To calculate the required sample size $n$ for a given margin of error $E$, we use the formula for the sample size in estimating a population mean:

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

Where:

• $Z_{\alpha/2}$ is the z-score corresponding to the desired confidence level (for a 95% confidence interval, $Z_{\alpha/2} = 1.96$).
• $s$ is the sample standard deviation.
• $E$ is the margin of error.

Given:

• Margin of error, $E = 8.5$ grams.
• Sample standard deviation, $s = 68.8$ grams.
• Confidence level is 95%, so $Z_{\alpha/2} = 1.96$.

Substitute the given values into the formula:

$n = \left( \frac{1.96 \cdot 68.8}{8.5} \right)^2$

Let's calculate that.The required sample size is 252.

Would you like further details on the calculation or any other clarification?

Here are 5 related questions you might explore next:

1. How does changing the margin of error affect the sample size?
2. What happens to the sample size if the confidence level is increased to 99%?
3. How do you interpret a confidence interval in terms of probability?
4. What are the assumptions needed for constructing a confidence interval for a population mean?
5. How can the sample size formula change if the population standard deviation is known?

Tip: Increasing the confidence level increases the required sample size, as higher confidence demands more precision.