Question 1205251
sample size is 6.
sample mean is 2.0016 millimeters.
sample standard deviation is .0012 millimeters.


since the sample mean and sample standard deviation are used, then the 99% confidence interval will be based on the t-score, rather than the z-score.


99% two tail confidence interval for t-score with 5 degrees of freedom (degrees of freedom equal sample size minus 1) is equal to t = plus or minus 4.032142983.


since you're looking at a distribution of sample means for samples of 6 elements each, the standard error is used.
standard error = standard deviation / sqrt(sample size) = .0012 / sqrt(6) = .0004898979486.


t-score formula is t = (x-m)/s
t is the critical t-score
x is the critical sample mean.
m is the sample mean.
s is the standard error.


on the low side of the confidence interval, the t-score formula becomes:
-4.032142983 = (x - 2.0016) / .0004898979486.
solve for x to get x = 1.999624661.


on the high side of the confidence interval, the t-score formula becomes:
4.032142983 = (x - 2.0016) / .0004898979486.
solve for x to get x = 2.003575339.


your two tail 99% confidence interval is from 1.999624661 to 2.003575339.