Estimates Made Using Sx Two different statistical estimates can be made using Sx. [1] the value of the next sample value, xi, where [2] the value of the true mean, x‘, where

Example 8.5 [1] p range that contains the next measurement with P = 95 % [2] same but for N = 5 [3] for N = 19 and P = 50 % [4] p range that contains the true mean

c2 and the c2 distribution In the same way we used the student t distribution to estimate the range containing the population mean… We can use the statistical variable c2 is and the c2 distribution to estimate the range containing the population standard deviation The statistical variable c2 is defined as we could calculate c2 directly from a sample if we assumed the population mean and standard deviation (e.g., assumed they equaled the sample mean and standard deviation) then since Thus, as N→∞,

The c2 distribution: how c2 behaves for normally distributed data c2 = f(n)  infinite number of c2 distributions (like for Student’s t). pdf PDF Figure 8.10 96 % 56 % 3.36 Figure 8.12 Determine Pr[c2≤10] for N = 11: consider a sample: N = 5, c2 = 11 (calculated) meaning? Determine Pr[c2≤10] for N = 5: Determine c2 for P = 50 % and N = 5:

subscript α often used: c2a “level of significance” P a P+a=1 α is directly related to the % confidence P used, e.g., with the student t distribution Figure 8.11

c2 Table For N = 13, find a when c2 = 21.0 a = 5 % For P = 5 %, find c2 if N = 20 Table 8.8 see also Excel or LabVIEW help files

Uses of the c2 distribution To infer s from Sx To establish a rejection criterion (e.g., when to stop making something and fix the equipment; see ex 8.9) To compare a sample to an assumed population (see ex 8.10) Back to the top bullet, the true variance, s2, estimated with P % confidence, is in the range noting a = 1 – P and n = N -1.

In-Class Example (x’ and s Inference) Given the mean and standard deviation are 10 and 1.5, respectively, for a sample of 16, estimate with 95 % confidence the ranges within which are the true mean and true standard deviation (assuming a normally distributed population). range for true mean = sample mean ± tn,PSx/√N = 10 ± (2.131)(1.5)/√16 >> between 9.2 and 10.8 range for true variance: nSx2/ca/22 ≤ s2 ≤ nSx2/c1-a/22 >> (15)(1.5)2/27.5 ≤ s2 ≤ (15 )(1.5)2/6.26 for a = 0.05 >> 1.23 ≤ s2 ≤ 5.39 >> 1.11 ≤ s ≤ 2.32

Using the c2 Table = 15 and P = 0.95 >> a (= 1 – P) = 0.05 gives c2 = 6.26 a/2 = 0.025 gives c2 = 27.5 = 15 and P = 0.95 >> a (= 1 – P) = 0.05

In-Class Example (cont’d) What happens to the range which contains the true standard deviation when P is reduced from 95 % to 90 %? To be less confident, we would expect the extent of the range to decrease. Let’s see what happens. P = 90 % → a = 10 % → a/2 = 0.05 and 1-(a/2) = 0.95 For n = 15: c20.05 (lower bound) = 25.0 (vs 27.5) and c20.95 (upper bound) = 7.26 (vs 6.26). So, the extent of the range does decrease. For the P = 90 % case: 1.16 (vs 1.11) ≤ s ≤ 2.16 (vs 2.32)