To determine the smallest sample size \( n \) required for Javon to achieve a margin of error (ME) of no more than 4 grams in constructing a 95% confidence interval for the mean weight of apples, we can use the formula for the margin of error in a confidence interval for the mean when the standard deviation is known:
\[ ME = z \cdot \frac{\sigma}{\sqrt{n}} \]
Where:
- \( ME \) is the margin of error,
- \( z \) is the z-score corresponding to the desired confidence level (1.96 for 95% confidence),
- \( \sigma \) is the standard deviation (12 grams), and
- \( n \) is the sample size.
We need to find \( n \) such that:
\[ ME \leq 4 \]
Substituting the values we have into the margin of error formula:
\[ 4 = 1.96 \cdot \frac{12}{\sqrt{n}} \]
To solve for \( n \), we first rearrange the equation to isolate \( \sqrt{n} \):
\[ \sqrt{n} = 1.96 \cdot \frac{12}{4} \]
Calculating the right-hand side:
\[ \sqrt{n} = 1.96 \cdot 3 = 5.88 \]
Next, we square both sides to find \( n \):
\[ n = (5.88)^2 \]
Calculating:
\[ n \approx 34.5744 \]
Since the sample size must be a whole number, we round up to the nearest whole number to ensure the margin of error requirement is met:
\[ n = 35 \]
Thus, the smallest approximate sample size required to obtain the desired margin of error is:
\[ \boxed{35} \]