ANALYSIS OF
VARIANCE (ANOVA)
ONE-WAY ANOVA
A method of statistical inference used to determine if the
means from more than two populations are different
PRODUCT COMPARISONS
• Say there are comparisons being done on the content (in milligrams) of a
product between three different product brands (A, B, C).
• To prevent any wonky suggestions, the testing is being done at two labs:
Lab Efficient and Lab Inefficient.
• It is assumed that both labs are following the same instructions for testing the brands.
• Each lab tested a random sample of 𝑛 = 6 products from each brand.
COMPARING LABS
Lab Efficient
Sample
1
2
3
4
5
6
Mean
A
10.21
10.25
10.24
9.8
9.77
9.73
10
B
11.32
11.2
11.4
10.5
10.68
10.9
11
Lab Inefficient
C
11.6
11.9
11.8
12.3
12.2
12.2
12
Sample
1
2
3
4
5
6
Mean
A
9.03
10.26
11.6
11.4
8.01
9.7
10
B
9.56
13.4
10.68
11.32
10.68
10.36
11
C
10.45
9.64
9.59
13.4
14.5
14.42
12
BETWEEN VS WITHIN VARIANCE
Lab Efficient
Sample
1
2
3
4
5
6
Mean
Variance
A
10.21
10.25
10.24
9.8
9.77
9.73
10
0.066
B
11.32
11.2
11.4
10.5
10.68
10.9
11
0.133
Lab Inefficient
C
11.6
11.9
11.8
12.3
12.2
12.2
12
0.076
Sample
1
2
3
4
5
6
Mean
Variance
A
9.03
10.26
11.6
11.4
8.01
9.7
10
1.916
B
9.56
13.4
10.68
11.32
10.68
10.36
11
1.710
C
10.45
9.64
9.59
13.4
14.5
14.42
12
5.569
USING ANOVA …
• Compares the between-sample variation to the within-sample variation.
• Lab Efficient has minimal between-sample and within-sample variation,
indicating there is a likely difference between Brands A, B, and C.
• Lab Inefficient, on the other hand, has large within-sample variation,
indicating there is not likely to be enough evidence to conclude there is a
difference between Brands A, B, and C.
WHAT WE MEAN BY “ONE-WAY”
• We are looking to see if there is a difference in the content
between brands, also known as an effect.
• The brand is considered one factor, and, within that one
factor, there are three levels (groups): Brands A, B, and C.
NULL HYPOTHESIS
There is no statistical difference between the means at each factor level
𝐻0 : 𝜇1 = 𝜇2 = ⋯ = 𝜇𝑡
where 𝑡 > 2
ALTERNATIVE HYPOTHESIS
At least one of the means is statistically different from another
𝐻1 : 𝜇𝑖 ≠ 𝜇𝑗
for some 𝑖 and 𝑗 where 𝑖 ≠ 𝑗
BACK TO THE PRODUCT COMPARISON
If we want to compare the product brands (A, B, C),
our statistical hypotheses are:
• 𝐻0 : 𝜇1 = 𝜇2 = 𝜇3
• 𝐻1 : 𝜇𝑖 ≠ 𝜇𝑗 where 𝑖 ≠ 𝑗
TEST STATISTIC
𝑏𝑒𝑡𝑤𝑒𝑒𝑛
𝑔𝑟𝑜𝑢𝑝
𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒
𝐹∗ =
𝑤𝑖𝑡ℎ𝑖𝑛 𝑔𝑟𝑜𝑢𝑝 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒
TEST STATISTIC
𝑀𝑒𝑎𝑛 𝑆𝑞𝑢𝑎𝑟𝑒 𝑇𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡 (𝑀𝑆𝑇)
𝐹 =
𝑀𝑒𝑎𝑛 𝑆𝑞𝑢𝑎𝑟𝑒 𝐸𝑟𝑟𝑜𝑟 (𝑀𝑆𝐸)
∗
MEAN SQUARE TREATMENT (MST)
• The total variation due to the differences in the factor levels (groups)
• If there is little to no difference the MST will be very small
• If there is a noticeable difference the MST will be large
MEAN SQUARE ERROR (MSE)
• The total variation due to random (unknown) error
• The larger the MSE, the less likely a difference will be
seen between the factor levels (groups)
Source
Sum of
Squares
Degrees of
Freedom
Treatment
𝑆𝑆𝑇
𝑡−1
Error
𝑆𝑆𝐸
𝑁 − 1 − (𝑡 − 1)
Total
𝑆𝑆𝑇𝑜𝑡𝑎𝑙
𝑁−1
Mean Square
F-Statistic
𝑆𝑆𝑇
𝑀𝑆𝑇 =
𝑡−1
𝑆𝑆𝐸
𝑀𝑆𝐸 =
𝑁 − 1 − (𝑡 − 1)
𝑀𝑆𝑇
𝑀𝑆𝐸
THE ONE-WAY ANOVA TABLE
Source
Sum Squares
Degrees of
Freedom
Mean Square
F-Statistic
Treatment
12.00
2
6.00
65.46
Error
1.38
15
0.092
Total
13.38
17
LAB EFFICIENT
ONE-WAY ANOVA TABLE
Source
Sum Squares
Degrees of
Freedom
Treatment
12.00
2
6
Error
45.98
15
3.07
Total
57.98
17
LAB INEFFICIENT
ONE-WAY ANOVA TABLE
Mean Square
F-Statistic
1.96
F-STATISTIC
𝑀𝑆𝑇
𝐹∗ =
~ 𝐹𝑑𝑓𝑇,𝑑𝑓𝐸
𝑀𝑆𝐸
BUT HOW DO WE DRAW A CONCLUSION?
