Let I, J > 5. Consider two-way ANOVA, where the observations satisfy the linear model

y_ij = α + β_i + γ_j + ε_ij, 1 ≤ i ≤ I, 1 ≤ j ≤ J 

E(ε_ij) = 0, Var(ε_ij) = σ², \(\sum_{i=1}^{!} β_i=\sum_{j=1}^j γ_j=0 .\)

In this set - up

β1 - β2 is estimable

Option (3) is correct

Given:-

Given set up is,

yij = α + βi + γj + εij, 1 ≤ i ≤ I, 1 ≤ j ≤ J 

E(εij) = 0, Var(εij) = σ2, 

\(\sum_{i=1}^{!} β_i=\sum_{j=1}^j γ_j=0 .\)

Concept Used:-

In the given setup, the requirement for a parameter to be estimable is that it can be estimated uniquely from the observed data. For a linear combination of parameters to be estimable, its coefficients must satisfy the condition of being a linearly independent combination.

Explanation:-

As, the constraint on the β parameters is only that their sum is zero, it is possible to estimate all β parameters except for one (which can be obtained by subtracting the sum of the other β parameters from zero).

So, β₁ is estimable.

As, there are no constraints on the γ parameters, their sum is not necessarily zero, and it is not possible to estimate any individual γ parameter without additional information.

Therefore, γ₁ is not estimable.

Now, β₁ - β₂ is not necessarily estimable because the difference between two specific βi parameters can only be estimated if I > 2 (i.e., there are at least three levels of the i factor).

Similarly, β₁ + γ₂ is not necessarily estimable because the sum of two specific parameters can only be estimated if both factors have at least two levels each (i.e., I > 1 and J > 1).

Whether β1 - β2 or β1 + γ2 are estimable depends on the values of I and J.

So, in the given set - up, β1 is estimable.

Hence, the correct option is 1.