Answer:
The sample mean is of 1925 calories.
The margin of error is of 75 calories.
The sample standard deviation is of 109.7992 calories.
Step-by-step explanation:
Sample mean:
The sample mean is the mean value of the two bounds of the confidence interval. So

The sample mean is of 1925 calories.
The margin of error
Difference between the bounds and the sample mean. So
2000 - 1925 = 1925 - 1850 = 75 calories.
The margin of error is of 75 calories.
Sample standard deviation:
Here, I am going to expand on the t-distribution.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 18 - 1 = 17
99% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 17 degrees of freedom(y-axis) and a confidence level of
. So we have T = 2.898
The margin of error is:

In which s is the standard deviation of the sample and n is the size of the sample.
Since 




The sample standard deviation is of 109.7992 calories.