100 observations needed for desired accuracy and confidence.
The formula for the confidence of a sampling is:
ME = z*d/sqrt(n)
where
ME = Margin of error
z = z score for desired level of confidence
d = standard deviation
n = number of samples
The z score desired is calculated as follows. If you want a 95% confidence, you calculate 1 - 0.95 = 0.05, then you divide the result by 2, getting 0.025, and finally you use a standard normal table to get the z score for the desired probability. So for this problem of 95.44% we get
(1 - 0.9544)/2 = 0.0456/2 = 0.0228
Looking up a standard normal table, the value of 0.0228 is found to have a z-score of 2.0, a.k.a. 2 standard deviations from the normal.
So let's substitute the known values into the formula and solve for n.
ME = z*d/sqrt(n)
1 = 2*5/sqrt(n)
sqrt(n) = 2*5
sqrt(n) = 10
n = 100
So the owner needs at least 100 samples to be 95.44% certain that his measurement error is within 1 second of the correct time.
