To answer this question, we need to calculate the required sample size $n$ using the formula for confidence intervals for a population mean with a known standard deviation. The formula is:

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

Where:

• $Z$ is the Z-score corresponding to the desired confidence level (for 95%, $Z = 1.96$),
• $\sigma$ is the population standard deviation (23.4 oz),
• $E$ is the margin of error (4 oz).

Step 1: Identify the known values

• $Z = 1.96$ (for 95% confidence level),
• $\sigma = 23.4$ oz,
• $E = 4$ oz.

Step 2: Plug the values into the formula

$n = \left( \frac{1.96 \times 23.4}{4} \right)^2$

$n = \left( \frac{45.864}{4} \right)^2$

$n = \left( 11.466 \right)^2$

$n = 131.4$

Step 3: Round up

Since the sample size must be a whole number, we round up to the nearest whole number, so:

$n = 132$

Conclusion:

To estimate the mean weight of lion cubs within 4 oz with 95% confidence, you need to survey 132 lion cubs.

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Would you like a more detailed explanation of this process? Here are 5 related questions for further practice:

1. How would the sample size change if the confidence level was increased to 99%?
2. What is the role of the population standard deviation in determining sample size?
3. How does changing the margin of error affect the sample size needed?
4. How would you approach this problem if the population standard deviation was unknown?
5. How does the central limit theorem relate to this problem?

Tip: When calculating sample sizes, always remember to round up to the nearest whole number to ensure that the sample size is sufficient.