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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

3 answers

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:

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

(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.

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.