Answer: 22.74

The Ljung-Box Q statistic formula is: \[Q(m) = n \sum_{j=1}^{h} \left( \frac{n + 2}{n - j} \right) \rho_j^2 = n(n + 2) \sum_{j=1}^{h} \left( \frac{\rho_j^2}{n - j} \right)\] In this case, time lag \( m = 3 \). Thus, \[Q(3) = 300(302) \left[ \frac{0.25^2}{299} + \frac{(-0.1)^2}{298} + \frac{(-0.05)^2}{297} \right] = 22.74\] **Step-by-step calculation:** - n = 300 data points - ρ₁ = 0.25, ρ₂ = -0.1, ρ₃ = -0.05 - Q(3) = 300 × 302 × [(0.0625/299) + (0.01/298) + (0.0025/297)] - Q(3) = 90,600 × [0.000209 + 0.0000336 + 0.00000842] - Q(3) = 90,600 × 0.00025102 = 22.74 The Ljung-Box Q statistic tests whether any of a group of autocorrelations of a time series are different from zero, and 22.74 is the correct calculated value.

Author: Tanishq Prabhu