To determine the smallest sample size \( n \) required for Javon to achieve a margin of error of no more than 4 grams with a 95% confidence interval, we can use the formula for the margin of error (ME) in estimating a population mean:
\[ ME = z \cdot \frac{\sigma}{\sqrt{n}} \]
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 population standard deviation,
- \( n \) is the sample size.
According to the problem, we have:
- Desired margin of error \( ME \) = 4 grams,
- Standard deviation \( \sigma \) = 12 grams,
- z-score \( z \) = 1.96.
Plugging these values into the margin of error formula:
\[ 4 = 1.96 \cdot \frac{12}{\sqrt{n}} \]
We want to solve for \( n \). First, rearrange the equation to isolate \( \sqrt{n} \):
\[ \sqrt{n} = 1.96 \cdot \frac{12}{4} \]
Calculating the right side:
\[ \sqrt{n} = 1.96 \cdot 3 = 5.88 \]
Next, squaring both sides to eliminate the square root yields:
\[ n = (5.88)^2 \]
Calculating \( n \):
\[ n = 34.5744 \]
Since the sample size \( n \) must be a whole number, we round up to the nearest whole number:
\[ n = 35 \]
Thus, the smallest approximate sample size required to achieve the desired margin of error of no more than 4 grams is
\[ \boxed{35}. \]