SOLUTION: Given the following observation in a simple random sample from a population that is approximately normally distributed. Construct and interpret a 90% confidence interval for the me

Question 1178025: Given the following observation in a simple random sample from a population that is approximately normally distributed. Construct and interpret a 90% confidence interval for the mean 67,79,71,98,74,70,59,102, 92,96
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
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.

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