SOLUTION: 2. Mr. Aboboy wants to estimate the average yearly number of eggs a hen lays in his poultry farm with 95% confidence. He allows a maximum error of 0.90. From past experience, he
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Question 1170232: 2. Mr. Aboboy wants to estimate the average yearly number of eggs a hen lays in his poultry farm with 95% confidence. He allows a maximum error of 0.90. From past experience, he knows that the standard deviation is equal to 20 eggs. How many hens should he use as his sample?
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Let's solve this problem step-by-step.
**Understanding the Problem**
Mr. Aboboy wants to determine the required sample size to estimate the average yearly number of eggs with a specific confidence level and margin of error.
**Given Information**
* Confidence level = 95%
* Maximum error (margin of error, E) = 0.90 eggs
* Population standard deviation (σ) = 20 eggs
**Key Concepts**
* **Confidence Interval:** The range within which we expect the true population mean to lie.
* **Margin of Error (E):** The maximum difference between the sample mean and the true population mean.
* **Z-score:** The number of standard deviations a value is from the mean in a standard normal distribution.
* **Sample Size Formula:** For estimating a population mean with a known standard deviation:
n = (Z * σ / E)²
**Calculations**
1. **Find the Z-score:**
* For a 95% confidence level, the alpha (α) is 1 - 0.95 = 0.05.
* Since it's a two-tailed test, we need the z-score for α/2 = 0.025.
* Using a standard normal distribution table or calculator, the z-score (Z) is approximately 1.96.
2. **Apply the Sample Size Formula:**
n = (Z * σ / E)²
n = (1.96 * 20 / 0.90)²
n = (39.2 / 0.90)²
n = (43.555555...)²
n ≈ 1896.9
3. **Round Up to the Nearest Whole Number:**
Since we cannot have a fraction of a hen, we must round up to the nearest whole number to ensure the desired confidence level and margin of error are met.
**Answer**
Mr. Aboboy should use a sample of 1897 hens.
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