1. The Question The Answer P = 94 %

2. Practical Uses of   To infer  from S x To compare a sample to an assumed population To establish a rejection criterion   = f(N) >> f( ). Thus, there are an infinite number of    distributions (like for Student’s t).   is the fractional difference between measured and expected frequencies of occurrence.

3. The   Variable The random variable,  , is defined as Thus, as N→∞,

4. Figure 8.12 The   Distribution Figure 8.10 pdf PDF chi2pdf(  2, ) chi2cdf(  2, ) Determine Pr[  2 ≤10] for N = 11: Determine Pr[  2 ≤10] for N = 5: Determine    for P = 50 % and N = 5:

5. Figure 8.11    Probability Density Function

6. Table 8.8    Table For N = 13, find  when  2 = 21.0 For P = 5 %, find  2 if N = 20

7. Estimating the True Variance The true variance,  2, estimated with P % confidence, is in the range noting  = 1 – P and = N -1.

8. In-Class Example (x’ and  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.

9. Using the    Table = 15 and P = 0.95 >>  (= 1 – P) = 0.05

10. Establishing a Rejection Criterion There is a probability  (= 1 – P), for a sample of size N with sample variance S x 2 drawn from a population with true variance  2, that the difference between S x 2 and  2 is solely due to random effects. For example, there is only a 5 % probability that a value of  2 = 25.0 would result solely due to random effects for a sample of N = 16 (>> = 15), as found from the  2 table.

11. 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 ?

12. Comparing a Sample and Population Figure 8.13 K (=1.15N 1/3 ) bins cover range ≥ ~±2  Procedure using chinormchk.m