Two Way ANOVA Lecture 47

Two Way ANOVA 

Lecture 47

Introduction

The two-way ANOVA technique is a statistical technique used to analyse the effect of several levels of two categorical independent variables on a quantitative dependent variable simultaneously. The quantitative dependent variable data are classified data on the basis of bi-criteria simultaneously. It is the extension of one-way ANOVA which studies the effect of several levels of one categorical variable or factor. Actually, the two-way ANOVA is used to study the bivariate classification of data on quantitative dependent variables at several levels of categorical independent variables.

Example

Suppose a researcher is interested in investigating the effect of three fertilisers on four varieties of potato yield and also wants to know whether the three fertilisers have a significant effect on the potato yield. The researcher further wants to know if the yield capabilities of four kinds of potatoes are significant.  

i. The application of two-way ANOVA is based on the following assumptions.

ii. The dependent variable should be measured on a continuous scale.

iii. The independent variable should be categorical.

iv. The collected data on the dependent variable should be independent.

v. The dependent variable should be following normal or approximately normal distribution.

vi. The variances of the sampled population should be identical.

Statistical Model of Two Way ANOVA

The two-way ANOVA is represented by the following linear model.

Where:

ρi: The effect of row-wise arranged categorical variables.

τj: The effect of row-wise arranged categorical variables. 

ϵij: Random error.

Analysis

Let a set of n observations be classified in “c” columns (called treatment 1) and “r” rows (called treatment 2). There are c columns and r rows in a table; then there will be r x c = n cells.

Let Yij denote an observation in the ith row and the jth column in a table consisting of r rows and c columns and containing samples from normal populations with means μij and the common variance of σ^2 classified according to two criteria of classification simultaneously.

The correction factor (CF) can be obtained as:

The sum of squares total can be obtained as:

The sum of square column treatment can be obtained as:

The sum of square row treatment can be obtained as:

The sum of square error can be obtained as:

Presentation in the ANOVA table:

Testing Procedure:

i. State null and alternative hypothesis

H01: μ•1 = μ•2 = ⋯ = μ•c vs. H11: μ•1 ≠ μ•2 ≠ ⋯ ≠ μ•c

H02: μ1•=μ2•=⋯=μr• Vs. H12: μ1•≠μ2•≠⋯≠μr•

ii. The significance level: α

iii. The test statistic:

iv. Critical Region:

Reject H01 when F1 > Fα (c-1, (r-1)(c-1))

Reject H02 when F2 > Fα (r-1, (r-1)(c-1))

v. Computation

vi. Remarks

Example 12.3: A supervisor is interested in knowing the performance of machines and the operators working on the machines. To perform the experiment, four machines and five operators are selected, and the following observations are collected at the end of the week.

Operator	Machine 1	Machine 2	Machine 3	Machine 4
1	46	56	55	47
2	54	55	51	56
3	48	56	50	58
4	46	60	51	59
5	51	53	53	58

Test the hypotheses at a 5% significance level that the performance of four machines and five operators are identical. Also write the appropriate statistical model for the above experiment.

Solution: The above experiment is represented by the following statistical model:

$$

Yij = μ + τ machines + τ operators + ϵij, i = 1, 2, 3, 4, 5, j = 1, 2, 3, 4

Where: 

μ is the average performance of machines and operators based on historical data.

τ machines is the performance of the machines in the current experiment.

τ operators is the performance of the operators in the current experiment.

ϵij is the random error.

i. State null and alternative hypothesis

H01: μ•1 = μ•2 = μ•3 = μ•4 vs. H11: μ•1 μ•2 ≠ μ•3 ≠ μ•4

H02: μ1• = μ2• = μ3• = μ4• = μ5•. H12: μ1• ≠ μ2• ≠ μ3• ≠ μ4• ≠ μ5•

ii. The significance level: α = 0.05

iii. The test statistic:

vi. Critical Region:

Reject H01 when FMachines > F0.05 (3, 12) = 3.49

Reject H02 when FOperators > F0.05 (4, 12) = 3.26

v. Computation

Presentation in the ANOVA Table:

vi. Remarks: The computed F value for four machines falls in the rejection region, and the F computed value for five operators falls in the acceptance region. Thus, it is concluded that the four machines are significant.

Post Hoc Test

The post hoc tests are used to investigate the pairwise comparison of treatment means.

• Read More: Post Hoc Tests