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Chapter 4. Inference about Process Quality
Random
Sample
Statistics
Chi-square (2) Distribution
mean   n
variance  2  2n

Let x1 , x2 , x3 ,..., xn be randome sample from N  , 
2
.
y has a chi-square distribution with n -1 degree of freedom.
t Distribution
y has a chi-square distribution.
x has standard normal distribution.
mean   0
variance  2 
k
for k  2.
k 2
y
degree of freedom 
k
k  , t distribution  N  0,1
F Distribution
w, y : independent chi-square distributions with degrees of freedom u, v.
w/u
Fu ,v 
y/v
Fu ,v is distributed with u numerator degrees of freedom, and v denominator.
Estimator: estimates probability parameter from samples
Good Characteristics for Estimators
• Unbiased
• Minimum variance
•As n gets large the bias goes to zero
Let x1 , x2 , x3 ,..., xn be random sample from N   ,  2  .
Relative range: W 
R

E (W )  d 2
Hypothesis Testing
Alternative
Hypothesis
Null
Hypothesis
•In this example, H1 is a two-sided alternative hypothesis
  P  type I error   P  reject H 0 | H 0 is true  : producer's risk
  P  type II error   P  fail to reject H 0 | H 0 is false  : consumer's risk
power =1    P  reject H 0 | H 0 is false 
•H1 is a two-sided alternative hypothesis.
•The procedure for testing this hypothesis is to:
 take a random sample of n observations on the random variable x,
 compute the test statistic, and
 reject H0 if |Z0| > Z/2, where Z/2 is the upper /2 percentage point
of the standard normal distribution.
One-Sided Alternative Hypotheses
• In some situations we may wish to reject H0 only if the
true mean is larger than µ0
– Thus, the one-sided alternative hypothesis is H1: µ>µ0, and we
would reject H0: µ=µ0 only if Z0>Zα
• If rejection is desired only when µ<µ0
– Then the alternative hypothesis is H1: µ<µ0, and we reject H0
only if Z0<−Zα
Confidence Interval
→
If P ( L ≤ μ ≤ U ) = 1- α
L ≤ μ ≤ U is 100 (1- α) % confidence interval.
If the variance is known.
• For the two-sided alternative hypothesis, reject H0 if |t0| > t/2,n-1,
where t/2,n-1, is the upper /2 percentage of the t distribution with n  1
degrees of freedom
• For the one-sided alternative hypotheses,
• If H1: µ1 > µ0, reject H0 if t0 > tα,n − 1, and
• If H1: µ1 < µ0, reject H0 if t0 < −tα,n − 1
• One could also compute the P-value for a t-test
t0.025, 14 = 2.145. Thus, we should accept H0.
• Section 3-3.4 describes hypothesis testing and confidence
intervals on the variance of a normal distribution
Suppose, out of n samples chosen, x samples belongs to a subclass
with probability p.
Confidence Intervals on a Population Proportion
For large n and p, use normal approximation.
For large n and small p, use Poisson approximation.
For small n, use binomial distribution.
1 n1
x1   x1i
n1 i 1
1
x2 
n2
n2
x
i 1
2i
E  x1  x2   E  x1   E  x2   1  2
Var  x1  x2   Var  x1   Var  x2  
Z
 12
n1
N (0,1)

 22
n2
Z0  Z / 2 or Z0  Z  / 2
Two independent samples of size n1 and n2.
Of them, x1 and x2 belong to the class of interest.
Estimators: pˆ1 
Z
N (0,1)
x1
n1
pˆ 2 
x2
n2
More Two Populations
Analysis of Variance (ANOVA)
Linear staticstical model
yij     i   ij
for i  1, 2,3,..., a and j=1,2,3,...,n
 yij : (ij ) th observation (random variable)
  : overall mean
  i : parameter for i th treatment (i     i )
  ij : random error component  ij  N (0,  2 )
If H0 is true:
If H1 is true:
Error mean square: MS E 
SS E
an unbiased estimator of  2
a(n  1)
For hypothesis H0 testing, use
with a-1 and a(n-1) degrees of freedom.
Alternative formulas for computing efficiency
residual: eij  yij  yi