Questions: 8.2 Homework Question 10 of 12 (1 point) Question Attempt 1 of Unlimited 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 <μ<.

8.2 Homework
Question 10 of 12 (1 point) Question Attempt 1 of Unlimited

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 <μ<.
Transcript text: 8.2 Homework Question 10 of 12 (1 point) Question Attempt 1 of Unlimited 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 $\square$ $<\mu<$ $\square$.
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Solution

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Solution Steps

Step 1: Calculate the Mean

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

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

Thus, the mean weight is \( \bar{x} = 13.396 \).

Step 2: Calculate the Sample Standard Deviation

The sample standard deviation \( s \) is computed as follows:

\[ s = 0.665 \]

Step 3: Calculate the Confidence Interval

To construct the 99.5% confidence interval for the mean, we use the formula:

\[ \bar{x} \pm t \frac{s}{\sqrt{n}} \]

Where:

  • \( \bar{x} = 13.396 \)
  • \( t \) is the t-value for \( n-1 = 13 \) degrees of freedom at a 99.5% confidence level, which is approximately \( 3.372 \).
  • \( s = 0.665 \)
  • \( n = 14 \)

Substituting the values, we have:

\[ 13.396 \pm 3.372 \cdot \frac{0.665}{\sqrt{14}} \]

Calculating the margin of error:

\[ \text{Margin of Error} = 3.372 \cdot \frac{0.665}{\sqrt{14}} \approx 0.599 \]

Thus, the confidence interval is:

\[ (13.396 - 0.599, 13.396 + 0.599) = (12.797, 13.995) \]

Final Answer

The 99.5% confidence interval for the mean weight is

\[ \boxed{(12.797, 13.995)} \]

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