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.(2 points)

Answer 1

To find the smallest sample size \( n \) required for a specified margin of error, we can use the formula for the margin of error (ME) in the context of estimating a population mean:

\[
\text{ME} = z \cdot \frac{\sigma}{\sqrt{n}}
\]

where:
- \( \text{ME} \) is the margin of error,
- \( z \) is the z-score corresponding to the confidence level,
- \( \sigma \) is the population standard deviation,
- \( n \) is the sample size.

Given:
- \( \text{ME} = 4 \) grams
- \( z = 1.96 \)
- \( \sigma = 12 \) grams

We want to find \( n \) such that:

\[
4 = 1.96 \cdot \frac{12}{\sqrt{n}}
\]

First, we will rearrange the formula to solve for \( n \):

1. Multiply both sides by \( \sqrt{n} \):

\[
4 \sqrt{n} = 1.96 \cdot 12
\]

2. Calculate \( 1.96 \cdot 12 \):

\[
1.96 \cdot 12 = 23.52
\]

So we have:

\[
4 \sqrt{n} = 23.52
\]

3. Divide both sides by 4:

\[
\sqrt{n} = \frac{23.52}{4} = 5.88
\]

4. Square both sides to solve for \( n \):

\[
n = (5.88)^2 = 34.5744
\]

Since the sample size \( n \) must be a whole number, we round up to the next whole number (since we can't sample a fraction of an apple), which gives:

\[
n = 35
\]

Thus, the smallest approximate sample size required to obtain the desired margin of error is \( \boxed{35} \).