Tutorial 8 – Analysis of Variance (ANOVA)
Presented by Eng. Alaa Zarif & Eng. Lobna el Seify
Introduction
Use the one-way ANOVA technique to determine if there is a significant difference among
three or more means of independent (unrelated) groups
ANOVA tells us if there’s a difference, but doesn’t tell us where exactly is the difference.
Determine which mean is different using the Scheffé or Tukey test if the null hypothesis is
rejected in the ANOVA
Assumptions of ANOVA
The following assumptions apply when to compare three or more means
The populations from which the samples were obtained must be normally or approximately
normally distributed
The samples must be independent of each other
The variances of the populations must be equal
Procedure
Solving Hypothesis-Testing Problems
Step 1 State the hypotheses and identify the claim.
• Ho : μ1 = μ2 = μ3 = ...... = μ𝑛
• Ha : At least one mean is different from the others
Step 2 Find the critical F value from the table.
Step 3 Compute the F - test value.
Step 4 Make the decision to reject or not reject the null hypothesis.
Step 5 Summarize the results.
Results of (a x n) = N runs
Treatment 1 2 Run (j) n Mean variance
1 𝑦11 𝑦12 …. 𝑦1𝑛 𝑌ത1 𝑆1 2
2 𝑦21 𝑦22 …. 𝑦2𝑛 𝑌ത2 𝑆2 2
.. .. .. …. .. .. ..
a 𝑦𝑎1 𝑦𝑎2 …. 𝑦𝑎𝑛 𝑌ത𝑎 𝑆𝑎 2
𝑦𝑖𝑗 = the jth observation under treatment i
ANOVA
𝑆𝑆𝐵 : Between-group variance
• It involves finding the variability between treatment averages and the grand
average.
𝑆𝑆𝑊 : Within-group variance (Error)
• Variability of individual observations within each treatment and the treatment
average
• It is made by computing the variance using all the data and is not affected by
differences in the means.
ANOVA Summary Table
Source Sum of Squares d.f. Mean of Squares F
Between 𝑆𝑆𝐵 i–1 MSB = 𝑠𝐵2 MS𝐵
2 MS𝑊
Within (error) 𝑆𝑆W N–i MSW = 𝑠𝑊
Total 𝑆𝑆𝐵 + 𝑆𝑆𝑊 a.N - 1
2 𝑆𝑆𝐵 σ 𝑛𝑖 (𝑋𝑖 −𝑋𝐺𝑀 )2
• 𝑆𝑆𝐵 = σ 𝑛𝑖 𝑋𝑖 − 𝑋𝐺𝑀 • 𝑠𝐵2 = =
𝑖 −1 𝑖−1
2 𝑆𝑆𝑊 σ 𝑛𝑖 −1 𝑠𝑖2
• 𝑆𝑆𝑊 = σ 𝑛𝑖 − 1 𝑆𝑖2 • 𝑠𝑊 = σ 𝑛𝑖 −1
= σ 𝑛𝑖 −1
Where,
• i is the no. of treatments/ Levels
• N is the sample size
Problem 1
A researcher wishes to try three different techniques to lower the blood pressure of
individuals diagnosed with high blood pressure. The subjects are randomly assigned to three
groups; the first group takes medication, the second group exercises, and the third group
follows a special diet. After four weeks, the reduction in each person’s blood pressure is
recorded. At α = 0.05, test the claim that there is no difference among the means.
