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Michael Brenndoerfer

ANOVA (Analysis of Variance) › Worked Example: Comparing Fertilizers

Michael BrenndoerferJanuary 5, 20265 min read

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Part 2 of 9Hypothesis Testing

Chapter 17ANOVA (Analysis of Variance) › Worked Example: Comparing Fertilizers

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A agricultural researcher tests three fertilizers to see if they produce different plant growth. She randomly assigns 10 plants to each fertilizer and measures growth (in cm) after 6 weeks.

In[4]:

Code

import numpy as np
from scipy import stats

# Plant growth data (cm) for three fertilizers
fertilizer_a = np.array([23, 25, 21, 24, 22, 26, 23, 24, 25, 22])
fertilizer_b = np.array([28, 31, 29, 30, 27, 32, 28, 29, 31, 30])
fertilizer_c = np.array([25, 27, 24, 26, 28, 25, 26, 27, 24, 26])

# Combine into groups
groups = [fertilizer_a, fertilizer_b, fertilizer_c]
group_names = ["Fertilizer A", "Fertilizer B", "Fertilizer C"]

# Step 1: Calculate basic statistics
k = len(groups)
ns = [len(g) for g in groups]
N = sum(ns)
means = [np.mean(g) for g in groups]
grand_mean = np.mean(np.concatenate(groups))

# Step 2: Calculate Sum of Squares Between (SSB)
ss_between = sum(n * (m - grand_mean) ** 2 for n, m in zip(ns, means))

# Step 3: Calculate Sum of Squares Within (SSW)
ss_within = sum(np.sum((g - m) ** 2) for g, m in zip(groups, means))

# Step 4: Calculate Total Sum of Squares (SST)
all_data = np.concatenate(groups)
ss_total = np.sum((all_data - grand_mean) ** 2)

# Verify decomposition: SST = SSB + SSW
assert np.isclose(ss_total, ss_between + ss_within)

# Step 5: Calculate degrees of freedom
df_between = k - 1
df_within = N - k
df_total = N - 1

# Step 6: Calculate Mean Squares
ms_between = ss_between / df_between
ms_within = ss_within / df_within

# Step 7: Calculate F-statistic
f_stat = ms_between / ms_within

# Step 8: Calculate p-value
p_value = stats.f.sf(f_stat, df_between, df_within)

# Step 9: Calculate effect size (eta-squared)
eta_squared = ss_between / ss_total

Let's trace through the calculation step by step:

Step 1: Basic Statistics

\bar{x}_A = 23.5, \quad \bar{x}_B = 29.5, \quad \bar{x}_C = 25.8

\bar{x} = \frac{23.5 \times 10 + 29.5 \times 10 + 25.8 \times 10}{30} = 26.27

Step 2: Sum of Squares Between

SS_{Between} = \sum_{j=1}^{3} n_j(\bar{x}_j - \bar{x})^2 = 10(23.5 - 26.27)^2 + 10(29.5 - 26.27)^2 + 10(25.8 - 26.27)^2

Step 3: Sum of Squares Within

SS_{Within} = \sum_{j=1}^{3}\sum_{i=1}^{10}(x_{ij} - \bar{x}_j)^2

Step 4: Mean Squares and F-Statistic

MS_{Between} = \frac{SS_{Between}}{k-1} = \frac{SS_{Between}}{2}

MS_{Within} = \frac{SS_{Within}}{N-k} = \frac{SS_{Within}}{27}

F = \frac{MS_{Between}}{MS_{Within}}

Out[5]:

Console

One-Way ANOVA: Effect of Fertilizer on Plant Growth
=================================================================

Group Statistics:
-----------------------------------------------------------------
  Fertilizer A: n = 10, mean = 23.50 cm
  Fertilizer B: n = 10, mean = 29.50 cm
  Fertilizer C: n = 10, mean = 25.80 cm
  Grand mean: 26.27 cm

ANOVA Table:
-----------------------------------------------------------------
Source                  SS     df         MS          F      p-value
-----------------------------------------------------------------
Between             183.27      2      91.63      40.83     6.87e-09
Within               60.60     27       2.24
Total               243.87     29
-----------------------------------------------------------------

Effect size (eta-squared): 0.752
  Interpretation: 75.2% of variance in growth is explained by fertilizer type

Decision (alpha = 0.05):
-----------------------------------------------------------------
F(2, 27) = 40.83, p < 0.001
Reject H_0: At least one fertilizer produces different growth

In[6]:

Code

# Verify our calculation with scipy
f_scipy, p_scipy = stats.f_oneway(fertilizer_a, fertilizer_b, fertilizer_c)
print("Verification with scipy.stats.f_oneway:")
print(f"  F = {f_scipy:.2f}, p = {p_scipy:.2e}")

Out[6]:

Console

Verification with scipy.stats.f_oneway:
  F = 40.83, p = 6.87e-09

Out[7]:

Visualization

Boxplots showing plant growth distributions for three fertilizers. — Boxplots showing the distribution of plant growth for each fertilizer, with group means marked. Fertilizer B clearly produces taller plants.

F-distribution plot with observed F-statistic in rejection region. — The F-distribution with the observed F-statistic far into the rejection region, confirming the significant difference.

The ANOVA reveals a highly significant difference among fertilizers ( $F(2, 27) = 34.62$ , $p < 0.001$ ). The effect size $\eta^2 = 0.72$ indicates that 72% of the variation in plant growth is explained by fertilizer type, a very large effect.