Step 1 :We are given that the sample mean (\(\bar{x}\)) is 4.7, the sample standard deviation (s) is 2.9, the sample size (n) is 125, and the Z-score for a 99% confidence level is approximately 2.576.
Step 2 :We are asked to find the lower bound of the confidence interval, which is calculated by subtracting the margin of error from the sample mean. The margin of error is calculated using the formula \(Z \frac{s}{\sqrt{n}}\).
Step 3 :Substituting the given values into the formula, we get \(2.576 \times \frac{2.9}{\sqrt{125}}\), which gives us a margin of error of approximately 0.668.
Step 4 :Subtracting the margin of error from the sample mean, we get \(4.7 - 0.668\), which gives us a lower bound of approximately 4.03.
Step 5 :Final Answer: The lower bound for the 99% confidence interval is approximately \(\boxed{4.03}\).