To construct a 95% confidence interval for the population mean $\mu$ with the given sample data, we'll use the t-distribution because the sample size is relatively small ($n = 10$) and the population standard deviation is unknown.

Given data:

• Sample size ($n$) = 10
• Sample mean ($\overline{x}$) = 14.4
• Sample standard deviation ($s$) = 4.3
• Confidence level = 95%

Steps:

1. The degrees of freedom (df) for the t-distribution: $df = n - 1 = 10 - 1 = 9$.
2. For a 95% confidence level and $df = 9$, we look up the critical t-value ($t_{\alpha/2}$) for a two-tailed test. From t-distribution tables or statistical software, $t_{0.025,9} \approx 2.262$.
3. The margin of error (ME) is calculated as: $ME = t_{\alpha/2} \times \frac{s}{\sqrt{n}}$ Substituting the given values: $ME = 2.262 \times \frac{4.3}{\sqrt{10}} \approx 2.262 \times 1.36 \approx 3.08$
4. The confidence interval is given by: $\text{Confidence Interval} = \overline{x} \pm ME$ $\text{Confidence Interval} = 14.4 \pm 3.08$ Therefore: $\text{Confidence Interval} = (14.4 - 3.08, 14.4 + 3.08) = (11.32, 17.48)$

Conclusion:

The 95% confidence interval for the population mean $\mu$ is approximately $(11.32, 17.48)$.

Would you like further explanation or have any other questions?

Relative Questions:

1. What is the importance of using the t-distribution for small sample sizes?
2. How does changing the confidence level affect the width of the confidence interval?
3. What is the role of degrees of freedom in determining the critical t-value?
4. Can a confidence interval include negative values for population means? In what cases?
5. How does increasing the sample size impact the margin of error and the confidence interval?

Tip:

Always ensure the underlying data's distribution is approximately normal when using t-distribution-based confidence intervals for small sample sizes.