To construct a 95% confidence interval for the mean sugar content of Alpha-Bits, we can use the formula for the confidence interval for the mean when the population standard deviation is unknown:
\[ \text{Confidence Interval} = \bar{x} \pm t^* \left( \frac{s}{\sqrt{n}} \right) \]
Where:
- \(\bar{x}\) is the sample mean
- \(t^*\) is the t-value from the t-distribution for a given confidence level and degrees of freedom
- \(s\) is the sample standard deviation
- \(n\) is the sample size
Given:
- Sample mean (\(\bar{x}\)) = 11.3 grams
- Sample standard deviation (\(s\)) = 2.45 grams
- Sample size (\(n\)) = 20
- Confidence level = 95%
Step 1: Calculate the Degrees of Freedom
The degrees of freedom (df) for this sample is: \[ df = n - 1 = 20 - 1 = 19 \]
Step 2: Find the t-value
For a 95% confidence interval and 19 degrees of freedom, we can look up the t-value in a t-distribution table or use a calculator. The t-value for 95% confidence level and 19 degrees of freedom is approximately 2.093.
Step 3: Calculate the Standard Error (SE)
\[ SE = \frac{s}{\sqrt{n}} = \frac{2.45}{\sqrt{20}} \approx \frac{2.45}{4.472} \approx 0.548 \]
Step 4: Calculate the Margin of Error (ME)
\[ ME = t^* \cdot SE = 2.093 \cdot 0.548 \approx 1.147 \]
Step 5: Construct the Confidence Interval
\[ \text{Confidence Interval} = \bar{x} \pm ME = 11.3 \pm 1.147 \]
Calculating the lower and upper bounds:
- Lower bound: \[ 11.3 - 1.147 \approx 10.153 \]
- Upper bound: \[ 11.3 + 1.147 \approx 12.447 \]
Conclusion
The 95% confidence interval for the mean sugar content for all single servings of Alpha-Bits is approximately \((10.15, 12.45)\) grams.