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To solve the problem in the image:

The problem requires finding the **95% confidence interval** for the mean (\(\mu\)) of the number of visitors to social networking sites, given the following data:

- Sample mean (\(\bar{x}\)) = 9.04 million
- Standard deviation (\(s\)) = 3.8 million
- Sample size (\(n\)) = 8
- Confidence level = 95%

### Steps:
1. **Calculate the standard error of the mean (SE):**
   \[
   SE = \frac{s}{\sqrt{n}}
   \]
   Substituting:
   \[
   SE = \frac{3.8}{\sqrt{8}} = \frac{3.8}{2.828} \approx 1.344
   \]

2. **Determine the critical value (\(t^*\)) for a 95% confidence level with \(n - 1 = 7\) degrees of freedom.**
   Using a \(t\)-distribution table or calculator:
   \[
   t^* \approx 2.365
   \]

3. **Compute the margin of error (ME):**
   \[
   ME = t^* \cdot SE
   \]
   Substituting:
   \[
   ME = 2.365 \cdot 1.344 \approx 3.177
   \]

4. **Calculate the confidence interval:**
   \[
   \text{Confidence Interval} = \bar{x} \pm ME
   \]
   Substituting:
   \[
   \text{Lower limit} = 9.04 - 3.177 \approx 5.86
   \]
   \[
   \text{Upper limit} = 9.04 + 3.177 \approx 12.22
   \]

### Final Answer:
The 95% confidence interval is approximately:
\[
5.86 \, \text{million} \leq \mu \leq 12.22 \, \text{million}
\]

Let me know if you'd like a detailed explanation or have further questions. Here are 5 related questions you might explore:
1. How does increasing the sample size affect the width of the confidence interval?
2. What is the difference between a \(t\)-distribution and a normal distribution?
3. How would the confidence interval change with a 99% confidence level?
4. What assumptions must hold true for this calculation to be valid?
5. How do you interpret the confidence interval in practical terms?

**Tip:** Always verify whether the population standard deviation is known to determine if you should use a \(z\)-distribution or \(t\)-distribution.

A recent survey of 8 social networking sites has a mean of 9.04 million visitors for a specific month. The standard deviation was 3.8 million. Find the 95% confidence interval of the true mean. Assume the variable is normally distributed. Round your answers to at least two decimal places.

Learn how to calculate the 95% confidence interval for the true mean number of visitors to social networking sites, given a mean of 9.04 million, standard deviation of 3.8 million, and a sample size of 8. Understand the use of t-distribution and margin of error in this statistical problem.

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