Final answer:
The 96% confidence interval for the average age of all students at this college is approximately (20.513 years to 21.487 years), calculated using the given sample mean, the Z-score corresponding to the 96% confidence level, the population standard deviation, and the sample size.
Step-by-step explanation:
To calculate the 96% confidence interval for the average age of all students at this college, we can use the formula for the confidence interval of the mean when the population standard deviation is known:
To create a confidence interval, the formula is:
Mean ± (Z * σ / √n), where:
- Mean is the sample mean, which is 21 years.
- Z is the Z-score that corresponds to the desired confidence level, which can be found in Z-tables or standard normal distribution tables. For a 96% confidence level, Z is approximately 2.05.
- σ (sigma) is the population standard deviation, which is the square root of the variance. Given the variance is 3.61, sigma is √3.61 = 1.9 years.
- n is the sample size, which is 64.
We then calculate the margin of error:
Margin of Error = 2.05 * (1.9 / √64) = 2.05 * (1.9 / 8) = 2.05 * 0.2375 ≈ 0.487
Therefore, we can calculate the confidence interval:
Lower bound = 21 - 0.487 = 20.513
Upper bound = 21 + 0.487 = 21.487
Hence, the 96% confidence interval for the average age of all students is approximately (20.513, 21.487).