A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 546 babies were born, and 273 of them were girls. Use the sample data to construct a 99\% confidence interval estimate of the percentage of girls born. Based on the result, does the method appear to be effective?
$< p< \square$ (Round to three decimal places as needed.)

Step 1 ：First, we calculate the sample proportion (p̂) of girls born, which is the number of girls born divided by the total number of babies born. In this case, we have 273 girls out of 546 babies, so \(p̂ = \frac{273}{546} = 0.5\).

Step 2 ：Next, we calculate the standard error, which is the square root of \(p̂(1 - p̂) / n\), where n is the total number of babies born. So, \(SE = \sqrt{0.5(1 - 0.5) / 546} = 0.0214\).

Step 3 ：We then use the Z-score for a 99% confidence interval, which is approximately 2.576, to calculate the margin of error. The margin of error is the Z-score times the standard error, so \(ME = 2.576 * 0.0214 = 0.0551\).

Step 4 ：Finally, we calculate the confidence interval by subtracting and adding the margin of error from/to the sample proportion. So, the confidence interval is \((p̂ - ME, p̂ + ME) = (0.5 - 0.0551, 0.5 + 0.0551) = (0.445, 0.555)\).

Step 5 ：Since the confidence interval includes 0.5, it suggests that the method does not appear to be effective in increasing the probability of conceiving a girl, because the percentage of girls born is not significantly different from 50%.

Step 6 ：Final Answer: The 99% confidence interval estimate of the percentage of girls born is approximately \(\boxed{(0.445, 0.555)}\).