Answer: [1.3997, 1.9002]

The confidence interval is given by: $$[\hat{\beta} - C_t \times \text{SEE}_{\hat{\beta}}, \hat{\beta} + C_t \times \text{SEE}_{\hat{\beta}}]$$ The critical value ($C_t$) at 1% (0.5% on each tail) level is 2.618. (From the standard normal table) Now, $\text{SEE}_{\hat{\beta}}$ is given by, $$\text{SEE}_{\hat{\beta}} = \sqrt{\frac{s^2}{n \hat{\sigma}^2_X}} = \sqrt{\frac{20.45}{120 \times 18.65}} = 0.0956$$ So, $$[\hat{\beta} - C_t \times \text{SEE}_{\hat{\beta}}, \hat{\beta} + C_t \times \text{SEE}_{\hat{\beta}}] = [1.65 - 2.618 \times 0.0956, 1.65 + 2.618 \times 0.0956] \\ = [1.3997, 1.9002]$$

Author: Nikitesh Somanthe