To calculate the 95% confidence interval for the population mean, we'll use the following formula for a normally distributed sample when the population standard deviation is unknown (which is typical when $n$ is small):

$\text{CI} = \bar{x} \pm t_{\frac{\alpha}{2}} \left( \frac{s}{\sqrt{n}} \right)$

Where:

• $\bar{x}$ is the sample mean,
• $t_{\frac{\alpha}{2}}$ is the critical value from the t-distribution for $\alpha = 0.05$ (95% confidence) and $n - 1$ degrees of freedom,
• $s$ is the sample standard deviation,
• $n$ is the sample size.

Step 1: Calculate the confidence interval with 18 in the sample.

1. Sample: 1, 2, 3, 4, 18
2. Sample Size: $n = 5$
3. Sample Mean: $\bar{x} = \frac{1 + 2 + 3 + 4 + 18}{5} = \frac{28}{5} = 5.6$
4. Sample Standard Deviation: $s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} = \sqrt{\frac{(1-5.6)^2 + (2-5.6)^2 + (3-5.6)^2 + (4-5.6)^2 + (18-5.6)^2}{4}} = 7.05 \, (\text{rounded})$
5. Degrees of freedom: $n - 1 = 4$
6. t-value for 95% confidence with 4 degrees of freedom: $t_{\frac{\alpha}{2}} = 2.776$

Now calculate the margin of error: $\text{Margin of Error} = t_{\frac{\alpha}{2}} \times \left( \frac{s}{\sqrt{n}} \right) = 2.776 \times \left( \frac{7.05}{\sqrt{5}} \right) = 2.776 \times 3.15 \approx 8.75$

Thus, the 95% confidence interval is: $\text{CI} = 5.6 \pm 8.75 = (-3.15, 14.35)$

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Step 2: Calculate the confidence interval with 18 replaced by 5.

1. Sample: 1, 2, 3, 4, 5
2. Sample Mean: $\bar{x} = \frac{1 + 2 + 3 + 4 + 5}{5} = \frac{15}{5} = 3$
3. Sample Standard Deviation: $s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} = \sqrt{\frac{(1-3)^2 + (2-3)^2 + (3-3)^2 + (4-3)^2 + (5-3)^2}{4}} = 1.58 \, (\text{rounded})$
4. Degrees of freedom: $n - 1 = 4$
5. t-value for 95% confidence with 4 degrees of freedom: $t_{\frac{\alpha}{2}} = 2.776$

Now calculate the margin of error: $\text{Margin of Error} = t_{\frac{\alpha}{2}} \times \left( \frac{s}{\sqrt{n}} \right) = 2.776 \times \left( \frac{1.58}{\sqrt{5}} \right) = 2.776 \times 0.71 \approx 1.97$

Thus, the 95% confidence interval is: $\text{CI} = 3 \pm 1.97 = (1.03, 4.97)$

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Step 3: Effect of the outlier (18) on the confidence interval.

Replacing 18 with 5 significantly reduces the standard deviation, which in turn makes the confidence interval narrower.

• With the outlier (18), the confidence interval was wide: (-3.15, 14.35).
• Without the outlier (replacing 18 with 5), the confidence interval was much narrower: (1.03, 4.97).

Effect of an outlier: Outliers increase the spread of the data, which increases the standard deviation and results in a wider confidence interval. This implies less precision in estimating the population mean. Removing outliers narrows the confidence interval, indicating a more precise estimate of the population mean.

Would you like further details on the calculations or have any questions?

Follow-up questions:

1. What happens to the confidence interval if we increase the sample size?
2. How does a higher confidence level (e.g., 99%) affect the confidence interval?
3. What is the impact of using the population standard deviation instead of the sample standard deviation?
4. How would a smaller outlier affect the confidence interval compared to a larger outlier?
5. What are the limitations of using a confidence interval with a small sample size?

Tip: Always check for outliers in your data, as they can distort statistical analyses and conclusions.