One-Way ANOVA vs. Two-Way ANOVA - Coding

In statistics, the Analysis of Variance (ANOVA) is a powerful tool used to analyze differences among group means and their associated procedures. ANOVA is essential for students and professionals in fields such as psychology, biology, education, and business, as it helps in understanding how different factors influence a particular outcome. This article aims to provide a comprehensive overview of two common types of ANOVA: one-way and two-way ANOVA.

Table of Content

One-Way ANOVA

When to Use One-Way ANOVA?
Assumptions of One-Way ANOVA
How to Perform One-Way ANOVA
One way ANOVA Example

Two-way ANOVA

When to Use Two-Way ANOVA
Assumptions of Two-Way ANOVA
How to Perform Two-Way ANOVA
Two-way ANOVA Example

Difference between One-way and Two-way ANOVA

One-Way ANOVA

One-way ANOVA is a statistical test used to determine whether there are any statistically significant differences between the means of three or more independent (unrelated) groups. It compares the means of the groups to see if at least one of them is significantly different from the others.

When to Use One-Way ANOVA?

One-way ANOVA is used when you have:

One independent variable (factor) with three or more levels (groups).
A continuous dependent variable.

For example, suppose a researcher wants to test the effect of three different diets on weight loss. The diets are labeled as Diet A, Diet B, and Diet C. The weight loss (in pounds) of participants on each diet is recorded, and one-way ANOVA is used to determine if there is a significant difference in weight loss among the diets.

Assumptions of One-Way ANOVA

Independence of Observations: The data collected from the groups should be independent of each other.
Normality: The data in each group should be approximately normally distributed.
Homogeneity: The variances among the groups should be approximately equal.

How to Perform One-Way ANOVA

Step 1: State the Hypothesis

Null Hypothesis([Tex]\Eta[/Tex]₀) : All group means are equal.

Alternative Hypothesis ([Tex]\Eta[/Tex]₁) : At least one group mean is different.

Step 2: Calculate the ANOVA table

Assume, there are k classes and each class k_i contains n_i number of elements, mean [Tex]\mu[/Tex]_i and n = n₁ + n₂ + … is the total number of elements.

The mean of all elements ie.,Global mean is given by

[Tex]\mu = \frac{\sum_{i=1}^{k}\mu_{i}}{k}[/Tex]

Now, Calculate the following,

Sum of squares between the Groups (SSB) = [Tex]\sum_{i=1}^{k}n_{i}(\mu_{i}-\mu)^2[/Tex]

SSB measures the variation among the means of the different groups (or levels) of the independent variable. It calculates how much each group mean differs from the overall mean of all observations combined.

Sum of squares within the Groups (SSW) = [Tex]\sum_{i=1}^{k} \sum_{j=1}^{n_i} (x_{ij} – \mu_{i})^2[/Tex]

where [Tex]x_{ij}[/Tex] is the [Tex]j[/Tex]th element in [Tex]i[/Tex]th group/class.

SSW measures the variation within each group (or level) of the independent variable. It calculates how much each individual observation deviates from its group mean.

Total sum of squares (SST) = [Tex]SSB +SSW[/Tex]

Mean sum of squares between the Groups (MSSB) = [Tex]\frac{\text{SSB}}{\text{k-1}}[/Tex]

Mean sum of squares within the Groups (MSSW) = [Tex]\frac{\text{SSW}}{\text{n-k}}[/Tex]

Source	Sum of Squares	Degree of Freedom	Mean Sum of Squares
Between the Groups	SSB	k-1	MSSB
Within the Groups	SSW	n-k	MSSW
Total	SST	(k-1)+(n-k) = n-1	–

Now, Find the F-ratio which is given by,

[Tex]\text{f-ratio} = \frac{\text{Varianve Between (MSSB)}}{\text{Varianve Within (MSSW)}}[/Tex]

Step 3: Find Critical f-value

Use the F-table to find the critical value for [Tex]f_{(k-1,n-k)}[/Tex]. Refer this article to know more about f-test.

Step 4: Make the Decision

Compare the calculated and critical value of f-ratio and make the decision as,

If critical f-value > calculated f-value, then Accept the null hypothesis which means all group means are equal.
If critical f-value < calculated f-value, then Reject the null hypothesis which means at least one group mean is different.

