the 4 different times?
12-69
Shall we use 1-way ANOVA or 2-way ANOVA ?
(both can compare several means)
Let’s try 1-way first.
12-70
Critical value= 3.239; p-value=0.098 (fail to reject H0)
12-71
Step 1: State the null and alternate hypotheses.
H0: µu = µw = µh = µr
H1: Not all treatment means are the same
Reject H0 if F > F,k-1,n-k
Step 2: State the level of significance.
The .05 significance level is stated in the problem.
Step 3: Use the F statistic
Step 4: State the decision rule.
Reject H0 if F > F,v1,v2
F > F.05,k-1,n-k
F > F.05,4-1,20-4
F > F.05,3,16
F > 3.24
12-72
Step 5. Compute the value of F and make a
decision.
The computed F is 2.483.
Critical value= 3.239.
We do not to reject H0.
12-73
Note: 1-way ANOVA (SSTotal = SST + SSE) does
not take drivers’ variation into account.
If we remove the effect of the drivers, at .05
significance level, is there a difference in the
mean travel time?
* 2-way ANOVA remove the variation due to drivers from SSE
12-74
Two-Way ANOVA Table
12-75
Two-Way Analysis of Variance (雙因子變異數分析)
SST: sum of square for treatment
 SSB: sum of squares for the blocks

SST  b ( x c  x G )
SSB  k( x b  x G )
2
2
Sum of Squared Errors: SSE = SS total – SST - SSB
12-76
Step 1: State the null and alternate hypotheses.
H0: µu = µw = µh = µr
H1: Not all treatment means are the same
Reject H0 if F > F,k-1,(k-1)(b-1)
Step 2: State the level of significance.
The .05 significance level is stated in the problem.
Step 3: Use the F statistic
Step 4: State the decision rule.
Reject H0 if F > F,v1,v2
F > F.05,k-1,(k-1)(b-1)
F > F.05,4-1,(3)(4)
F > F.05,3,12
F > 3.49
12-77
10:0012:00
12:0014:00
14:0016:00
16:0018:00
Driver
means
Andy
18
17
21
22
19.5
Ben
16
23
23
22
21
David
21
21
26
22
22.5
George
23
22
29
25
24.75
Jack
25
24
28
28
26.25
12-78
4 nc
SSTrt    ( X c  X G ) 2
c 1 j 1
4
 b (X c  X G)
2
c 1
 b( X 1  X G ) 2  b( X 2  X G ) 2  b( X 3  X G ) 2  b( X 4  X G ) 2
 5 * [( 20.6  22.8) 2  ( 21.4  22.8) 2  ( 25.4  22.8) 2  ( 23.8  22.8) 2 ]
 72.8
12-79
The computed F is 24.27/3.06=7.93
Critical value= 3.49.
Decision: We reject H0 at significance level 0.05;
Conclusion: The mean times for the routes are not all the same.
12-80
Compare 1-way and 2-way ANOVA tables
F.05,3,16=3.24
F.05,3,12=3.49
F.05,4,12=3.26
12-81
Example (continued)
With 1-way ANOVA to test the mean time for the
routes, we conclude:
The mean times for the routes are the same.
With 2-way ANOVA, we conclude:
(1)The mean time is not the same for all drivers
(2)The mean times for the routes are not all the
same
12-82
Back to the example of values of residential homes
Treat it as two independent
samples:
It does not consider the
variation among homes.
Treat it as two dependent
samples:
It subtracts variation among
homes from SSE.
12-83
Supplementary materials
on two-sided tests and
confidence intervals
12-84
Relation between 95 % confidence interval and a
2-tailed test H 0 :   0 at significance level α
not reject H 0 if Z  Z  / 2
Note :
{X :
X  0
/ n
 Z / 2 }


 { X : X  Z / 2
 0  X  Z  / 2
}
n
n
What does this mean?
10-85
Relation between 95 % confidence interval and a
2-tailed test H 0 :   0 at significance level α
not reject H 0 if Z  Z  / 2
Note :
{X :
X  0
/ n
 Z / 2 }


 { X : X  Z / 2
 0  X  Z  / 2
}
n
n
What does this mean?
Ans: we would not reject H0 at level α if
 0 lies in the (1- α)100% confidence interval
10-86
Relation between (1- α)100% confidence interval
and a 2-tailed test H0: μ1 = μ2 at significance level α
not reject H 0 if Z  Z  / 2
Note :
{X1  X 2 :
X1  X 2
 Z }
1 1
2
MSE (  )
n1 n2
What does this mean?1
 {X1  X 2 : X1  X 2  Z
2
1
1 1
MSE (  )  0  X 1  X 2  Z  MSE (  ) }
n1 n2
n1 n2
2
We would not reject H0 at level α if 0 lies in the (1α)100% confidence interval. In above case, 0 is not in
(10.46, 26.04), so we reject H0 at level α .
10-87