Questions: Eat your cereal: Boxes of cereal are labeled as containing 14 ounces. Following are the weights, in ounces, of a sample of 14 boxes. It is reasonable to assume that the population is approximately normal. 13.09, 14.96, 13.18, 13.19, 13.17, 13.09, 13.22, 14.96, 13.12, 13.11, 13.18, 13.19, 13.05, 13.04 Part 1 of 2 (a) Construct a 99.5% confidence interval for the mean weight. Round the answers to at least three decimal places. A 99.5% confidence interval for the mean weight is <mu<.

Solution Steps

The mean weight of the cereal boxes is calculated using the formula:

\[ \mu = \frac{\sum_{i=1}^N x_i}{N} = \frac{187.55}{14} = 13.396 \]

Thus, the mean weight is \( \mu = 13.396 \).

The variance is calculated as follows:

\[ \sigma^2 = \frac{\sum (x_i - \mu)^2}{n-1} = 0.442 \]

The standard deviation is then:

\[ \sigma = \sqrt{0.442} = 0.665 \]

Thus, the standard deviation is \( \sigma = 0.665 \).

For a confidence level of \( 99.5\% \), the Z critical value is determined using the formula:

\[ Z = \Phi^{-1}\left(1 - \frac{\alpha}{2}\right) \]

The calculated Z critical value is:

\[ Z = 2.807 \]

The margin of error (ME) is calculated using the formula:

\[ \text{ME} = Z \cdot \left(\frac{\sigma}{\sqrt{n}}\right) \]

Substituting the values:

\[ \text{ME} = 2.807 \cdot \left(\frac{0.665}{\sqrt{14}}\right) \approx 0.4989 \]

The confidence interval for the mean weight is given by:

\[ \left( \mu - \text{ME}, \mu + \text{ME} \right) \]

Calculating the lower and upper bounds:

\[ 12.8971 < \mu < 13.8949 \]

Final Answer