To determine the sample size needed to construct a confidence interval with a given margin of error, we can use the formula for the sample size for a population mean when the population standard deviation (σ) is known:

$n = \left( \frac{Z_{\alpha/2} \cdot \sigma}{E} \right)^2$

Where:

• $n$ is the sample size.
• $Z_{\alpha/2}$ is the critical value corresponding to the desired confidence level.
• $\sigma$ is the population standard deviation.
• $E$ is the margin of error (the maximum error allowed).

Step-by-step solution:

1. Critical value for 95% confidence level: For a 95% confidence interval, the Z-score ($Z_{\alpha/2}$) is 1.96 (this comes from standard normal distribution tables).

2. Given values:

   • $\sigma = 8.8 \, \text{lbs}$
   • $E = 2.62 \, \text{lbs}$

3. Plug values into the formula:

$n = \left( \frac{1.96 \cdot 8.8}{2.62} \right)^2$

4. Calculate:

$n = \left( \frac{17.328}{2.62} \right)^2 = (6.61)^2 = 43.7$

Since we need to round up to the nearest whole number, the sample size should be:

$n = 44$

Conclusion:

The zoologist must randomly select 44 elephants to construct the 95% confidence interval with an error of at most 2.62 lbs.

---

Do you need further clarification on any step or details about confidence intervals?

Here are some follow-up questions:

1. What happens if we used a different confidence level, say 90% or 99%? How does that affect the sample size?
2. How does the margin of error influence the sample size?
3. What if the population standard deviation was not known? How would the sample size formula change?
4. Can we use the same formula for constructing a confidence interval for proportions?
5. How would the calculation change if the sample size was already partially determined?

Tip: Increasing the confidence level or decreasing the margin of error will both require a larger sample size to achieve the same level of precision.