One way ANOVA Example

Let’s take the example of effect of three different diets on weight loss. The diets are labeled as Diet A, Diet B, and Diet C. The weight loss (in pounds) of participants on each diet is recorded.

Patients	Diet A	Diet B	Diet C
1	25	31	24
2	30	39	30
3	36	38	28
4	38	42	25
5	31	35	28

Step 1: State the Hypothesis

Null Hypothesis([Tex]\Eta_0[/Tex]) : [Tex]\mu_1 = \mu_2 = \mu_3[/Tex]

Alternate Hypothesis([Tex]\Eta_1[/Tex]) : [Tex]\mu_1 \neq \mu_2 \neq \mu_3[/Tex]

Step 2: Calculate the ANOVA table

For Diet A, mean([Tex]\mu_1[/Tex]) = [Tex]\frac{25+30+36+38+31}{5} = \frac{160}{5} = 32[/Tex]

For Diet B, mean([Tex]\mu_2[/Tex]) = [Tex]\frac{31+39+38+42+35}{5} = \frac{185}{5} = 37[/Tex]

For Diet C, mean([Tex]\mu_3[/Tex]) = [Tex]\frac{24+30+28+25+28}{5} = \frac{135}{5} = 27[/Tex]

Therefore the Global mean([Tex]\mu[/Tex]) = [Tex]\frac{32+37+27}{3} = \frac{96}{3} = 32[/Tex]

Now, Calculate:

SSB = [Tex]\sum_{i=1}^{k}n_{i}(\mu_{i}-\mu)^2 = 5(32-32)^2 + 5(37-32)^2 + 5(27-32)^2 = 250[/Tex]

SSW = [Tex]\sum_{i=1}^{k} \sum_{j=1}^{n_i} (x_{ij} – \mu_{i})^2[/Tex]

= [Tex](25-32)^2 + (30-32)^2+(36-32)^2+(38-32)^2+(31-32)^2+(31-37)^2+(39-37)^2+(38-37)^2+(42-37)^2+(35-37)^2+(24-27)^2+(30-27)^2+(28-27)^2+(25-27)^2+(28-27)^2[/Tex]

= [Tex]200[/Tex]

Create the ANOVA table.

Source	Sum of Square	Degree of Freedom	Mean Sum of Squares
Between the Groups	250	k-1 = 3-1 = 2	250/2 = 125
Within the Groups	200	n-k = 15-3 = 12	200/12 = 16.667
Total	450	n-1 = 15-1 = 14	–

The value of f-ratio = [Tex]\frac{MSSB}{MSSW} = \frac{125}{16.667}=7.4998[/Tex]

Step 3: Find Critical f-value

Find the critical f-ratio from the f-table where df1=2, df2=12 for [Tex]\alpha = 0.05[/Tex] (If not given consider 5%).

F-table

Therefore Critical f-ratio = 3.89

Step 4: Make the Decision

Compare the calculated and critical value of f-ratio.

Calculated f-ratio > Critical f-ratio (7.4998 > 3.89)

Hence, we reject the Null Hypothesis stating that the group means are not equal and there is a significant difference in weight loss among the diets.

Two-way ANOVA

Two-way ANOVA is used to examine the influence of two different categorical independent variables on one continuous dependent variable. It also helps in understanding if there is an interaction between the two independent variables on the dependent variable.

When to Use Two-Way ANOVA

One-way ANOVA is used when you have:

Two independent variables (factors), each with two or more levels (groups).
A continuous dependent variable.

For example, Consider a study evaluating the effects of two different fertilizers and two different watering frequencies on plant growth. Here, the two fertilizers and the watering frequencies are the independent variables, and the plant growth is the dependent variable. Two-way ANOVA can determine if there are significant effects of fertilizers, watering frequencies, and their interaction on plant growth.

Assumptions of Two-Way ANOVA

Independence of Observations: The data collected from the groups should be independent of each other.
Normality: The data in each group should be approximately normally distributed.
Homogeneity: The variances among the groups should be approximately equal.
Additivity and Linearity: The combined effect of the two independent variables should be equal to the sum of their individual effects plus any interaction effect.
Interaction Effect: The effect of one independent variable on the dependent variable might depend on the level of the other independent variable.

