To construct a 95% confidence interval for the population mean $\mu$, we can follow the steps outlined below:

Step 1: Identify the Given Information

• Sample mean $\bar{x} = 17.94$
• Sample standard deviation $s = 7.37$
• Sample size $n = 20$
• Confidence level = 95%, so the significance level $\alpha = 0.05$, and the critical value $t_{\alpha/2}$ is based on the t-distribution with $n - 1 = 19$ degrees of freedom.

Step 2: Find the Critical Value

For a 95% confidence level with 19 degrees of freedom, we need the critical value $t_{\alpha/2}$ from the t-distribution table.

From standard tables, the critical value $t_{\alpha/2}$ for $\alpha = 0.05$ and 19 degrees of freedom is approximately 2.093.

Step 3: Calculate the Margin of Error (ME)

The margin of error is given by the formula:

$\text{ME} = t_{\alpha/2} \times \frac{s}{\sqrt{n}}$

Substituting the values:

$\text{ME} = 2.093 \times \frac{7.37}{\sqrt{20}} = 2.093 \times 1.648 = 3.448$

Step 4: Construct the Confidence Interval

The confidence interval is calculated as:

$\text{CI} = \bar{x} \pm \text{ME}$

Substitute the values:

$\text{CI} = 17.94 \pm 3.45$

So the confidence interval is:

$(17.94 - 3.45, 17.94 + 3.45) = (14.49, 21.39)$

Conclusion

The 95% confidence interval for the population mean $\mu$ is (14.49, 21.39).

Would you like more details on the t-distribution or any other part of this calculation?

Here are five related questions to expand your understanding:

1. How would the confidence interval change if the sample size were larger?
2. What is the effect of increasing the confidence level (e.g., from 95% to 99%) on the interval?
3. Why do we use the t-distribution instead of the z-distribution for small sample sizes?
4. How is the critical value of the t-distribution calculated?
5. What assumptions must be met to use the t-distribution for this confidence interval?

Tip: Increasing the sample size decreases the margin of error, leading to a narrower confidence interval for the mean.