Question 1170718: It is known that the weights of apples from a farm are normally distributed. In order to estimate the mean weight, a random sample of 150 apples is considered and the sample mean and population standard deviation are 6 kg and 0.8 kg respectively.
(a) Construct a 95% confidence interval estimate for the population mean weight of apples.
(b) The researcher suggests doing the study again so that 98% confidence interval estimate for the population mean weight of apples is (5.8835,6.1165) kg. How large should the sample size be?
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let's solve this problem step-by-step.
**Given:**
* Sample size (n) = 150
* Sample mean (x̄) = 6 kg
* Population standard deviation (σ) = 0.8 kg
**(a) Construct a 95% Confidence Interval**
1. **Find the critical z-value (z_c):**
* For a 95% confidence interval, z_c = 1.96 (from a z-table or calculator).
2. **Calculate the margin of error (E):**
* E = z_c * (σ / √n)
* E = 1.96 * (0.8 / √150)
* E ≈ 1.96 * (0.8 / 12.247)
* E ≈ 1.96 * 0.0653
* E ≈ 0.128 kg
3. **Construct the confidence interval:**
* Confidence Interval = x̄ ± E
* Confidence Interval = 6 ± 0.128
* Confidence Interval = (6 - 0.128, 6 + 0.128)
* Confidence Interval = (5.872, 6.128) kg
4. **Answer:**
* The 95% confidence interval for the population mean weight of apples is (5.872, 6.128) kg.
**(b) Find the sample size for a 98% Confidence Interval**
1. **Given Confidence Interval:**
* (5.8835, 6.1165) kg
2. **Calculate the margin of error (E):**
* E = (Upper Bound - Lower Bound) / 2
* E = (6.1165 - 5.8835) / 2
* E = 0.233 / 2
* E = 0.1165 kg
3. **Find the critical z-value (z_c):**
* For a 98% confidence interval, z_c ≈ 2.33 (from a z-table or calculator).
4. **Use the margin of error formula to solve for n:**
* E = z_c * (σ / √n)
* 0.1165 = 2.33 * (0.8 / √n)
* √n = (2.33 * 0.8) / 0.1165
* √n = 1.864 / 0.1165
* √n ≈ 16.00
* n = (16.00)²
* n = 256
5. **Answer:**
* The sample size should be 256 apples.
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