To determine the smallest sample size required for Javon to achieve a specified margin of error with a given confidence level, we can use the formula for the margin of error (ME) for a mean:
\[ ME = z \times \frac{\sigma}{\sqrt{n}} \]
Where:
- \( ME \) is the margin of error
- \( z \) is the z-score corresponding to the desired confidence level
- \( \sigma \) is the population standard deviation
- \( n \) is the sample size
In this case:
- \( ME = 4 \) grams
- \( z = 1.96 \) (for a 95% confidence level)
- \( \sigma = 12 \) grams
We can rearrange the formula to solve for \( n \):
\[ n = \left( \frac{z \cdot \sigma}{ME} \right)^2 \]
Substituting in the given values:
\[ n = \left( \frac{1.96 \cdot 12}{4} \right)^2 \]
Calculating the numerator:
\[ 1.96 \cdot 12 = 23.52 \]
Now substituting this back into the equation:
\[ n = \left( \frac{23.52}{4} \right)^2 \]
Calculating the division:
\[ \frac{23.52}{4} = 5.88 \]
Finally, squaring this result:
\[ n = (5.88)^2 \approx 34.5744 \]
Since sample sizes must be whole numbers, we round up to the nearest whole number:
\[ n \approx 35 \]
Thus, the smallest approximate sample size required to obtain the desired margin of error is 35.