Questions: You measure 41 watermelons' weights, and find they have a mean weight of 43 ounces. Assume the population standard deviation is 11.2 ounces. Based on this, construct a 95% confidence interval for the true population mean watermelon weight. Round your answers to two decimal places.

You measure 41 watermelons' weights, and find they have a mean weight of 43 ounces. Assume the population standard deviation is 11.2 ounces. Based on this, construct a 95% confidence interval for the true population mean watermelon weight. Round your answers to two decimal places.
Transcript text: You measure 41 watermelons' weights, and find they have a mean weight of 43 ounces. Assume the population standard deviation is 11.2 ounces. Based on this, construct a $95 \%$ confidence interval for the true population mean watermelon weight. Round your answers to two decimal places.
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Solution

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Solution Steps

Step 1: Given Information

We have measured the weights of 41 watermelons, resulting in a sample mean (\(\bar{x}\)) of 43 ounces. The population standard deviation (\(\sigma\)) is 11.2 ounces. We aim to construct a 95% confidence interval for the true population mean (\(\mu\)).

Step 2: Determine the Z-Score

For a 95% confidence level, the Z-score corresponding to the critical value is \(z = 1.96\).

Step 3: Calculate the Standard Error

The standard error (SE) is calculated using the formula: \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{11.2}{\sqrt{41}} \approx 1.743 \]

Step 4: Calculate the Margin of Error

The margin of error (ME) is given by: \[ ME = z \cdot SE = 1.96 \cdot 1.743 \approx 3.42 \]

Step 5: Construct the Confidence Interval

The confidence interval is calculated as: \[ \bar{x} \pm ME = 43 \pm 3.42 \] This results in: \[ (43 - 3.42, 43 + 3.42) = (39.58, 46.42) \]

Final Answer

The 95% confidence interval for the true population mean watermelon weight is: \[ \boxed{39.58 < \mu < 46.42} \]

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