To construct a 90% 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-score corresponding to the desired level of confidence 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 = 90%
-
Determine the degrees of freedom: \[ df = n - 1 = 20 - 1 = 19 \]
-
Look up the t-score (\(t^*\)) for a 90% confidence level with 19 degrees of freedom. You can find this value using a t-table or a calculator. For a 90% confidence interval, the t-score (two-tailed) is approximately:
\[ t^* \approx 1.729 \]
-
Calculate the standard error (SE): \[ SE = \frac{s}{\sqrt{n}} = \frac{2.45}{\sqrt{20}} \approx \frac{2.45}{4.472} \approx 0.549 \]
-
Calculate the margin of error: \[ \text{Margin of Error} = t^* \cdot SE = 1.729 \cdot 0.549 \approx 0.950 \]
-
Construct the confidence interval: \[ \text{Confidence Interval} = \bar{x} \pm \text{Margin of Error} \] \[ = 11.3 \pm 0.950 \]
- Lower limit: \(11.3 - 0.950 = 10.35\)
- Upper limit: \(11.3 + 0.950 = 12.25\)
Therefore, the 90% confidence interval for the mean sugar content of all single servings of Alpha-Bits is approximately:
\[ \boxed{(10.35, 12.25)} \]