SOLUTION: A bottled water distributor wants to estimate the amount of water contained in 1-gallon bottles purchased from a nationally known water bottling company. The water bottling c
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Question 1169828: A bottled water distributor wants to estimate the amount of water contained in 1-gallon bottles purchased from a nationally known water bottling company. The water bottling company's specifications state that the standard deviation of the amount of water is equal to 0.04 gallon. A random sample of 50 bottles is selected, and the sample mean amount of water per 1-gallon bottle is 0.971 gallon. Complete parts (a) through (d).
a. Construct a 99 % confidence interval estimate for the population mean amount of water included in a 1-gallon bottle.
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
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.
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