How to Perform Two-Way ANOVA

Step 1: State the Hypothesis

Null Hypothesis ([Tex]\Eta_0[/Tex]) :

There is no significant difference between the groups of the first variable.
There is no significant difference between the groups of the second variable.
There is no interaction between both variables.

Alternate Hypothesis ([Tex]\Eta_1[/Tex]) :

There is a significant difference between the groups of the first variable.
There is a significant difference between the groups of the second variable.
There is an interaction between both variables.

Step 2: Calculate the ANOVA table

Calculate Correction Factor (CF), which is given by,

[Tex]\text{CF} = \frac{(\Sigma{x})^2}{n}[/Tex]

Where [Tex]n[/Tex] is the total number of observations in the given data.

Now, Calculate the following,

Total Sum of Squares (SST) = [Tex]\sum_{i}^{} \sum_{j}^{} x_{ij}^2 – CF[/Tex]

For variation between groups of first variable,

Sum of Squares of Column (SSB_c) = [Tex]\frac{\sum_{i=1}^{p}n_i^2}{a}-CF[/Tex]

[Tex]p[/Tex] is the number of groups in first independent variable.
[Tex]n_i[/Tex] is the sum of observation in [Tex]p_i[/Tex] group of first independent variable.
[Tex]a[/Tex] is the number of observation in [Tex]p_i[/Tex] group of first independent variable.

For variation between groups of second variable,

Sum of Squares of Row (SSB_r) = [Tex]\frac{\sum_{j=1}^{q}m_j^2}{b}-CF[/Tex]

[Tex]q[/Tex] is the number of groups in first independent variable.
[Tex]m_i[/Tex] is the sum of observation in [Tex]q_i[/Tex] group of first independent variable.
[Tex]b[/Tex] is the number of observation in [Tex]q_i[/Tex] group of first independent variable.

For the error between the Groups,

Sum of Squares within the Groups (SSE) = [Tex]\sum_{i=1}^{p} \sum_{j=1}^{q} \sum_{k=1}^{n_{ij}} (X_{ijk} – \bar{X}_{ij})^2[/Tex]

[Tex]p[/Tex] is the number of levels (groups) in the first independent variable.
[Tex]q[/Tex] is the number of levels (groups) in the second independent variable.
[Tex]n_{ij}[/Tex] is the number of observations in the cell corresponding to the [Tex]i[/Tex]-th group of the first variable and the [Tex]j[/Tex]-th group of the second variable.
[Tex]X_{ijk}[/Tex] is the [Tex]k[/Tex]-th observation in [Tex]i,j[/Tex]-th cell.
[Tex]\bar{X}_{ij}[/Tex] is the mean of observations in [Tex]i,j[/Tex]-th cell.

For Interaction between Both variables
Sum of Squares Interaction (SSI) = [Tex]SST – SSB_c – SSB_r – SSE[/Tex]

The table will be formed as follows:

Source	Sum of Squares	Degree of Freedom	Mean Sum of Squares	F-ratio
Column	SSB_c	p – 1	MSSBc = [Tex]\frac{SSB_c}{p-1}[/Tex]	[Tex]\frac{MSSB_c}{MSSE}[/Tex]
Row	SSB_r	q – 1	MSSBr = [Tex]\frac{SSB_r}{q-1}[/Tex]	[Tex]\frac{MSSB_r}{MSSE}[/Tex]
Interaction	SSI	(p – 1)(q – 1)	MSSI = [Tex]\frac{SSI}{(p-1)(q-1)}[/Tex]	[Tex]\frac{MSSI}{MSSE}[/Tex]
Residual	SSE	vdf = p x q x (n-1)	MSSE = [Tex]\frac{SSE}{pq(n-1)}[/Tex]	–
Total	SST	N – 1	–	–

Here,

[Tex]p [/Tex] is the number of groups in first independent variable.
[Tex]q[/Tex] is the number of groups in second independent variable.
[Tex]n[/Tex] is the number of observations in one group.
[Tex]N[/Tex] is the total number of observations.

Step 3: Find Critical f-value

Now find Critical f-value for Column, Row and Interaction from the F table at [Tex]\alpha = 0.05[/Tex].

[Tex]f_{c(p-1,vdf)}[/Tex]
[Tex]f_{r(q-1,vdf)}[/Tex]
[Tex]f_{i((p-1)(q-1),vdf)}[/Tex]

Step 4: Make the Decision

Compare the calculated and critical value of f-ratio for all sources.

