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
![M = \frac{1850 + 2000}{2} = 1925](https://tex.z-dn.net/?f=M%20%3D%20%5Cfrac%7B1850%20%2B%202000%7D%7B2%7D%20%3D%201925)
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:
![M = T\frac{s}{\sqrt{n}}](https://tex.z-dn.net/?f=M%20%3D%20T%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D)
In which s is the standard deviation of the sample and n is the size of the sample.
Since ![M = 75, T = 2.898, n = 18](https://tex.z-dn.net/?f=M%20%3D%2075%2C%20T%20%3D%202.898%2C%20n%20%3D%2018)
![M = T\frac{s}{\sqrt{n}}](https://tex.z-dn.net/?f=M%20%3D%20T%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D)
![75 = 2.898\frac{s}{\sqrt{18}}](https://tex.z-dn.net/?f=75%20%3D%202.898%5Cfrac%7Bs%7D%7B%5Csqrt%7B18%7D%7D)
![s = \frac{75\sqrt{18}}{2.898}](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B75%5Csqrt%7B18%7D%7D%7B2.898%7D)
![s = 109.7992](https://tex.z-dn.net/?f=s%20%3D%20109.7992)
The sample standard deviation is of 109.7992 calories.