Question

You measure 37 watermelons' weights, and find they have a mean weight of 57 ounces. Assume the population standard deviation is 12.4 ounces. Based on this, construct a 99% confidence interval for the true population mean watermelon weight.

Give your answers as decimals, to two places

Answer 1

To construct a 99% confidence interval for the true population mean watermelon weight, we can use the formula for a confidence interval based on the mean of a sample:

\[
\text{Confidence Interval} = \bar{x} \pm z \cdot \left(\frac{\sigma}{\sqrt{n}}\right)
\]

Where:
- \(\bar{x}\) is the sample mean,
- \(z\) is the z-score corresponding to the desired confidence level,
- \(\sigma\) is the population standard deviation,
- \(n\) is the sample size.

Given the following values:
- Sample mean (\(\bar{x}\)) = 57 ounces
- Population standard deviation (\(\sigma\)) = 12.4 ounces
- Sample size (\(n\)) = 37
- Confidence level = 99% (for which the z-score is approximately 2.576)

First, we calculate the standard error (SE):

\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{12.4}{\sqrt{37}} \approx \frac{12.4}{6.083} \approx 2.04
\]

Next, we compute the margin of error (ME):

\[
ME = z \cdot SE = 2.576 \cdot 2.04 \approx 5.25
\]

Now, we can find the confidence interval:

\[
\text{Lower Limit} = \bar{x} - ME = 57 - 5.25 = 51.75
\]
\[
\text{Upper Limit} = \bar{x} + ME = 57 + 5.25 = 62.25
\]

Thus, the 99% confidence interval for the true population mean watermelon weight is:

\[
\text{Confidence Interval} = (51.75, 62.25)
\]

Final answer: The 99% confidence interval is (51.75, 62.25).