Answer: $0.194 < b < 1.006$

The 95% confidence interval for the slope coefficient b is calculated using the formula: $$CI = \hat{b} \pm t_{\frac{\alpha}{2}, n-2} s_{\hat{b}}$$ Where: - $\hat{b} = 0.6$ (point estimate) - $s_{\hat{b}} = 0.2$ (standard error) - $\alpha = 5\%$, so $\alpha/2 = 0.025$ - Degrees of freedom = n - 2 = 36 - 2 = 34 - $t_{0.025,34} = \pm 2.03$ (from t-distribution table) Calculation: $$CI = 0.6 \pm 2.03 \times 0.2 = 0.6 \pm 0.406 = [0.194, 1.006]$$ Therefore, the 95% confidence interval is $0.194 < b < 1.006$. **Note:** The degrees of freedom are n-2 because we're estimating two parameters (intercept and slope) in simple linear regression.

Author: Nikitesh Somanthe