To determine the smallest sample size \( n \) needed to achieve a specified margin of error for a confidence interval for the mean, we use the following formula for the margin of error \( E \):
\[ E = z \cdot \frac{\sigma}{\sqrt{n}} \]
Where:
- \( E \) is the margin of error,
- \( z \) is the z-value corresponding to the desired confidence level,
- \( \sigma \) is the estimated population standard deviation, and
- \( n \) is the sample size.
Given:
- Desired margin of error \( E = 4 , \text{grams} \)
- Z-score for a 95% confidence interval \( z = 1.96 \)
- Estimated standard deviation \( \sigma = 12 , \text{grams} \)
We can rearrange the equation to solve for \( n \):
\[ 4 = 1.96 \cdot \frac{12}{\sqrt{n}} \]
First, isolate \( \sqrt{n} \):
\[ 4 \cdot \sqrt{n} = 1.96 \cdot 12 \]
Calculating \( 1.96 \cdot 12 \):
\[ 1.96 \cdot 12 = 23.52 \]
Now substitute back into the equation:
\[ 4 \cdot \sqrt{n} = 23.52 \]
Dividing both sides by 4:
\[ \sqrt{n} = \frac{23.52}{4} = 5.88 \]
Now, square both sides to solve for \( n \):
\[ n = (5.88)^2 \]
Calculating \( (5.88)^2 \):
\[ n \approx 34.5744 \]
Since sample size \( n \) must be a whole number, we round up to the nearest whole number (always round up to ensure the margin of error requirement is satisfied):
\[ n \approx 35 \]
Thus, the smallest approximate sample size required to obtain the desired margin of error is:
\[ \boxed{35} \]