Problem Breakdown

The problem asks to calculate confidence intervals for $\theta$ (mean of Gaussian random variables $X_1, \ldots, X_N$) with known variance $\sigma^2 = 1$. $N = 30$ and confidence level $1 - \alpha$.

The formulas provided are:

1. (a) $Z = \hat{\theta} - \frac{2.14}{\sqrt{N}} \leq \theta \leq \hat{\theta} + \frac{2.14}{\sqrt{N}}$
2. (b) $Z = \hat{\theta} - \frac{1.85}{\sqrt{N}} \leq \theta \leq \hat{\theta} + \frac{1.85}{\sqrt{N}}$

Now, let’s compute these step by step.

Step 1: Identify parameters

• $N = 30$
• $\sigma^2 = 1 \Rightarrow \sigma = 1$
• $\sqrt{N} = \sqrt{30} = 5.477$ (approximately)

Step 2: Confidence Interval (a)

The formula is: $\hat{\theta} - \frac{2.14}{\sqrt{N}} \leq \theta \leq \hat{\theta} + \frac{2.14}{\sqrt{N}}$

1. Calculate the margin of error: $\frac{2.14}{\sqrt{N}} = \frac{2.14}{5.477} \approx 0.3907$

2. Substitute the value of the margin of error: $\theta \in \left[ \hat{\theta} - 0.3907, \hat{\theta} + 0.3907 \right]$

Thus, the confidence interval is: $\left[ \hat{\theta} - 0.3907, \hat{\theta} + 0.3907 \right]$

Step 3: Confidence Interval (b)

The formula is: $\hat{\theta} - \frac{1.85}{\sqrt{N}} \leq \theta \leq \hat{\theta} + \frac{1.85}{\sqrt{N}}$

1. Calculate the margin of error: $\frac{1.85}{\sqrt{N}} = \frac{1.85}{5.477} \approx 0.3378$

2. Substitute the value of the margin of error: $\theta \in \left[ \hat{\theta} - 0.3378, \hat{\theta} + 0.3378 \right]$

Thus, the confidence interval is: $\left[ \hat{\theta} - 0.3378, \hat{\theta} + 0.3378 \right]$

Final Results

1. For (a): Confidence interval is: $\left[ \hat{\theta} - 0.3907, \hat{\theta} + 0.3907 \right]$

2. For (b): Confidence interval is: $\left[ \hat{\theta} - 0.3378, \hat{\theta} + 0.3378 \right]$

Do you want further clarifications, or should I assist with anything else?

Follow-Up Questions

1. What is the significance of using different constants (e.g., 2.14 and 1.85) in the formulas?
2. How does increasing $N$ affect the width of the confidence intervals?
3. Can you explain why the variance $\sigma^2$ affects the margin of error?
4. How does the confidence level $1-\alpha$ relate to the values 2.14 and 1.85?
5. Would the confidence interval change if $\sigma^2$ were unknown?

Tip:

The width of a confidence interval is inversely proportional to $\sqrt{N}$, so increasing the sample size $N$ decreases the interval width, providing a more precise estimate.