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
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.(2 points)

Answer 1

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:

\[
\boxed{35}
\]