Final answer:
To construct a 99% confidence interval for the population mean with a sample mean of 40.15, sample size of 54, and population standard deviation of 6.7, we find a z-value of 2.576 and calculate the margin of error. The resulting 99% confidence interval for the population mean is approximately (37.80, 42.50).
Step-by-step explanation:
The question asks us to construct a 99% confidence interval for the population mean (μ) given that the sample size (n) is 54, the sample mean (μ) is 40.15, and the population standard deviation (σ) is 6.7. To calculate the confidence interval for μ, we use the formula: μ = x ± (z* × (σ/√n)), where z* is the z-value that corresponds to the desired level of confidence. Since we are constructing a 99% confidence interval, we look up the z-value for a 99% confidence level, which is approximately 2.576.
We then calculate the margin of error (E) as follows: E = z* × (σ/√n) = 2.576 × (σ/√n) = 2.576 × (6.7/√54) ≈ 2.576 × 0.91287 ≈ 2.35. Now, we can find the confidence interval by adding and subtracting the margin of error from the sample mean: Lower limit = x - E = 40.15 - 2.35 ≈ 37.80, Upper limit = x + E = 40.15 + 2.35 ≈ 42.50. Therefore, the 99% confidence interval for μ is approximately (37.80, 42.50).