In a sample of 10 CEOs, they spent an average of 12.9 hours each week looking into new product opportunities with a sample standard deviation of 3.8 hours. Find the 95% confidence interval. Assume the times are normally distributed.

*Homework Help:*

5VE. Confidence intervals with sample standard deviationLinks to an external site. (1:37)

Group of answer choices

(9.1, 16.7)

(10.2, 15.6)

(11.1, 14.7)

(10.5, 15.3)

To find the 95% confidence interval for the mean when the sample standard deviation is known, we use the formula:

\[
\text{CI} = \bar{x} \pm t \times \frac{s}{\sqrt{n}}
\]

Where:
- \(\bar{x}\) = sample mean
- \(t\) = t-value from the t-distribution for the desired confidence level and degrees of freedom (\(df = n - 1\))
- \(s\) = sample standard deviation
- \(n\) = sample size

Given:
- \(\bar{x} = 12.9\) hours
- \(s = 3.8\) hours
- \(n = 10\)
- Confidence level = 95%

For a 95% confidence level with \(df = 9\) (since \(df = n - 1 = 10 - 1 = 9\)), we look up the t-value from the t-distribution table. The \(t\) value for 9 degrees of freedom at 95% confidence is approximately **2.262**.

Now we can plug the values into the formula:

\[
\text{CI} = 12.9 \pm 2.262 \times \frac{3.8}{\sqrt{10}}
\]

Let's calculate the margin of error:

\[
\text{Margin of Error} = 2.262 \times \frac{3.8}{\sqrt{10}} = 2.262 \times 1.202 = 2.72
\]

So the confidence interval is:

\[
\text{CI} = 12.9 \pm 2.72
\]

Calculating the lower and upper bounds:

\[
\text{Lower bound} = 12.9 - 2.72 = 10.18
\]
\[
\text{Upper bound} = 12.9 + 2.72 = 15.62
\]

The 95% confidence interval is approximately **(10.2, 15.6)**.

So, the correct answer is **(10.2, 15.6)**.

Would you like any further details or have questions?

---

Here are 8 related questions for further exploration:
1. How does the sample size affect the width of the confidence interval?
2. What is the difference between using a t-distribution and a z-distribution for confidence intervals?
3. How would the confidence interval change if the confidence level was increased to 99%?
4. Why is the t-distribution used instead of the z-distribution when the sample size is small?
5. What assumptions must be met to construct a valid confidence interval?
6. How would a larger sample standard deviation affect the confidence interval?
7. Can the confidence interval be used to make predictions about individual data points?
8. What are the implications of a wide confidence interval?

**Tip:** Always verify the degrees of freedom when using the t-distribution, as it directly affects the critical t-value.

Learn how to calculate a 95% confidence interval for the mean using sample standard deviation. Example problem: In a sample of 10 CEOs, they spent an average of 12.9 hours each week looking into new product opportunities with a sample standard deviation of 3.8 hours. Find the 95% confidence interval. Suitable for advanced high school students studying statistics.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation