To determine the smallest sample size \( n \) required to achieve a specified margin of error in estimating the mean, you can use the formula for the margin of error (ME) in a confidence interval, which is given by:
\[ ME = z \times \frac{\sigma}{\sqrt{n}} \]
Where:
- \( ME \) is the desired margin of error (in grams),
- \( z \) is the z-score corresponding to the desired confidence level,
- \( \sigma \) is the population standard deviation (in grams),
- \( n \) is the sample size we wish to determine.
For this problem:
- \( ME = 4 \) grams (the maximum allowable margin of error),
- \( z = 1.96 \) (for a 95% confidence interval),
- \( \sigma = 12 \) grams.
We need to rearrange the formula to solve for \( n \):
\[ 4 = 1.96 \times \frac{12}{\sqrt{n}} \]
Now, isolate \( \sqrt{n} \):
\[ \sqrt{n} = \frac{1.96 \times 12}{4} \]
Calculating the right-hand side:
\[ \sqrt{n} = \frac{23.52}{4} = 5.88 \]
Now, square both sides to find \( n \):
\[ n = (5.88)^2 \]
Calculating \( (5.88)^2 \):
\[ n \approx 34.5744 \]
Since the sample size must be a whole number, 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 of no more than 4 grams is 35 apples.