• The p-value!
• We need to use the F-distribution to calculate the p-value…
𝑃(𝐹 > 𝐹 ∗ ) ~ 𝐹𝑑𝑓𝑇,𝑑𝑓𝐸
Source
Sum Squares
Degrees of
Freedom
Mean Square
F-Statistic
Treatment
12.00
2
6
65.46
Error
1.38
15
0.092
Total
13.38
17
LAB EFFICIENT
ONE-WAY ANOVA TABLE
CONCLUSION FOR LAB EFFICIENT
We reject the null hypothesis 𝐻0 : 𝜇1 = 𝜇2 = 𝜇3 and
conclude it is more likely one of the product brands
is different from another.
Source
Sum Squares
Degrees of
Freedom
Treatment
12
2
6
Error
45.98
15
3.07
Total
57.98
17
LAB INEFFICIENT
ONE-WAY ANOVA TABLE
Mean Square
F-Statistic
1.96
CONCLUSION FOR LAB INEFFICIENT
We fail to reject the null hypothesis 𝐻0 : 𝜇1 = 𝜇2 = 𝜇3 and
conclude there is not enough evidence to demonstrate
the product brands are different from another.
HOW MUCH DO YOU SLEEP?
Mean (𝑦ത𝑖. )
Sample Size
Lion
7.56
32
Badger
6.59
16
Raven / Eagle
6.83
27
Snake
6.57
7
Total / Grand
6.89
82
𝐹𝑎𝑐𝑡𝑜𝑟 (𝑇𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡) 𝐿𝑒𝑣𝑒𝑙
𝑂𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛𝑠
𝑇𝑜𝑡𝑎𝑙𝑠
𝐴𝑣𝑒𝑟𝑎𝑔𝑒𝑠
1
𝑦11
𝑦12
…
𝑦1𝑛1
𝑦1.
𝑦ത1.
2
𝑦21
𝑦22
…
𝑦2𝑛2
𝑦2.
𝑦ത2.
⋮
⋮
⋮
…
⋮
⋮
⋮
𝑡
𝑦𝑡1
𝑦𝑡2
…
𝑦𝑡𝑛𝑡
𝑦𝑡.
𝑦ത𝑡.
𝑦..
𝑦ത..
Total Observations
Combined Average
*also known as “Factor”
LET’S BREAK “TREATMENT” DOWN…
Source
Sum of Squares
Degrees of Freedom
Mean Square
𝑡−1
𝑆𝑆𝑇
𝑀𝑆𝑇 =
𝑡−1
𝑡
Treatment*
𝑆𝑆𝑇 = ෍ 𝑛𝑖 𝑦ത𝑖. − 𝑦ത.. 2
𝑖=1
𝑆𝑆𝑇 = 14.606
SUM OF SQUARES TREATMENT (SST)
Known as the “Between Group Sum of Squares”
𝑡
𝑆𝑆𝑇 = ෍ 𝑛𝑖 𝑦ത𝑖. − 𝑦ത.. 2
𝑖=1
LET’S BREAK “TOTAL” DOWN…
Source
Sum of Squares
Degrees of Freedom
Total
𝑆𝑆𝑇𝑜𝑡𝑎𝑙 = ෍(𝑦𝑖𝑗 − 𝑦ത.. )2
𝑁−1
𝑖,𝑗
𝑆𝑆𝑇𝑜𝑡𝑎𝑙 = 122.305
SUM OF SQUARES TOTAL (SSTOTAL)
𝑆𝑆𝑇𝑜𝑡𝑎𝑙 = ෍(𝑦𝑖𝑗 − 𝑦ത.. )2
𝑖,𝑗
*also known as “Residuals”
LET’S BREAK “ERROR” DOWN…
Source
Sum of Squares
Error*
𝑆𝑆𝐸 = ෍(𝑦𝑖𝑗 − 𝑦ത𝑖. )2
𝑖,𝑗
Degrees of Freedom
Mean Square
𝑁−1 − 𝑡−1
𝑆𝑆𝐸
𝑀𝑆𝐸 =
𝑁−1 − 𝑡−1
𝑆𝑆𝐸 = 107.699
SUM OF SQUARES ERROR (SSE)
Known as the “Within Group Sum of Squares”
𝑆𝑆𝐸 = ෍(𝑦𝑖𝑗 − 𝑦ത𝑖. )
𝑖,𝑗
2
Source
Sum of Squares
Degrees of Freedom
Treatment
Error
Total
THE ONE-WAY ANOVA TABLE
Mean Square
F-Statistic