SOLUTION: (1 point) The weights of a random sample of cereal boxes that are supposed to weigh 1 pound are given below. Estimate the standard deviation of the entire population with 90% confi

Question 1188788: (1 point) The weights of a random sample of cereal boxes that are supposed to weigh 1 pound are given below. Estimate the standard deviation of the entire population with 90% confidence.
0.96, 0.98, 1.03, 1.04, 0.99, 1.02, 1.02, 1.03
LCL =
UCL =

Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Here's how to estimate the population standard deviation with 90% confidence:
**1. Calculate the sample standard deviation (s):**
First, we need to calculate the standard deviation of the given sample data. Here's how:
* **Calculate the mean (x̄):** (0.96 + 0.98 + 1.03 + 1.04 + 0.99 + 1.02 + 1.02 + 1.03) / 8 = 1.01
* **Calculate the variance:** Find the squared difference between each value and the mean, sum them, and divide by (n-1) where n is the sample size.
* **Calculate the standard deviation (s):** Take the square root of the variance.
Doing these calculations, we find s ≈ 0.027
**2. Find the degrees of freedom (df):**
df = n - 1 = 8 - 1 = 7
**3. Find the chi-square values:**
For a 90% confidence level, alpha (α) = 1 - 0.90 = 0.10. We need to find the chi-square values for α/2 and 1-α/2.
* α/2 = 0.10 / 2 = 0.05
* 1 - α/2 = 1 - 0.05 = 0.95
Using a chi-square table or calculator, look up the values for df = 7:
* χ²(0.05, 7) ≈ 14.067
* χ²(0.95, 7) ≈ 2.167
**4. Calculate the lower confidence limit (LCL):**
LCL = sqrt[ (n-1) * s² / χ²(α/2, df) ]
LCL = sqrt[ (7 * 0.027²) / 14.067 ]
LCL ≈ sqrt(0.000357)
LCL ≈ 0.0189
**5. Calculate the upper confidence limit (UCL):**
UCL = sqrt[ (n-1) * s² / χ²(1-α/2, df) ]
UCL = sqrt[ (7 * 0.027²) / 2.167 ]
UCL ≈ sqrt(0.00233)
UCL ≈ 0.0483
**Answer:**
LCL ≈ 0.0189
UCL ≈ 0.0483
Therefore, we are 90% confident that the population standard deviation of the weights of the cereal boxes is between approximately 0.0189 and 0.0483 pounds.

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