Example

Regression Line given by Y_ij = 41.13 + 2.4801 IQ_ij + 1.029 IQ (mean_j) + U_oj + U_1j IQ_ij + R_ij
Standard Deviation of random slope “Language Score” = √0.195 = 0.44
Average slope = 2.480

Correlation between random slope and random intercept = -0.83/√(8.88×0.195) = -0.63
The negative correlation means that classes with a higher performance for a pupil of average intelligence have a lower within-class effect of intelligence. Thus, the higher average performance tends to be achieved more by higher language scores of the less intelligent, than by higher scores of the more intelligence students.

Variance of Language Score, given IQ_ij = -4 = 8.88 + 2x(-0.835)x(-4) + (-4)² x0.195 + 39.69 = 58.37
Variance of Language Score, given IQ_ij = 4 = 8.88 + 2x(-0.835)x(4) + (4)² x0.195 + 39.69 = 45.01
Covariance for Language Score between two different individual in the same group = 8.88 – 4² x 0.195 = 5.76
Correlation Coefficient = 5.76/√(58.37×45.01) = 0.11

---

Explanation to random intercepts and slopes

Can be achieve by

• explaining variability between individuals
• explaining variability between group
  • group level one explain variability of slope & intercepts

β_0j = γ₀₀ + γ₀₁ z_j + U_0j , intercepts as outcome model

β_1j = γ₁₀ + γ₁₁ z_j + U_1j , slope as outcome model

Both β_0j and β_1j is latent regression because cannot be observed without error.