Question

Javon wants to estimate the mean weight of apples in his orchard. He’ll sample n apples and make a 95% confidence interval for the mean weight, so z=1.96 He is willing to use a standard deviation of 12 grams as an estimate, and he wants the margin of error to be no more than 4 grams. What is the smallest approximate sample size required to obtain the desired margin of error? Show your work.

Answer 1

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:

\[
\boxed{35}
\]