Answer: 0.4169

The standard error of the intercept $\hat{\beta}_0$ is calculated using the formula: $$ \text{SEE}_{\beta_0} = \sqrt{\frac{s^2(\hat{\mu}_X^2 + \hat{\sigma}_X^2)}{n\hat{\sigma}_X^2}} $$ Given: - $s^2 = 20.45$ (estimated error variance) - $\hat{\mu}_X = 0.61$ (sample mean of X) - $\hat{\sigma}_X^2 = 18.65$ (sample variance of X) - $n = 120$ (10 years × 12 months = 120 observations) Plugging in the values: $$ \text{SEE}_{\beta_0} = \sqrt{\frac{20.45(0.61^2 + 18.65)}{120 \times 18.65}} = \sqrt{\frac{20.45(0.3721 + 18.65)}{120 \times 18.65}} = \sqrt{\frac{20.45 \times 19.0221}{2238}} = \sqrt{\frac{389.002}{2238}} = \sqrt{0.1738} = 0.4169 $$ Thus, the standard error estimate of $\hat{\beta}_0$ is 0.4169, which corresponds to option D.

Author: Nikitesh Somanthe