Question 1169828
Alright, let's construct the confidence interval for the population mean amount of water in the 1-gallon bottles.

**Understanding the Problem**

We're given the sample mean, standard deviation, and sample size. We need to construct a 99% confidence interval for the population mean.

**Given Information**

* Sample size (n) = 50
* Sample mean (x̄) = 0.971 gallon
* Population standard deviation (σ) = 0.04 gallon
* Confidence level = 99%

**a. Constructing the Confidence Interval**

Since we know the population standard deviation, we'll use the z-distribution.

1.  **Find the z-critical value:**
    * For a 99% confidence interval, the area in the middle is 0.99.
    * The remaining area in the tails is 1 - 0.99 = 0.01.
    * The area in each tail is 0.01 / 2 = 0.005.
    * We need to find the z-value that corresponds to an area of 0.995 (0.99 + 0.005) in the cumulative standard normal distribution.
    * Using a z-table or calculator, the z-critical value for a 99% confidence interval is approximately 2.576.

2.  **Calculate the standard error:**
    * The standard error (SE) is calculated as:
        * SE = σ / √n = 0.04 / √50 ≈ 0.04 / 7.071 ≈ 0.005657

3.  **Calculate the margin of error (ME):**
    * The margin of error is calculated as:
        * ME = z * SE = 2.576 * 0.005657 ≈ 0.01457

4.  **Construct the confidence interval:**
    * The confidence interval is calculated as:
        * x̄ ± ME = 0.971 ± 0.01457

5.  **Calculate the interval endpoints:**
    * Lower bound: 0.971 - 0.01457 ≈ 0.95643
    * Upper bound: 0.971 + 0.01457 ≈ 0.98557

Therefore, the 99% confidence interval is approximately (0.95643, 0.98557).

**Final Answer**

a. The 99% confidence interval estimate for the population mean amount of water is approximately (0.9564, 0.9856) gallons.