15 The precision of a method is being established and the fo
Solution
Getting the mean and standard deviation of the data,
mean = 22.184
standard deviation = 0.062289646
Note that
Lower Bound = X - t(alpha/2) * s / sqrt(n)
Upper Bound = X + t(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.025
X = sample mean = 22.184
t(alpha/2) = critical t for the confidence interval = 2.776445105
s = sample standard deviation = 0.062289646
n = sample size = 5
df = n - 1 = 4
Thus,
Lower bound = 22.10665719
Upper bound = 22.26134281
Thus, the confidence interval is
( 22.10665719 , 22.26134281 )
As 22.09 is not in this interval, then NO: IT IS NOT AN ACCEPTABLE MEASUREMENT AT 95% CONFIDENCE LEVEL.
