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

The answer is not (51.75, 62.25)

Answer 1

To construct a 99% confidence interval for the true population mean watermelon weight, we will use the formula:

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

Where:
- \(\bar{x}\) is the sample mean.
- \(z^*\) is the critical value for 99% confidence.
- \(\sigma\) is the population standard deviation.
- \(n\) is the sample size.

Given:
- \(\bar{x} = 57\) ounces
- \(\sigma = 12.4\) ounces
- \(n = 37\)

**Step 1: Find the critical value \(z^*\) for 99% confidence.**

For a 99% confidence level, the critical z-value can be looked up in a z-table or calculated. Typically for 99%, \(z^*\) is approximately 2.576.

**Step 2: Calculate the standard error (SE).**

\[
\text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{12.4}{\sqrt{37}}
\]

Calculating \(\sqrt{37}\):

\[
\sqrt{37} \approx 6.083
\]

Now calculating the SE:

\[
\text{SE} = \frac{12.4}{6.083} \approx 2.04
\]

**Step 3: Calculate the margin of error (ME).**

\[
\text{ME} = z^* \times \text{SE} \approx 2.576 \times 2.04 \approx 5.26
\]

**Step 4: Construct the confidence interval.**

\[
\text{Confidence Interval} = \bar{x} \pm \text{ME} = 57 \pm 5.26
\]

Calculating the lower and upper bounds:

Lower bound:

\[
57 - 5.26 \approx 51.74
\]

Upper bound:

\[
57 + 5.26 \approx 62.26
\]

**Final Answer:**

The 99% confidence interval for the true population mean watermelon weight is approximately:

\[
(51.74, 62.26)
\]

Thus, the resultant confidence interval rounded to two decimal places is:

\((51.74, 62.26)\)