Question 1178447: A particular type of vacuum-packed coffee packet contains an average of 16 ounces. It has been observed that the number of ounces of coffee in these packets is normally distributed with σ = 1.41 ounce. A random sample of 15 of these coffee packets is selected, and the observations are used to calculate s. Find the numbers a and b such that P (a ≤ S^2 ≤ b) = 0.90.
thank you
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's break down how to find the values 'a' and 'b' such that P(a ≤ S² ≤ b) = 0.90, where S² is the sample variance.
**Understanding the Problem**
* **Population Distribution:** The ounces of coffee are normally distributed with a standard deviation (σ) of 1.41 ounces.
* **Sample Size:** A random sample of 15 packets is taken (n = 15).
* **Sample Variance (S²):** We're interested in the distribution of the sample variance.
* **Chi-Square Distribution:** The quantity (n - 1)S² / σ² follows a chi-square distribution with (n - 1) degrees of freedom.
* **Goal:** Find 'a' and 'b' such that the probability that S² falls between them is 0.90.
**Steps to Solve**
1. **Degrees of Freedom:**
* The degrees of freedom (df) are n - 1 = 15 - 1 = 14.
2. **Chi-Square Distribution:**
* We know that (14 * S²) / (1.41²) ~ χ²(14).
3. **Finding Chi-Square Percentiles:**
* We want to find values χ²_lower and χ²_upper such that P(χ²_lower ≤ χ²(14) ≤ χ²_upper) = 0.90.
* Since we want the middle 90%, we need to exclude the lower 5% and the upper 5%.
* Therefore, we need to find the 5th and 95th percentiles of the chi-square distribution with 14 degrees of freedom.
4. **Calculating 'a' and 'b':**
* Let χ²_0.05 be the 5th percentile and χ²_0.95 be the 95th percentile.
* We have:
* χ²_0.05 = (14 * b) / (1.41²)
* χ²_0.95 = (14 * a) / (1.41²)
* Therefore:
* a = (14 * 1.41²) / χ²_0.95
* b = (14 * 1.41²) / χ²_0.05
5. **Using scipy.stats:**
* Using python and the scipy.stats library, we can compute the answer.
```python
import scipy.stats as stats
# Sample size
n = 15
# Population standard deviation
sigma = 1.41
# Degrees of freedom
df = n - 1
# Find the chi-square values for the 5th and 95th percentiles
a = (df * sigma**2) / stats.chi2.ppf(0.95, df)
b = (df * sigma**2) / stats.chi2.ppf(0.05, df)
print(f"a: {a:.4f}")
print(f"b: {b:.4f}")
```
6. **Results**
* a ≈ 1.1752
* b ≈ 4.2360
**Conclusion**
The values of 'a' and 'b' are approximately 1.1752 and 4.2360, respectively. This means that there is a 90% probability that the sample variance (S²) will fall between these two values.
|
|
|
