To construct a 98% confidence interval for the mean weight of the cereal boxes, we can follow these steps:
- Calculate the sample mean (\(\bar{x}\)).
- Calculate the sample standard deviation (s).
- Determine the critical value from the t-distribution for a 98% confidence level.
- Use the formula for the confidence interval.
Step 1: Calculate the Sample Mean
\[ \bar{x} = \frac{\sum x_i}{n} \]
Where \(n\) is the number of boxes and \(x_i\) are the weights.
\[ \bar{x} = \frac{14.07 + 13.99 + 14.16 + 14.17 + 14.15 + 14.07 + 14.13 + 13.99 + 14.03 + 14.02 + 14.09 + 14.10 + 14.20 + 14.18 + 14.04 + 14.03}{16} \] \[ = \frac{226.33}{16} = 14.145625 \approx 14.146 \]
Step 2: Calculate the Sample Standard Deviation
\[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \]
Calculating each \( (x_i - \bar{x})^2 \):
- \( (14.07 - 14.146)^2 = 0.00576 \)
- \( (13.99 - 14.146)^2 = 0.024576 \)
- \( (14.16 - 14.146)^2 = 0.000196 \)
- \( (14.17 - 14.146)^2 = 0.000576 \)
- \( (14.15 - 14.146)^2 = 0.000016 \)
- \( (14.07 - 14.146)^2 = 0.00576 \)
- \( (14.13 - 14.146)^2 = 0.000256 \)
- \( (13.99 - 14.146)^2 = 0.024576 \)
- \( (14.03 - 14.146)^2 = 0.013056 \)
- \( (14.02 - 14.146)^2 = 0.015876 \)
- \( (14.09 - 14.146)^2 = 0.003136 \)
- \( (14.10 - 14.146)^2 = 0.002116 \)
- \( (14.20 - 14.146)^2 = 0.002976 \)
- \( (14.18 - 14.146)^2 = 0.001176 \)
- \( (14.04 - 14.146)^2 = 0.011556 \)
- \( (14.03 - 14.146)^2 = 0.013056 \)
Now summing these squared differences:
\[ \sum (x_i - \bar{x})^2 = 0.024576 + 0.00576 + 0.000196 + 0.000576 + 0.000016 + 0.00576 + 0.000256 + 0.024576 + 0.013056 + 0.015876 + 0.003136 + 0.002116 + 0.002976 + 0.001176 + 0.011556 + 0.013056 \approx 0.16996 \]
Calculating the sample standard deviation:
\[ s = \sqrt{\frac{0.16996}{15}} \approx \sqrt{0.011331} \approx 0.1065 \]
Step 3: Determine the Critical Value from the t-distribution
For a 98% confidence interval with \( n - 1 = 15 \) degrees of freedom, we check the t-distribution table or use a calculator.
The critical value \( t_{0.01, 15} \) (two-tailed) is approximately \( 2.602 \).
Step 4: Calculate the Confidence Interval
The confidence interval is calculated as follows:
\[ \text{CI} = \bar{x} \pm t \cdot \frac{s}{\sqrt{n}} \]
Calculating the standard error:
\[ \text{SE} = \frac{s}{\sqrt{n}} = \frac{0.1065}{\sqrt{16}} = \frac{0.1065}{4} = 0.026625 \]
Now, calculating the margin of error:
\[ ME = t \cdot \text{SE} = 2.602 \cdot 0.026625 \approx 0.0693 \]
Finally, the confidence interval:
\[ \text{CI} = 14.146 \pm 0.0693 \]
Thus:
\[ \text{Lower bound} = 14.146 - 0.0693 \approx 14.077 \] \[ \text{Upper bound} = 14.146 + 0.0693 \approx 14.215 \]
Final Result
The 98% confidence interval for the mean weight of the cereal boxes is approximately:
\[ (14.077, 14.215) \]
Rounded to three decimal places, the interval is:
\[ (14.077, 14.215) \]