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fb4_16 - PROBLEM 4.16 KN OWN Data of Table 4.8 Column 1 N =...

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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) that-p = 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. Zia-16 L—m..__ ...
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