Construct a confidence interval around the sample mean X=25 for the following conditions:
Sx = 4.5, a =0.05, df = 20


To construct the confidence interval for the sample mean, we'll use the **t-distribution** since the sample size is relatively small, and the population standard deviation is unknown. Here's how to construct the confidence interval:

### Given Information
- Sample mean (\(\bar{X}\)): 25
- Sample standard deviation (\(S_x\)): 4.5
- Significance level (\(\alpha\)): 0.05
- Degrees of freedom (\(df\)): 20

### Step 1: Find the Critical Value
For a confidence interval with \(\alpha = 0.05\) and \(df = 20\), we need to find the critical value (\(t_{\alpha/2}\)) from the t-distribution table. Since the confidence level is \(1 - \alpha = 0.95\), the critical value for a two-tailed test at the 95% confidence level and 20 degrees of freedom is:

\[
t_{\alpha/2, df=20} \approx 2.086
\]

### Step 2: Calculate the Margin of Error
The margin of error (ME) is calculated using the formula:

\[
\text{ME} = t_{\alpha/2} \cdot \frac{S_x}{\sqrt{n}}
\]

However, we don’t have the sample size (\(n\)) explicitly mentioned, so we assume that the degrees of freedom are \(n - 1 = 20\), which implies \(n = 21\). Substituting the values:

\[
\text{ME} = 2.086 \cdot \frac{4.5}{\sqrt{21}} \approx 2.046
\]

### Step 3: Construct the Confidence Interval
The confidence interval for the mean is:

\[
\bar{X} \pm \text{ME}
\]

\[
25 \pm 2.046
\]

\[
(25 - 2.046, 25 + 2.046) \approx (22.954, 27.046)
\]

### Final Answer
The 95% confidence interval for the sample mean is approximately:

\[
(22.95, 27.05)
\]

Would you like further details or have any questions about the process?

### Related Questions
1. How would the confidence interval change if the sample size increased?
2. What happens to the confidence interval if we use a 99% confidence level instead of 95%?
3. Why do we use the t-distribution instead of the normal distribution here?
4. How can you interpret the meaning of this confidence interval in a practical context?
5. How would the interval change if the sample standard deviation (\(S_x\)) was higher?

**Tip:** Always check the degrees of freedom and the corresponding critical value from the t-distribution table for accurate results.

Construct a confidence interval around the sample mean X=25 for the following conditions: Sx = 4.5, a =0.05, df = 20

Learn how to construct a 95% confidence interval for a sample mean of 25 with a standard deviation of 4.5, using a significance level of 0.05 and 20 degrees of freedom. This problem involves statistical concepts such as the t-distribution and margin of error.

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