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:
\((51.74, 62.26)\)
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