How to answer one way anova questions

1. Approach to ANOVA Questions
One way ANOVA
1. State the one-way ANOVA model (estimates only one effect)
𝑦𝑖𝑗 = 𝜇 + 𝛼𝑖 + 𝜀𝑖𝑗
𝑤ℎ𝑒𝑟𝑒
2. State the hypotheses
By the treatment effects Treatment means
𝐻0: 𝛼1 = 𝛼2 = ⋯ = 𝛼𝐾 = 0 𝜇1 = 𝜇2 = ⋯ = 𝜇𝑖 = 𝜇
𝐻𝐴: some 𝛼𝑖 ≠ 0 some 𝜇𝑖 ≠ 𝜇
3. Rejection region /Criteria
Reject 𝐻0 , the null hypothesis if 𝐹𝑐 ≥ 𝐹𝑇 = 𝐹∝, 𝑘−1, 𝑛−𝑘
Where;
∝ 𝑖𝑠 𝑡ℎ𝑒 𝑙𝑒𝑣𝑒𝑙 𝑜𝑓 𝑠𝑖𝑔𝑛𝑖𝑓𝑖𝑐𝑎𝑛𝑐𝑒
𝑘 − 1 = 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚 𝑓𝑜𝑟 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡
𝑁 − 𝑘 = 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑒𝑟𝑟𝑜𝑟 𝑡𝑒𝑟𝑚
4. Computations
We compute
(1) The total of the values of individual items in the samples, 𝑇 = ∑ ∑ 𝑦𝑖𝑗
𝑛
𝑖=1
𝑘
𝑗=1
where i = 1 , 2, . . . n and j = 1, 2, . . . k
(2) Correction factor =
𝑇2
𝑁
, where 𝑁 = ∑ 𝑛𝑗
(3) 𝑆𝑆𝑇 = ∑ ∑ 𝑦𝑖𝑗
2
𝑖
𝑗 −
𝑇2
𝑁
(4) 𝑆𝑆𝐵 =
1
𝑛
∑ 𝑇𝑗
2
−
𝑇2
𝑁
(5) 𝑆𝑆𝐸 = 𝑆𝑆𝑇 − 𝑆𝑆𝐵
(6) Draw ANOVA Table to compute the F-statistic

2. 5. Compare 𝐹𝑐 with 𝐹𝑇 to take Decision and Conclude.
Exercise 2
The data in the table below gives the number of hours of pain relief provided by 4 different
types of headache tablets administered to 24 people. The 24 experimental units were randomly
divided into 4 groups and each group was treated with a different brand/type. Do the different
drug types give significantly different hours of pain relief?
Typical Solution.
1. State the one-way ANOVA model (estimates only one effect)
𝑦𝑖𝑗 = 𝜇 + 𝛼𝑖 + 𝜀𝑖𝑗
𝑤ℎ𝑒𝑟𝑒
𝑦𝑖𝑗 - jth observation under the ith drug type
µ - general mean
𝛼𝑖 - ith treatment effect (ith drug type effect)
∈𝑖𝑗 - error term
2. Hypothesis
Ho: α1 = α2 = α3 = α4 = 0 OR µ1 = µ2 = µ3 = µ4 = µ(say)
HA: Some αi ≠ 0 Some µio ≠ µ
Source of variation Sum of
square
Degrees of
freedom
Mean Squares Computed F – ratio
Between samples or
categories
SSB k - 1
MSB =
1

k
SSB
= SB
2
Fk-1, n-k = 2
2
W
B
S
S
Within samples or
categories
SSW N - k
MSW =
k
n
SSW

=SW
2
Compare with Fα, (k-1), (n-k)
from F tables
Total SST N- 1
Brands
1 2 3 4
12.2 4.9 8.0 4.6
9.5 10.6 12.1 6.1
11.6 7.0 5.7 5.0
13.0 8.3 8.6 3.8
10.1 5.5 7.2 8.2
9.6 11.7 12.4 7.7

3. 3. Rejection region
Reject Ho if Fc ≥ FT (F0.05, 3, 20) = 3.10
Accept Ho if Fc < F0.05, 3, 20 = 3.10
4. Computations
We compute
(1). The total of the values of individual items in the samples, 𝑇 = ∑ ∑ 𝑦𝑖𝑗
𝑛
𝑖=1
𝑘
𝑗=1 =
where i = 1 , 2, . . . n and j = 1, 2, . . . k
(2). Correction factor =
𝑇2
𝑁
, where 𝑁 = ∑ 𝑛𝑗
(3). 𝑆𝑆𝑇 = ∑ ∑ 𝑦𝑖𝑗
2
𝑖
𝑗 −
𝑇2
𝑁
= 1905.02 −
203 .42
24
= 1905.02 − 1723.815
= 181.205
(4) 𝑆𝑆𝐵 =
1
𝑛
∑ 𝑇𝑗
2
−
𝑇2
𝑁
𝑆𝑆𝐵 =
1
6
(662
+ 482
+ 542
+ 35.42) −
203.42
24
𝑆𝑆𝐵 = (1804.86)− 1723.815
𝑆𝑆𝐵 = 81.045
(5). 𝑆𝑆𝐸 = 𝑆𝑆𝑇 − 𝑆𝑆𝐵
𝑆𝑆𝐸 = 181.205 − 81.045
𝑆𝑆𝐸 = 100.160
(6) ANOVA Table
Source of variation Df SS MS F-ratio
Between treatments 3 81.045 MSB = 27.015 Fc =
27.015
5.008
= 5.1638
Within Samples Errors 20 100.160 MSE = 5.008
Total 23 181.205
5. Decision and Conclusion
Fc = 5.1638 > FT = 3.10 ; Reject Ho implying;
Drug types are significantly different in terms of the hours of pain relief they give.
Note: The sums of squares are NEVER negative.

4. Source of variation Sum of
square
Degrees of
freedom
Mean Squares Computed F – ratio
Between samples or
categories
SSB k - 1
MSB =
1

k
SSB
= SB
2
Fk-1, n-k = 2
2
W
B
S
S
Within samples or
categories
SSW n - k
MSW =
k
n
SSW

=SW
2
Compare with Fα, (k-1), (n-k)
from F tables
Total SST n - 1