Question
Regular consumption of pre-sweetened cereals contributes to tooth decay, heart disease as well as other diseases according to research by certain institutes. In a random sample of 20 similar servings of Alpha-Bits the average sugar content was 11.3 grams with standard deviation of 2.45 grams. Assuming the sugar contents are normally distributed, construct a 95% confidence interval for the mean sugar content for all single servings of Alpha-Bits.
Answers
GPT-4o mini
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.
\[
\text{Confidence Interval} = \bar{x} \pm t^* \left( \frac{s}{\sqrt{n}} \right)
\]
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.