Question 22 The Archie Carr National Wildlife Refuge near Melbourne Beach was established to protect habitat at the most significant area for loggerhead sea turtle nesting in the world, and the most significant area for green turtle nesting in North America. Biologists working at the refuge need to estimate the clutch size (number of eggs per nest) for this year's loggerhead sea turtle nesting season. A recent survey of 10 nests yielded the following data: \begin{tabular}{|r|r|} \hline 110 & 109 \\ \hline 92 & 99 \\ \hline 102 & 98 \\ \hline 116 & 128 \\ \hline 98 & 130 \\ \hline \end{tabular} Determine the point estimate, $\bar{x}$, and the sample standard deviation, $s$, for these data. Round the solutions to four decimal places, if necessary. \[ \begin{array}{l} \bar{x}= \\ s= \end{array} \] Using a 99.98 confidence level, determine the margin of error, $E$, and a $99.9 \%$ confidence interval for the average clutch size of loggerhead sea turtles at the Archie Carr National Wildlife Refuge this season. Report the confidence interval using interval notation. Round solutions to two decimal places, if necessary. The margin of error is given by $E=$ A $99.9 \%$ confidence interval is given by

Step 1 ：Calculate the point estimate, which is the mean of the data. The mean is calculated by adding all the data points and dividing by the number of data points. The data points are: 110, 109, 92, 99, 102, 98, 116, 128, 98, 130. So, the mean (point estimate) is: \[\bar{x} = \frac{110 + 109 + 92 + 99 + 102 + 98 + 116 + 128 + 98 + 130}{10} = 108.2\]

Step 2 ：Calculate the sample standard deviation. This is done by subtracting each data point from the mean, squaring the result, adding all these squared results together, dividing by the number of data points minus 1, and then taking the square root. The sample standard deviation is: \[s = \sqrt{\frac{(110-108.2)^2 + (109-108.2)^2 + (92-108.2)^2 + (99-108.2)^2 + (102-108.2)^2 + (98-108.2)^2 + (116-108.2)^2 + (128-108.2)^2 + (98-108.2)^2 + (130-108.2)^2}{10-1}} = 13.2810\]

Step 3 ：Calculate the margin of error. The margin of error is given by the formula: \[E = z \cdot \frac{s}{\sqrt{n}}\] Where z is the z-score corresponding to the desired confidence level (for a 99.9% confidence level, z = 3.291), s is the sample standard deviation, and n is the number of data points. So, the margin of error is: \[E = 3.291 \cdot \frac{13.2810}{\sqrt{10}} = 14.53\]

Step 4 ：Calculate the confidence interval. The confidence interval is given by the formula: \[\bar{x} \pm E\] So, the 99.9% confidence interval is: \[108.2 \pm 14.53 = [93.67, 122.73]\]

Step 5 ：The point estimate is 108.2, the sample standard deviation is 13.2810, the margin of error is 14.53, and the 99.9% confidence interval is [93.67, 122.73]. The final answer is \[\boxed{[93.67, 122.73]}\]