Problem 1
Step 1: State the hypotheses and identify the claim
H0: μ1 = μ2 = μ3 (claim)
H1: At least one mean is different from the others
Problem 1
Step 2: Find the critical F- value
Since i = 3, N = 15, and α = 0.05
d.f.N. = i – 1 = 3 – 1 = 2
d.f.D. = N – i = 15 – 3 = 12
The critical value is 3.89, obtained from Table F
Problem 1
2
Step 3: Compute the F- test value. • 𝑆𝑆𝐵 = σ 𝑛𝑖 𝑋𝑖 − 𝑋𝐺𝑀
• 𝑆𝑆𝑊 = σ 𝑛𝑖 − 1 𝑆𝑖2
𝑀𝑆𝐵 𝑠𝐵 2 𝑆𝑆 𝑆𝑆W
F= = 2= 𝐵 /
𝑀𝑆𝑊 𝑠𝑊 𝑖 −1 𝑖(𝑛 −1)
σ𝑋 10+12+9+⋯+4 116
• ሜ
𝑋𝐺𝑀 = = = = 7.73
𝑁 15 15
2 𝑆𝑆𝐵 5(11.8 −7.73)2 + 5(3.8 −7.73)2 + 5(7.6 −7.73)2 160
• 𝑀𝑆𝐵 = 𝑠𝐵 = = = = 80.07
𝑖 −1 3−1 2
𝑆𝑆 4 5.7 + 4 10.2 + 4(10.3) 104.8
• 𝑀𝑆𝑊 = 𝑠𝑊 2 = 𝑖(𝑛 −1)
W
= = = 8.37
3(5 −1) 12
𝑀𝑆𝐵 80.07
• F= = = 9.17
𝑀𝑆𝑊 8.37
Problem 1
Source Sum of Squares d.f. Mean of Squares F
Between 160.13 2 80.07
9.17
Within (error) 104.80 12 8.73
Total 264.93 14
Step 4: Make the decision
Reject the null hypothesis, since 9.17 > 3.89
Step 5: Summarize the results
There is enough evidence to reject the claim and conclude that at least one mean is
different from the others
Problem 2
A state employee wishes to see if there is a significant difference in the number of employees
at the interchanges of three state toll roads. At α = 0.05, can it be concluded that there is a
significant difference in the average number of employees at each interchange?
Problem 2
Step 1: State the hypotheses and identify the claim
H0: μ1 = μ2 = μ3
H1: At least one mean is different from the others (claim)
Problem 2
Step 2: Find the critical value
Since i = 3, N = 18, and α = 0.05,
d.f.N. = 2, d.f.D. = 15
The critical value is 3.68, obtained from Table F
Problem 2
2
Step 3: Compute the F- test value. • 𝑆𝑆𝐵 = σ 𝑛𝑖 𝑋𝑖 − 𝑋𝐺𝑀
• 𝑆𝑆𝑊 = σ 𝑛𝑖 − 1 𝑆𝑖2
𝑀𝑆𝐵
F=
𝑀𝑆𝑊
σ𝑋 7+14+32+⋯+11 152
• ሜ
𝑋𝐺𝑀 = = = = 8.4
𝑁 15 18
2 𝑆𝑆𝐵 6(15.5 −8.4)2 + 6(4 −8.4)2 + 6(5.8 −8.4)2 459.19
• 𝑀𝑆𝐵 = 𝑠𝐵 = = = = 229.6
𝑖 −1 3−1 2
2 𝑆𝑆W 5 81.9 + 5 25.6 + 5(29.0) 682.5
• 𝑀𝑆𝑊 = 𝑠𝑊 = = = = 45.5
𝑖(𝑛 −1) 3(6 −1) 15
𝑀𝑆𝐵 229.6
• F= = = 5.05
𝑀𝑆𝑊 45.5
Problem 2
Source Sum of Squares d.f. Mean of Squares F
Between 459.18 2 229.6
5.05
Within (error) 682.5 15 45.5
Total 1141.68 17
Step 4: Make the decision
Reject the null hypothesis, since 5.05 > 3.68
Step 5: Summarize the results.
There is enough evidence to support the claim that there is a difference among the means
Problem 3
To examine the effects of pets and friends in stressful situations, researchers recruited 45 people
to participate in an experiment. Fifteen of the subjects were randomly assigned to each of three
groups to perform a stressful task alone (control group), with a good friend present, or with their
dog present. Each subject’s mean heart rate during the task was recorded. Test the appropriate
hypotheses at the α = 0.05 level to decide if the mean heart rate differs between the groups
Problem 4
Suppose that a random sample of n = 5 was selected from the vineyard properties for sale in
Sonoma County, California, in each of three years. The following data are consistent with
summary information on price per acre for disease-resistant grape vineyards in Sonoma County.
Carry out an ANOVA to determine whether there is evidence to support the claim that the mean
price per acre for vineyard land in Sonoma County was not the same for each of the three years
considered. Test at the 0.05 level and at the 0.01 level
1996: 30000 34000 36000 38000 40000
1997: 30000 35000 37000 38000 40000
1998: 40000 41000 43000 44000 50000
