Question

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

Construct a 99.5% confidence interval. round to 3 decimal places

Answer 1

To construct a 99.5% confidence interval for the mean weight of the cereal boxes, we can follow these steps:

### Step 1: Calculate the Sample Mean and Sample Standard Deviation

1. **Data:** The weights of the boxes are:
- 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

2. **Sample Size (n)**: 14

3. **Sample Mean (\(\bar{x}\))**:
\[
\bar{x} = \frac{\sum x_i}{n} = \frac{13.09 + 14.96 + 13.18 + 13.19 + 13.17 + 13.09 + 13.22 + 14.96 + 13.12 + 13.11 + 13.18 + 13.19 + 13.05 + 13.04}{14}
\]
\[
= \frac{184.36}{14} \approx 13.5986
\]

4. **Sample Standard Deviation (s)**:
\[
s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}
\]
First, calculate each \((x_i - \bar{x})^2\):
- (13.09 - 13.5986)² ≈ 0.2605
- (14.96 - 13.5986)² ≈ 1.8829
- (13.18 - 13.5986)² ≈ 0.1752
- (13.19 - 13.5986)² ≈ 0.1668
- (13.17 - 13.5986)² ≈ 0.1826
- (13.09 - 13.5986)² ≈ 0.2605
- (13.22 - 13.5986)² ≈ 0.1430
- (14.96 - 13.5986)² ≈ 1.8829
- (13.12 - 13.5986)² ≈ 0.2300
- (13.11 - 13.5986)² ≈ 0.2380
- (13.18 - 13.5986)² ≈ 0.1752
- (13.19 - 13.5986)² ≈ 0.1668
- (13.05 - 13.5986)² ≈ 0.3025
- (13.04 - 13.5986)² ≈ 0.3102

Sum of squared deviations ≈ 4.2035
\[
s = \sqrt{\frac{4.2035}{14-1}} = \sqrt{\frac{4.2035}{13}} \approx 0.5486
\]

### Step 2: Find the Critical Value

For a 99.5% confidence interval with \( n - 1 = 13 \) degrees of freedom, we can find the critical t-value (using a t-distribution table or calculator).

The critical t-value for \( \alpha = 0.005 \) (0.5% in each tail) and 13 degrees of freedom is approximately \( t_{0.005, 13} \approx 3.012 \).

### Step 3: Calculate the Margin of Error (ME)

\[
\text{Margin of Error (ME)} = t^* \cdot \frac{s}{\sqrt{n}}
\]
\[
= 3.012 \cdot \frac{0.5486}{\sqrt{14}} \approx 3.012 \cdot 0.1464 \approx 0.4414
\]

### Step 4: Construct the Confidence Interval

The confidence interval is given by:

\[
\bar{x} \pm \text{ME}
\]

Calculating:

\[
(13.5986 - 0.4414, 13.5986 + 0.4414) = (13.1572, 14.0400)
\]

### Step 5: Round to 3 Decimal Places

Thus, the 99.5% confidence interval for the mean weight of the cereal boxes is:

\[
\boxed{(13.157, 14.040)}
\]

Answer 2

(b) The quality control manager is concerned that the mean weight is actually less than 14 ounces. Based on the confidence interval, is there a reason to be concerned? Explain.
The quality control manager should [be or not be?] concerned that the mean weight is actually less than 14 ounces.

Answer 3

To evaluate the concern of the quality control manager regarding whether the mean weight of the cereal boxes is actually less than 14 ounces, we can analyze the 99.5% confidence interval we calculated:

\[
\text{Confidence Interval: } (13.157, 14.040)
\]

### Interpretation of the Confidence Interval

1. **Range of Values**: The interval \( (13.157, 14.040) \) indicates that, based on the sample data, we are 99.5% confident that the true mean weight of the cereal boxes falls between approximately 13.157 ounces and 14.040 ounces.

2. **Concern About Mean Being Less Than 14 Ounces**:
- The upper limit of the interval (14.040) is slightly above 14 ounces, but the lower limit (13.157) is well below 14 ounces.
- This confidence interval includes values that are less than 14 ounces (as indicated by the lower limit).

### Conclusion

Because the confidence interval includes values below 14 ounces, the quality control manager **should be concerned** that the mean weight is actually less than 14 ounces.

Thus, to summarize:

The quality control manager should **be concerned** that the mean weight is actually less than 14 ounces, as the confidence interval includes values that are below 14 ounces.