If [Tex]f_{c(p-1,vdf)} > f_{c \ cal.}[/Tex], then accept the Null Hypothesis, indicating no significant difference between the groups of the first variable, or vice versa.
If [Tex]f_{r((q-1),vdf)} > f_{r \ cal.}[/Tex], then accept the Null Hypothesis, indicating no significant difference between the groups of the second variable, or vice versa.
If [Tex]f_{i((p-1)(q-1),vdf)} > f_{i \ cal.}[/Tex], then accept the Null Hypothesis, indicating no significant interaction between the variables, or vice versa.

Two-way ANOVA Example

Let’s take the example of effect of three different drugs on weight loss of different group of people.

Group of People

Drug X

Drug Y

Drug Z

Step 1: State the Hypothesis

Null Hypothesis ([Tex]\Eta_0[/Tex]) :

There is no significant difference between the effect of drugs.
There is no significant difference of weight loss between the group of people.
There is no interaction between both variables.

Alternate Hypothesis ([Tex]\Eta_1[/Tex]) :

There is a significant difference between the effect of drugs.
There is a significant difference of weight loss between the group of people.
There is an interaction between both variables.

Step 2: Calculate the ANOVA table

Calculate Correction Factor (CF) as,

[Tex]\text{CF} = \frac{(\Sigma{x})^2}{n} \newline = \frac{(25+27+7+8+13+18+21+24+16+11+19+14+29+31+19+21+30+27)^2}{18}\newline =\frac{(360)^2}{18}\newline =7200 [/Tex]

Now calculate the below values:

[Tex]\text{SST}=\sum_{i}^{} \sum_{j}^{} x_{ij}^2 – CF\newline = (25^2+27^2+21^2+24^2+…)-7200\newline = 8144-7200\newline =944[/Tex]

[Tex]\text{SSB}_c=\frac{\sum_{i=1}^{p}n_i^2}{a}-CF\newline = (\frac{(25+27+21+24+29+31)^2}{6}+\frac{(7+8+11+16+19+21)^2}{6}+\frac{(13+18+19+14+30+27)^2}{6})-7200\newline =469[/Tex]

[Tex]\text{SSB}_r=\frac{\sum_{j=1}^{q}m_j^2}{a}-CF\newline = (\frac{(25+27+7+8+13+18)^2}{6}+\frac{(21+24+11+16+19+14)^2}{6}+\frac{(29+31+19+21+30+27)^2}{6})-7200\newline =346.333[/Tex]

For Residual, find mean of each cell. For example, [Tex]\frac{25+27}{2}=26[/Tex]

[Tex]\text{SSE}=\sum_{i=1}^{p} \sum_{j=1}^{q} \sum_{k=1}^{n_{ij}} (X_{ijk} – \bar{X}_{ij})^2\newline = (25-26)^2+(27-26)^2+(7-7.5)^2+(8-7.5)^2+(13-15.5)^2+(18-15.5)^2+…+(30-28.5)^2+(27-28.5)^2\newline =53[/Tex]

The Interaction will be,

[Tex]\text{SSI}=SST – SSB_c – SSB_r – SSE\newline =944-469-346.333-53\newline =75.667 [/Tex]

The table will be formed as follows:

Source	Sum of Squares	Degree of Freedom	Mean Sum of Squares	F-ratio (Calculated)
Drugs (Column)	469	3-1=2	[Tex]\frac{469}{2}=234.5[/Tex]	[Tex]\frac{234.5}{5.889}=39.82[/Tex]
Group of People (Row)	346.333	3-1=2	[Tex]\frac{346.333}{2}=173.1665[/Tex]	[Tex]\frac{173.1665}{5.889}=29.405[/Tex]
Interaction	75.667	(3-1)(3-1)=4	[Tex]\frac{75.667}{4}=18.9167[/Tex]	[Tex]\frac{18.9167}{5.889}=3.2122[/Tex]
Residual	53	3×3(2-1)=9	[Tex]\frac{53}{9}=5.889[/Tex]	–
Total	944	18-1=17	–	–

Step 3: Find Critical f-values

Find Critical values of f-ratio from F-table for,

[Tex]f_{c(p-1,vdf)} = f_{c(2,9)}=4.26[/Tex]
[Tex]f_{r(q-1,vdf)}=f_{r(2,9)}=4.26[/Tex]
[Tex]f_{i((p-1)(q-1),vdf)}=f_{i(4,9)}=3.63[/Tex]

Step 4: Make the Decision

Compare all the calculated and critical f-values.

