Anova Test Practice Problems

The Analysis of Variance (ANOVA) test is a statistical method used to determine whether there are significant differences between the means of three or more groups.

Example 1: Three different kinds of food are tested on three groups of rats for 5 weeks. The objective is to check the difference in mean weight (in grams) of the rats per week. Apply one-way ANOVA using a 0.05 significance level to the following data:

Solution:

H₀: μ₁= μ₂=μ₃
H₁: The means are not equal

Since, X̄₁ = 5, X̄₂ = 9, X̄₃ = 10

Total mean = X̄ = 8

• SSB = 6(5 - 8)² + 6(9 - 8)² + 6(10 - 8)² = 84
• SSE = 68
• MSB = SSB/df₁ = 42
• MSE = SSE/df₂ = 4.53
• f = MSB/MSE = 42/4.53 = 9.33

Since f > F, the null hypothesis stands rejected.

Example 2: Calculate the ANOVA coefficient for the following data:

Solution:

Plant	n	x	s	s2
Hibiscus	5	12	2	4
Marigold	5	16	1	1
Rose	5	20	4	16

p = 3
n = 5
N = 15
x̄ = 16

SST = Σn(x−x̄)²

• SST= 5(12 − 16)² + 5(16 − 16)² + 11(20 − 16)² = 160
• MST = SST/p-1 = 160/3-1 = 80
• SSE = ∑ (n−1) = 4 (4 + 1) + 4(16) = 84
• MSE = 7
• F = MST/MSE = 80/7
• F = 11.429

Example 3: The following data show the number of worms quarantined from the GI areas of four groups of muskrats in a carbon tetrachloride anthelmintic study. Conduct a two-way ANOVA test.

Solution:

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square
Between the groups	62111.6	8	9078.067
Within the groups	98787.8	16	4567.89
Total	167771.4	24

Since F = MST / MSE
= 9.4062 / 3.66
F = 2.57

Example 4: Enlist the results in APA format after performing ANOVA on the following data set:

\begin{bmatrix}  \textbf{n} & \textbf{mean} & \textbf{sd} \\  30 & 50.26 & 10.45 \\  30 & 45.32 & 12.76 \\  30 & 53.67 & 11.47 \\ \end{bmatrix}

Solution:

• Variance of first set = (10.45)² = 109.2
• Variance of second set = (12.76)² = 162.82
• Variance of third set = (11.47)² = 131.56

MS_error = {109.2 + 162.82 + 131.56} / {3}
= 134.53

MS_between = (17.62)(30) = 528.75
F = MS_between  /  MS_error 
  = 528.75 / 134.53
F = 4.86

APA writeup: F(2, 87)=3.93, p >=0.01, η²=0.08.

Practice Problem

Question 1. Method A = {80, 85, 90, 87}, Method B = {75, 78, 72, 74}, and Method C = {88, 85, 90, 92} are given. State the null and alternative hypotheses for performing a One-Way ANOVA test.
Question 2. Calculate the F-statistic for the given data using One-Way ANOVA. Group 1 = {5, 6, 7, 8}, Group 2 = {4, 5, 6, 5}, and Group 3 = {7, 7, 6, 8}.
Question 3. Interpret the significance of the p-value for the interaction effect in a Two-Way ANOVA, where the p-value for the interaction effect is 0.02 and the significance level (α) is 0.05.
Question 4. Group A = {10, 12, 14, 13}, Group B = {15, 17, 16, 18}, and Group C = {20, 22, 21, 23} are given. Interpret the p-value of the ANOVA test and explain whether the null hypothesis is rejected, where F-statistic = 4.86 and p-value = 0.01.

Answer:-

1. Null Hypothesis (H₀): μ₁ = μ₂ = μ₃ (The means of all groups are equal). Alternative Hypothesis (H₁): At least one mean is different.
2. F-statistic = 4.58.
3. If the p-value (0.02) is less than the significance level (0.05), reject the null hypothesis and conclude that there is a significant interaction effect.
4. Since the p-value (0.01) is less than 0.05 and the F-statistic is significant, we reject the null hypothesis, indicating a significant difference between the group means.