Establishing a Rejection Criterion There is a probability a (= 1 – P), for a sample of size N with sample variance Sx2 drawn from a population with true variance s2, that the difference between Sx2 and s2 is solely due to random effects. For example, there is only a 5 % probability that a value of c2 = 25.0 would result solely due to random effects for a sample of N = 16 (n = 15), as found from the c2 table.

In-Class Example A property sales group claims that they are 95 % confident that the standard deviation in the horizontal displacement of a floor in their new ‘safe’ apartment during an earthquake will be less than 6 inches. The data that they use to support their claim consists of a sample size of 21 and a measured standard deviation of 5 inches. Does their data support their claim? NOTE: Here, the claimed 95 % confidence means that the sales group is 95 % confident that the difference between the sample standard deviation of 5 in. and the ‘population’ standard deviation of 6 in. is solely due to random effects. That is, a = 0.95. What is a for this c2 value? Using the c2 table for n = 20 and interpolating gives a = 0.84 = 84 % → P = 1 – a = 16 %

In-Class Example (cont’d) So, the actually is an 84 % confidence that the standard deviation will be less than 6 inches. What would be the value of the standard deviation to be 95 % confident in the claim? in.

In-Class Example X c2 of data For 95 % confidence:

In-Class Example (Rejection Criterion) The sample standard deviation of the length of 12 widgets taken off of an assembly lines is 0.20 mm. What must be the widget population’s standard deviation to support the conclusion that the probability is 50 % for any difference between the sample’s and population’s standard deviations to be the result of random effects ? c2 ≡ nSx2/s2 = ca2 = c0.502 = 10.3 for n = 11 >> s2 = (11)(0.20)2/10.3 = 0.0427 >> s = 0.21 mm

Comparing a Sample and Population using chinormchk.m Procedure → a (for n = K-3) K (=1.15N1/3) bins cover range ≥ ~±2s Figure 8.13