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One-Way ANOVA (Analysis of Variance) is a statistical test used to determine if there are any statistically significant differences between the means of three or more independent (unrelated) groups.
Let $Y_{ij}$ be the $j$ th observation of $i$ th group . We can write the model as
Model
\[Y_{ij} = \mu + \tau_i + \epsilon_{ij}\]- $Y_{ij}$: The observation from group $i$ and individual $j$
- $\mu$: The overall mean
- $\tau_i$: The effect of the $i$-th group
- $\epsilon_{ij}$: The random error term
Now using LSE of $\mu , \alpha_i,\epsilon_{ij} $ The reduced model assumes that there are no differences between group means:
\[Y_{ij} = \mu + \epsilon_{ij}\]Required ANOVA Table
The ANOVA summary table partitions the total variation into components due to the model (between groups) and error (within groups).
| Source | Sum of Squares (SS) | Degrees of Freedom (df) | Mean Square (MS) | F-statistic |
|---|---|---|---|---|
| Between Groups | $SSB$ | $k-1$ | $MSB = \frac{SSB}{k-1}$ | $F = \frac{MSB}{MSW}$ |
| Within Groups | $SSW$ | $N-k$ | $MSW = \frac{SSW}{N-k}$ | |
| Total | $SST$ | $N-1$ |
- $SSB$: Between-group sum of squares
- $SSW$: Within-group sum of squares
- $SST$: Total sum of squares ($SST = SSB + SSW$)
- $k$: Number of groups
- $N$: Total number of observations
Example
Consider the following data:
- Group 1: $1, 2, 5$
- Group 2: $2, 4, 2$
- Group 3: $2, 3, 4$
Calculate Group Means and Overall Mean
- Group 1 mean: $\mu_1 = \frac{1 + 2 + 5}{3} = 2.67$
- Group 2 mean: $\mu_2 = \frac{2 + 4 + 2}{3} = 2.67$
- Group 3 mean: $\mu_3 = \frac{2 + 3 + 4}{3} = 3.00$
- Overall mean: $\mu = \frac{1 + 2 + 5 + 2 + 4 + 2 + 2 + 3 + 4}{9} = 2.78$
Compute the Sum of Squares
\[SSB = \sum_{i=1}^{k} n_i (\mu_i - \mu)^2\]where $n_i$ is the number of observations in group $i$.
\[SSB = 3(2.67 - 2.78)^2 + 3(2.67 - 2.78)^2 + 3(3.00 - 2.78)^2\] \[SSB = 3 \times (-0.11)^2 + 3 \times (-0.11)^2 + 3 \times 0.22^2\] \[SSB = 0.3636\]Within-group sum of squares
\[SSW = \sum_{i=1}^{k} \sum_{j=1}^{n_i} (Y_{ij} - \mu_i)^2\]For Group 1:
\[SSW_1 = (1 - 2.67)^2 + (2 - 2.67)^2 + (5 - 2.67)^2 = 2.7889 + 0.4489 + 5.4489 = 8.6867\]For Group 2:
\[SSW_2 = (2 - 2.67)^2 + (4 - 2.67)^2 + (2 - 2.67)^2 = 0.4489 + 1.7689 + 0.4489 = 2.6667\]For Group 3:
\[SSW_3 = (2 - 3.00)^2 + (3 - 3.00)^2 + (4 - 3.00)^2 = 1 + 0 + 1 = 2\] \[SSW = SSW_1 + SSW_2 + SSW_3 = 8.6867 + 2.6667 + 2 = 13.3534\]Now calculate Mean squares and F statistics in ANOVA table
ANOVA Summary Table
| Source | Sum of Squares (SS) | Degrees of Freedom (df) | Mean Square (MS) | F-statistic |
|---|---|---|---|---|
| Between Groups | 0.3636 | 2 | SSB/2= 0.1818 | 0.0817 |
| Within Groups | 13.3534 | 6 | SSW/6 2.2256 | |
| Total | 13.717 | 8 |
Conclusion
Compare the F-statistic to the critical value from the F-distribution table at a chosen significance level (e.g., $\alpha = 0.05$). For $df_1 = 2$ and $df_2 = 6$, the critical value is approximately 5.14. Since 0.0817 < 5.14, we fail to reject the null hypothesis and conclude that there are no significant differences between the group means.
Ref:
R code
data <- data.frame(
group1 = c(1, 2, 5),
group2 = c(2, 4, 2),
group3 = c(2, 3, 4)
)
# Calculate group means
group_means <- c(mean(data$group1), mean(data$group2), mean(data$group3))
# Calculate overall mean
overall_mean <- mean(unlist(data))
# Number of groups and total number of observations
k <- ncol(data)
N <- length(unlist(data))
# Calculate Sum of Squares Between (SSB)
ssb <- 0
for (i in 1:k) {
ssb <- ssb + length(data[, i]) * (group_means[i] - overall_mean)^2
}
# Calculate Sum of Squares Within (SSW)
ssw <- 0
for (i in 1:k) {
ssw <- ssw + sum((data[, i] - group_means[i])^2)
}
# Calculate Total Sum of Squares (SST)
sst <- ssw + ssb
# Degrees of Freedom
df_between <- k - 1
df_within <- N - k
df_total <- df_between + df_within
# Mean Squares
msb <- ssb / df_between
msw <- ssw / df_within
# F-statistic
f_statistic <- msb / msw
