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To solve this confidence interval problem, let's go through the steps:

### Given Information:
- **Sample size (n)** = 17
- **Sample mean (\(\bar{x}\))** = 5.405 kg
- **Sample standard deviation (s)** = 1.866 kg
- **Confidence level** = 99%, which has a **critical value (t\(_{\alpha/2}\))** = 2.921 for a sample size of 17

### Steps to Calculate the 99% Confidence Interval:

1. **Point Estimate**: 
   - The point estimate is the **sample mean**, which is already given as \( \bar{x} = 5.405 \).

2. **Sample Standard Deviation**: 
   - This value is also given directly as **1.866**.

3. **Standard Error (SE)**:
   - The formula for the standard error of the mean is:
     \[
     \text{SE} = \frac{s}{\sqrt{n}}
     \]
   - Plugging in the values:
     \[
     \text{SE} = \frac{1.866}{\sqrt{17}} \approx 0.4524
     \]

4. **Margin of Error (ME)**:
   - The margin of error for a confidence interval is calculated as:
     \[
     \text{ME} = t_{\alpha/2} \times \text{SE}
     \]
   - Using the critical value for 99% confidence (2.921):
     \[
     \text{ME} = 2.921 \times 0.4524 \approx 1.3219
     \]

5. **Confidence Interval**:
   - The confidence interval is given by:
     \[
     \bar{x} \pm \text{ME}
     \]
   - Plugging in the values:
     \[
     5.405 \pm 1.3219
     \]
   - This results in:
     \[
     \text{Lower limit} = 5.405 - 1.3219 \approx 4.0831
     \]
     \[
     \text{Upper limit} = 5.405 + 1.3219 \approx 6.7269
     \]

### Final Answer:
- **99% Confidence Interval**: \( (4.0831, 6.7269) \)

Please let me know if you want more details or have questions.

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Here are five follow-up questions you might consider:
1. How would the confidence interval change if the sample size were larger?
2. What would the confidence interval be if we used a 95% confidence level instead?
3. How does the critical value affect the width of the confidence interval?
4. Why do we use the t-distribution instead of the z-distribution for this problem?
5. How does the margin of error reflect the variability in the sample?

**Tip**: Confidence intervals become narrower with larger sample sizes, as the standard error decreases.

You are a researcher studying the health habits of people living in a large rural area. You are interested in the amount of weight gained early in life. You want to estimate the population mean of all weight gains for the area’s baby boys in their first year. To estimate the population mean of all such weight gains, you select a random sample of 17 baby boys from this area and you record the weight (in kg) each gained in their first year. Assume the population is approximately normally distributed. Based on your sample, follow the steps below to construct a 99% confidence interval of the weight gains in the first year for all the area’s baby boys.

Learn how to construct a 99% confidence interval to estimate the population mean weight gain of baby boys in their first year. Using a sample of 17, a mean weight gain of 5.405 kg, and a standard deviation of 1.866 kg, we calculate the margin of error and the confidence interval.

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