Answer: F-statistic = 12.96; At least one of the 5 independent variables is significantly different from zero.

## Explanation ### F-Statistic Calculation The F-statistic is calculated using the formula: $$F = \frac{[\frac{\text{ESS}}{k}]}{[\frac{\text{SSR}}{(n-k-1)}]}$$ Where: - ESS (Explained Sum of Squares) = 270 - SSR (Sum of Squared Residuals) = 250 - k (number of independent variables) = 5 - n (number of observations) = 66 Substituting the values: $$F = \frac{[\frac{270}{5}]}{[\frac{250}{(66-5-1)}]} = \frac{[54]}{[\frac{250}{60}]} = \frac{54}{4.1667} = 12.96$$ ### Hypothesis Testing The hypotheses are: - **H₀**: B₁ = B₂ = B₃ = B₄ = B₅ = 0 (all coefficients are zero) - **H₁**: At least one Bⱼ ≠ 0 (at least one coefficient is non-zero) ### Critical Value Comparison The critical F-value at 10% significance level with: - Numerator degrees of freedom = k = 5 - Denominator degrees of freedom = n - k - 1 = 66 - 5 - 1 = 60 Critical F-value ≈ 1.946 Since 12.96 > 1.946, we **reject the null hypothesis**. ### Conclusion We conclude that **at least one of the five independent variables is significantly different from zero**. Therefore, option A is correct with F-statistic = 12.96 and the correct conclusion.

Author: Tanishq Prabhu