Unformatted text preview: PROBLEM 4.16 KN OWN: Data of Table 4.8, Column 1
N = 20 FIND: 5i tS; (95%) SOLUTION For the 20 data points in Column 1, we ﬁnd using equations (4.14a and b) thatp = 4.77
mm and SD = 0.273 m. From equation 4.16, the standard deviation of the means is found to be
Sp “2
S; = N132 = {1273/20 = 0.061 m with v z N  l = 19. Then from Table 4.4, t19,95 = 2.093. Then, we can expect that the true
mean value would lie within the precision interval deﬁned by (4.17) as p'=13itS; =4.77i(2.093)(0.061)24.77:0.13 (95%) This gives the range of mean values we would expect to ﬁnd from a given data set. A 95%
probability indicates that 19 out of every 20 complete data sets would show a sample mean
value within the range deﬁned for p‘. Compare the meaning of this statement to that found in Problem 4.11. They are very different! COMMENT The reasoning behind this precision interval for the mean value lies within the limitations of
ﬁnite statistics. While the sample mean value deﬁnes the mean value of the 20 data points
exactly, it is not necessarily the true mean value of the measured variable (compare the
results for these three real data sets in columns 1,2, and 3 — each has a different sample
mean). Different data sets of the same variable will give somewhat different mean values. As
N becomes large, the sample mean will approach the true mean and the precision interval will go to zero. Remember this assumes that there is no bias error acting on the measurement. Zia16 L—m..__ ...
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 Fall '08
 Singh,V

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