Questions: You measure 31 watermelons' weights, and find they have a mean weight of 33 ounces. Assume the population standard deviation is 3.3 ounces. Based on this, what is the maximal margin of error associated with a 90% confidence interval for the true population mean watermelon weight. Give your answer as a decimal, to two places

Solution Steps

For a 90% confidence level, the Z-score is found to be \( Z = 1.64 \).

The formula for the margin of error (ME) is given by:

\[ \text{Margin of Error} = \frac{Z \times \sigma}{\sqrt{n}} \]

Substituting the known values:

• \( Z = 1.64 \)
• \( \sigma = 3.3 \)
• \( n = 31 \)

We can compute the margin of error as follows:

\[ \text{Margin of Error} = \frac{1.64 \times 3.3}{\sqrt{31}} \]

Calculating the denominator:

\[ \sqrt{31} \approx 5.57 \]

Now substituting back into the equation:

\[ \text{Margin of Error} = \frac{1.64 \times 3.3}{5.57} \approx \frac{5.412}{5.57} \approx 0.97 \]

The maximal margin of error associated with a 90% confidence interval for the true population mean watermelon weight is:

\[ \pm 0.97 \]

Final Answer