To find the smallest sample size \( n \) required for a specified margin of error, we can use the formula for the margin of error (ME) in the context of estimating a population mean:
\[ \text{ME} = z \cdot \frac{\sigma}{\sqrt{n}} \]
where:
- \( \text{ME} \) is the margin of error,
- \( z \) is the z-score corresponding to the confidence level,
- \( \sigma \) is the population standard deviation,
- \( n \) is the sample size.
Given:
- \( \text{ME} = 4 \) grams
- \( z = 1.96 \)
- \( \sigma = 12 \) grams
We want to find \( n \) such that:
\[ 4 = 1.96 \cdot \frac{12}{\sqrt{n}} \]
First, we will rearrange the formula to solve for \( n \):
- Multiply both sides by \( \sqrt{n} \):
\[ 4 \sqrt{n} = 1.96 \cdot 12 \]
- Calculate \( 1.96 \cdot 12 \):
\[ 1.96 \cdot 12 = 23.52 \]
So we have:
\[ 4 \sqrt{n} = 23.52 \]
- Divide both sides by 4:
\[ \sqrt{n} = \frac{23.52}{4} = 5.88 \]
- Square both sides to solve for \( n \):
\[ n = (5.88)^2 = 34.5744 \]
Since the sample size \( n \) must be a whole number, we round up to the next whole number (since we can't sample a fraction of an apple), which gives:
\[ n = 35 \]
Thus, the smallest approximate sample size required to obtain the desired margin of error is \( \boxed{35} \).