You measure 49 watermelons' weights, and find they have a mean weight of 60 ounces. Assume the population standard deviation is 5 ounces. Based on this, construct a 99% confidence interval for the true population mean watermelon weight.

Give your answers as decimals, to two places

1 answer

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

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

Where:

  • \(\bar{x}\) is the sample mean.
  • \(z^*\) is the z-value that corresponds to the desired confidence level (99% in this case).
  • \(\sigma\) is the population standard deviation.
  • \(n\) is the sample size.

Step 1: Identify the values

  • Sample mean (\(\bar{x}\)) = 60 ounces
  • Population standard deviation (\(\sigma\)) = 5 ounces
  • Sample size (\(n\)) = 49

Step 2: Find the z-value for 99% confidence level

For a 99% confidence level, the z-value (from the z-table) is approximately:

\[ z^* \approx 2.576 \]

Step 3: Calculate the standard error of the mean (SEM)

The standard error of the mean is calculated as follows:

\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{5}{\sqrt{49}} = \frac{5}{7} \approx 0.7143 \]

Step 4: Calculate the margin of error

The margin of error (ME) is given by:

\[ ME = z^* \cdot SE = 2.576 \cdot 0.7143 \approx 1.8387 \]

Step 5: Calculate the confidence interval

Now we can calculate the confidence interval:

\[ CI = \bar{x} \pm ME = 60 \pm 1.8387 \]

Calculating the lower and upper bounds:

  • Lower bound: \(60 - 1.8387 \approx 58.16\)
  • Upper bound: \(60 + 1.8387 \approx 61.84\)

Final Result

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

\[ \boxed{(58.16, 61.84)} \]