Answer: 24.5

The Box-Pierce Q-statistic is given by: $$Q_{BP} = n \sum \rho_k^2$$ Where $n$ is the sample size and $\rho$ represents the autocorrelations. Therefore, $$Q_{BP} = 200(0.3^2 + (-0.15)^2 + (-0.10)^2) = 24.5$$ Calculation breakdown: - $0.3^2 = 0.09$ - $(-0.15)^2 = 0.0225$ - $(-0.10)^2 = 0.01$ - Sum = 0.09 + 0.0225 + 0.01 = 0.1225 - $Q_{BP} = 200 \times 0.1225 = 24.5$

Author: Tanishq Prabhu