Question 1082720
mean with is $210,558 with s=$102,878
99%CI is ($118,320, $302,800)
without the outlier mean is $227,891 with s=$87,618
99%CI is ($144,170, $311,620)
This is done using t.995, df=11 * s/sqrt(12) for the first and 1 less in n and df for the second. That is the interval, and it is added to and subtracted from the sample mean.
The interval is narrower, because the variability is less.  The mean is larger 
 because a small number outlier was removed.