```Question 298823
A.)  Ok, we know our confidence interval, it is a 95% interval.

Our variable of interest X : the amount of time a component moves from one station to another.
s = 5.1 sec  (Standard deviation)
We want to find n: the number of measurements to make such that our error doesn't exceed 1.5

I will assume this is a Normal distribution
Error is X - u.   The Z variable is  z = (x-u)/[s/sqrt(n)]
We want the Probability(X-u <= 1.5 ) >= 0.95

So:   P((x-u)/[s/sqrt(n)]  <= 1.5/[5.1*sqrt(n)] ) >= 0.95

But since the normal distribution is symmetric about mean, I will look for a z-value corresponding to 0.95/2 = 0.475, which is z= 1.96

from the z-score chart this corresponds to :  0 < z <= 1,96

Z = (x-u)/[s/sqrt(n)]  <= 1.5/[5.1*sqrt(n)]

therefore  sqrt(n)* 1.5/5.1 = 1.96

n = (5.1*1.96/1.5)^2
n = 44.408

so I would say 44 or more tests.
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The estimator for standard deviation  is  s/sqrt(n)  not s/sqrt(n -1)
that is why I have Z = (x-u)/[s/sqrt(n)],  if I used s/sqrt(n-1) as my estimator, then Z = (x-u)/[s/sqrt(n - 1)] would be my Z statistic  and then n would be 45 or more tests.

I assumed we used a normal distribution because the number of measurements was large.  If the number of samples is small, we would be forced to use the t-distribution.

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