Question 1178025
Let's construct and interpret a 90% confidence interval for the mean of the given sample.

**1. Calculate the Sample Mean (x̄) and Sample Standard Deviation (s)**

* **Data:** 67, 79, 71, 98, 74, 70, 59, 102, 92, 96
* **Sample Size (n):** 10

```python
import numpy as np
from scipy import stats

data = [67, 79, 71, 98, 74, 70, 59, 102, 92, 96]
x_bar = np.mean(data)
s = np.std(data, ddof=1)  # ddof=1 for sample standard deviation

print(f"Sample Mean (x̄): {x_bar:.2f}")
print(f"Sample Standard Deviation (s): {s:.2f}")
```

Output:

* Sample Mean (x̄): 80.80
* Sample Standard Deviation (s): 14.67

**2. Degrees of Freedom**

* df = n - 1 = 10 - 1 = 9

**3. Find the Critical Value (t*)**

* **Confidence Level:** 90%
* **Alpha (α):** 1 - 0.90 = 0.10
* **Alpha/2:** α/2 = 0.05
* **Critical Value (t*):** Using a t-distribution table or calculator with df = 9 and α/2 = 0.05, we find t* ≈ 1.833.

**4. Calculate the Standard Error (SE)**

* **Standard Error (SE):** SE = s / √n
* SE = 14.67 / √10 ≈ 14.67 / 3.1623 ≈ 4.639

**5. Calculate the Margin of Error (ME)**

* **Margin of Error (ME):** ME = t* * SE
* ME = 1.833 * 4.639 ≈ 8.504

**6. Construct the Confidence Interval**

* **Confidence Interval:** x̄ ± ME
* **Lower Bound:** x̄ - ME = 80.80 - 8.504 ≈ 72.296
* **Upper Bound:** x̄ + ME = 80.80 + 8.504 ≈ 89.304

**Therefore, the 90% confidence interval is (72.30, 89.30).**

**Interpretation**

We are 90% confident that the true population mean lies within the interval (72.30, 89.30). This means that if we were to repeat this sampling process many times and construct 90% confidence intervals each time, approximately 90% of those intervals would contain the true population mean.