A botanist wishes to estimate the typical number of seeds for a certain fruit. She samples 38 specimens and counts the number of seeds in each. Use her sample results (mean $=36.9$, standard deviation $=6.3$ ) to find the $99 \%$ confidence interval for the number of seeds for the species. Enter your answer as an open-interval (i.e., parentheses) accurate to one decimal place (because the sample statistics are reported accurate to one decimal place). \[ 99 \% \text { C.I. }=(34.1,39.7) \]

Step 1 ：The botanist has collected a sample of 38 specimens and counted the number of seeds in each. The mean number of seeds is 36.9 and the standard deviation is 6.3. She wants to estimate the typical number of seeds for this fruit species with a 99% confidence interval.

Step 2 ：The formula for a confidence interval is \(\text{mean} \pm \left( Z \times \frac{\text{standard deviation}}{\sqrt{\text{sample size}}}\right)\), where Z is the Z-score for the desired confidence level.

Step 3 ：For a 99% confidence level, the Z-score is approximately 2.576.

Step 4 ：Substituting the given values into the formula, we get \(36.9 \pm \left( 2.576 \times \frac{6.3}{\sqrt{38}}\right)\).

Step 5 ：Calculating the margin of error, we get approximately 2.632658999246919.

Step 6 ：Subtracting and adding this margin of error from the mean, we get the confidence interval as \((36.9 - 2.632658999246919, 36.9 + 2.632658999246919)\), which simplifies to \((34.3, 39.5)\).

Step 7 ：Final Answer: The 99% confidence interval for the number of seeds for the species is \(\boxed{(34.3, 39.5)}\).