Assuming that the population is normally​ distributed, construct a 95 % confidence interval for the population​ mean, based on the following sample size of n equals 5.
​1, 2,​ 3, 4​, and 18  
In the given​ data, replace the value 18 with 5 and recalculate the confidence interval. Using these​ results, describe the effect of an outlier​ (that is, an extreme​ value) on the confidence​ interval, in general.
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Part 1
Find a 95 % confidence interval for the population​ mean, using the formula or technology.
  
enter your response hereless than or equalsmuless than or equals
  
enter your response here ​(Round to two decimal places as​ needed.)


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:
\[
\text{CI} = 5.6 \pm 8.75 = (-3.15, 14.35)
\]

---

### **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:
\[
\text{CI} = 3 \pm 1.97 = (1.03, 4.97)
\]

---

### **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.

Assuming that the population is normally distributed, construct a 95% confidence interval for the population mean, based on the following sample size of n = 5: 1, 2, 3, 4, and 18. In the given data, replace the value 18 with 5 and recalculate the confidence interval. Using these results, describe the effect of an outlier (that is, an extreme value) on the confidence interval, in general.

Learn how to construct a 95% confidence interval for the population mean of a normally distributed sample and understand the effect of outliers. This problem involves calculating the confidence interval with and without an outlier in the sample. The key statistical concepts include confidence intervals, sample mean, standard deviation, and the impact of extreme values.

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