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)
Give your answers as decimals, to two places
The answer is not (51.75, 62.25)
Answers
GPT-4o mini
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:
\((51.74, 62.26)\)
\[
\text{Confidence Interval} = \bar{x} \pm z^* \left( \frac{\sigma}{\sqrt{n}} \right)
\]
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:
\((51.74, 62.26)\)