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

Question

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=
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.
make it simple in a small paragraph and like a kid wrote it.

Answer 1

To find out how many apples Javon needs to weigh, we can use a formula for finding the sample size for a confidence interval. The formula is:

\[ n = \left( \frac{z \times \sigma}{E} \right)^2 \]

Where:

• \( n \) is the sample size,
• \( z \) is the z-value (which is 1.96 for 95% confidence),
• \( \sigma \) is the estimated standard deviation (12 grams),
• \( E \) is the margin of error (4 grams).

Now, let's plug in the numbers:

\[ n = \left( \frac{1.96 \times 12}{4} \right)^2 \]

First, we calculate \( 1.96 \times 12 \), which equals 23.52.

Now divide that by 4:

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

Now we square that:

\[ 5.88^2 \approx 34.5744 \]

Since we can't have a fraction of an apple, we round up to the nearest whole number. So, Javon needs at least 35 apples to estimate the mean weight with his desired accuracy!