Question 1174397: A sample of 60 grade 9 students' ages was obtained to estimate the mean age of all grade 9 students'.sample mean is equal to 15.3 years and the population variance is 16. What is the point estimate for mean and Find the 95% confidence interval for mean.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! **Point Estimate for the Mean:**
* The sample mean (x̄) is the best point estimate for the population mean (μ).
* Therefore, the point estimate for the mean age of all grade 9 students is **15.3 years**.
**95% Confidence Interval for the Mean:**
**1. Determine the Critical Value:**
* Since the population variance is known and the sample size is relatively large (n > 30), we can use the z-distribution.
* For a 95% confidence interval, the alpha level (α) is 1 - 0.95 = 0.05.
* Since we're constructing a two-tailed confidence interval, we divide alpha by 2: α/2 = 0.025.
* The z-score corresponding to a cumulative probability of 0.975 (1 - 0.025) in a standard normal distribution table or using a calculator is approximately 1.96.
**2. Calculate the Margin of Error:**
* The margin of error (E) is calculated as:
E = z * (σ / √n)
where:
* z is the critical value (1.96)
* σ is the population standard deviation (√population variance = √16 = 4)
* n is the sample size (60)
* E = 1.96 * (4 / √60) ≈ 1.01
**3. Construct the Confidence Interval:**
* The confidence interval is calculated as:
(x̄ - E, x̄ + E)
* (15.3 - 1.01, 15.3 + 1.01)
* (14.29, 16.31)
**Interpretation:**
We are 95% confident that the true mean age of all grade 9 students falls between 14.29 years and 16.31 years.
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