[Tex]f_{c(p-1,vdf)} < f_{c \ cal.}(4.26<39.82)[/Tex], therefore the Null Hypothesis is rejected and there is a significant difference between the effect of drugs.
[Tex]f_{r((q-1),vdf)} < f_{r \ cal.}(4.26<29.405)[/Tex], therefore the Null Hypothesis is rejected and there is a significant difference of weight loss between the group of people.
[Tex]f_{i((p-1)(q-1),vdf)} > f_{i \ cal.}(3.63>3.2122)[/Tex], there the Null Hypothesis is accepted and there is no interaction between both variables.

Difference between One-Way ANOVA and Two-Way ANOVA

Parameters	One-way ANOVA	Two-way ANOVA
Definition	A statistical test used to determine if there are any statistically significant differences between the means of three or more independent (unrelated) groups.	A statistical test used to determine the effect of two independent variables on a dependent variable and whether there is an interaction between the two variables.
Number of Factors	One independent variable (factor) with multiple levels.	Two independent variables (factors), each with multiple levels.
Hypotheses	Tests if all group means are equal.	Tests main effects of each factor and their interaction effect.
Complexity	Simpler design with one factor.	More complex design with two factors.
Degrees of Freedom	Fewer degrees of freedom required.	More degrees of freedom required due to additional factor and interaction term.
Interaction Effects	Does not account for interaction effects.	Examines interaction between two factors, determining if the effect of one factor depends on the level of the other factor.
Data Requirements	Requires data to be categorized into groups based on a single factor.	Requires data to be categorized into groups based on two factors, and ideally, the sample size should be balanced across groups.
Assumptions	Assumes independence of observations, normality, and homogeneity of variances within groups.	Assumes independence of observations, normality, homogeneity of variances within groups, and additionally, that the interaction effects are consistent across levels.
Example	Comparing mean test scores of students from three teaching methods.	Comparing mean test scores based on teaching method and gender.

Conclusion – One-Way ANOVA vs. Two-Way ANOVA

Understanding the differences between one-way and two-way ANOVA is crucial for conducting effective statistical analyses in research and data interpretation. One-way ANOVA is suitable for comparing the means of multiple groups defined by a single factor, while two-way ANOVA expands this capability by assessing the effects of two independent variables and their interaction. Each method has specific applications and considerations, such as interpreting interaction effects in two-way ANOVA and addressing assumptions like normality and homogeneity of variances. By choosing the appropriate ANOVA method based on your research design and hypotheses, you can effectively analyze and draw meaningful conclusions from your data.

FAQs on One-Way ANOVA vs. Two-Way ANOVA

What is the primary difference between one-way ANOVA and two-way ANOVA?

The primary difference lies in the number of independent variables. One-way ANOVA involves a single independent variable with multiple levels, while two-way ANOVA involves two independent variables, allowing for the examination of interaction effects between them.

When should I use a one-way ANOVA instead of a t-test?

Use a one-way ANOVA when comparing the means of three or more groups. A t-test is limited to comparing the means of only two groups. If you have more than two groups, ANOVA is the appropriate choice to avoid inflating the Type I error rate.

Can I use ANOVA if my data is not normally distributed?

ANOVA assumes normality of the residuals within each group. If this assumption is violated, you might need to transform your data or use a non-parametric alternative such as the Kruskal-Wallis test (for one-way ANOVA) or the Friedman test (for two-way ANOVA).

How do I interpret interaction effects in two-way ANOVA?

Interaction effects occur when the effect of one independent variable depends on the level of the other independent variable. If the interaction term is significant, it suggests that the relationship between the independent variables and the dependent variable is not additive, and you should interpret the main effects with caution, considering the interaction.

What are the post-hoc tests, and when should they be used in ANOVA?

Post-hoc tests, such as Tukey’s HSD, Bonferroni, etc., are used after finding a significant ANOVA result to determine which specific group means differ from each other. They help control for Type I error when making multiple comparisons.

Reffered: https://www.geeksforgeeks.org

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