Problem
Let $Y_1 \sim N(μ_1,1),Y_2 \sim N(μ_2,1), \ \mathrm{and} \ Y_3 \sim N(μ_3,1)$ and also assume that these three random variables are mutually independent. The observed sample values are $y_1 = 2, y_2 = 0 \ \mathrm{and} \ y_3 = −1$, respectively. We are interested in doing inference on the parameter $$\theta = 0.1\mu_1 + 0.2\mu_2 + 0.7\mu_3,$$ where $\theta$ is a weighted average of the population means.
$(a)\quad$ Derive an exact lower range $90\%$ confidence interval for $\theta$ and calculate this interval for the given sample.
$(b)\quad$ Now suppose that $\mu_2 = 0$ and $\mu_3 = 0$ are known. Derive an exact upper range $90\%$ confidence interval for $\theta$ taking this additional information into account. Calculate this interval for the given sample.
My answer
$(a)$
We know that $$\overline{Y} = 0.1 Y_1 + 0.2 Y_2 + 0.7 Y_3 \sim N\left(0.1\mu_1 + 0.2\mu_2 + 0.7\mu_3 , \frac{27}{50}\right).$$
Thus, the exact lower range $90\%$ confidence interval for $\theta$ is $$\begin{aligned} \theta & = \left(- \infty, [0.1(2) + 0.2(0) + 0.7(-1)] + 1.282 \sqrt{\frac{27}{50}}\right)\\ & = \left(- \infty, -0.5 + 1.282 \sqrt{\frac{27}{50}}\right) \end{aligned}$$
$(b)$
Similarly, the exact upper range 90% confidence interval for $\theta$ is $$\begin{aligned} \theta & = \left(0.1(2) - 1.282 \sqrt{\frac{27}{50}}, \infty \right)\\ & = \left(0.2 - 1.282 \sqrt{\frac{27}{50}}, \infty \right) \end{aligned}$$
As I have just covered confidence intervals, I am uncertain on my calculations and approaches. Any intuitive explanations will be highly appreciated!