1. STATISTICAL INFERENCE PART VI
HYPOTHESIS TESTING

2. Inference About the Difference of Two Population Proportions
Independent populations
Population Population 2
PARAMETERS: p1
PARAMETERS: p2
Sample size: n2
Sample size: n1
Statistics:
Statistics:

3. SAMPLING DISTRIBUTION OF
A point estimator of p1-p2 is
The sampling distribution of is
if nipi  5 and niqi  5, i=1,2.

4. STATISTICAL TESTS Two-tailed test Ho:p1=p2 HA:p1p2
Reject H0 if z < -z/2 or z > z/2.
One-tailed tests
HA:p1 > p2
Reject H0 if z > z
Ho:p1=p2
HA:p1 < p2
Reject H0 if z < -z

5. EXAMPLE
A manufacturer claims that compared with his closest competitor, fewer of his employees are union members. Of 318 of his employees, 117 are unionists. From a sample of 255 of the competitor’s labor force, 109 are union members. Perform a test at  = 0.05.

6. SOLUTION H0: p1 = p2 HA: p1 < p2
and , so pooled sample proportion is
Test Statistic:

7. Decision Rule: Reject H0 if z < -z0.05=-1.96.
Conclusion: Because z = > -z0.05=-1.96, not reject H0 at =0.05. Manufacturer is wrong. There is no significant difference between the proportions of union members in these two companies.

8. Example
In a study, doctors discovered that aspirin seems to help prevent heart attacks. Half of 22,000 male participants took aspirin and the other half took a placebo. After 3 years, 104 of the aspirin and 189 of the placebo group had heart attacks. Test whether this result is significant.
p1: proportion of all men who regularly take aspirin and suffer from heart attack.
p2: proportion of all men who do not take aspirin and suffer from heart attack

9. Test of Hypothesis for p1-p2
H0: p1-p2 = 0
HA: p1-p2 < 0
Test Statistic:
Conclusion: Reject H0 since p-value=P(z<-5.02)  0

10. Confidence Interval for p1-p2
A 100(1-C.I. for pp is given by:

11. Inference About Comparing Two Population Variances
Independent populations
Population Population 2
PARAMETERS:
PARAMETERS:
Sample size: n2
Sample size: n1
Statistics:
Statistics:

12. SAMPLING DISTRIBUTION OF
For independent r.s. from normal populations
(1-α)100% CI for

13. STATISTICAL TESTS Two-tailed test Ho: (or ) HA:
Reject H0 if F < F/2,n1-1,n2-1 or F > F 1- /2,n1-1,n2-1.
One-tailed tests
Ho:
Reject H0 if F > F 1- , n1-1,n2-1
Ho:
HA:
Reject H0 if F < F,n1-1,n2-1

14. Example
A scientist would like to know whether group discussion tends to affect the closeness of judgments by property appraisers. Out of 16 appraisers, 6 randomly selected appraisers made decisions after a group discussion, and rest of them made their decisions without a group discussion. As a measure of closeness of judgments, scientist used variance.
Hypothesis: Groups discussion will reduce the variability of appraisals.

15. Example, cont.
Appraisal values (in thousand $)
Statistics
With discussion
97, 101,102,95,98,103
n1=6 s1²=9.867
Without discussion
118, 109, 84, 85, 100, 121, 115, 93, 91, 112
n2=10 s2²=194.18
Ho: versus H1:
Reject Ho. Group discussion reduces the variability in appraisals.

16. HOW TO DERIVE AN APPROPRIATE TEST
TEST OF HYPOTHESIS
HOW TO DERIVE AN APPROPRIATE TEST
Definition: A test which minimizes the Type II error (β) for fixed Type I error (α) is called a most powerful test or best test of size α.

17. MOST POWERFUL TEST (MPT)
H0:=0 Simple Hypothesis
H1:=1  Simple Hypothesis
Reject H0 if (x1,x2,…,xn)C
The Neyman-Pearson Lemma:
Reject H0 if
Proof: Available in text books (e.g. Bain & Engelhardt, 1992, p.g.408)

18. EXAMPLES where  0 >  1. X~N(, 2) where 2 is known. H0:  =  0
Find the most powerful test of size .

19. Solution

20. Solution, cont. What is c?: It is a constant that satisfies
since X~N(, 2).
For a pre-specified α, most powerful test says,
Reject Ho if

21. Examples Example2: See Bain & Engelhardt, 1992, p.g.410 Find MPT of
Ho: p=p0 vs H1: p=p1 > p0
Example 3: See Bain & Engelhardt, 1992, p.g.411 Find MPT of
Ho: X~Unif(0,1) vs H1: X~Exp(1)

22. UNIFORMLY MOST POWERFUL (UMP) TEST
If a test is most powerful against every possible value in a composite alternative, then it will be a UMP test.
One way of finding UMPT is find MPT by Neyman-Pearson Lemma for a particular alternative value, and then show that test does not depend the specific alternative value.
Example: X~N(, 2), we reject Ho if
Note that this does not depend on particular value of μ1, but only on the
fact that  0 >  1. So this is a UMPT of H0:  = 0 vs H1:  <  0.

23. UNIFORMLY MOST POWERFUL (UMP) TEST
To find UMPT, we can also use Monotone Likelihood Ratio (MLR).
If L=L(0)/L(1) depends on (x1,x2,…,xn) only through the statistic y=u(x1,x2,…,xn) and L is an increasing function of y for every given 0>1, then we have a monotone likelihood ratio (MLR) in statistic y.
If L is a decreasing function of y for every given 0>1, then we have a monotone likelihood ratio (MLR) in statistic −y.

24. UNIFORMLY MOST POWERFUL (UMP) TEST
Theorem: If a joint pdf f(x1,x2,…,xn;) has MLR in the statistic Y, then a UMP test of size  for H0:0 vs H1:>0 is to reject H0 if Yc where P(Y c0)=.
for H0:0 vs H1:<0 is to reject H0 if Yc where P(Y  c0)=.

25. EXAMPLE
X~Exp()
H0:0
H1:>0
Find UMPT of size .

26. GENERALIZED LIKELIHOOD RATIO TEST (GLRT)
GLRT is the generalization of MPT and provides a desirable test in many applications but it is not necessarily a UMP test.

27. GENERALIZED LIKELIHOOD RATIO TEST (GLRT)
H0:0 H1: 1
and
Let
MLE of 
MLE of  under H0

28. GENERALIZED LIKELIHOOD RATIO TEST (GLRT)
GLRT: Reject H0 if 0

29. EXAMPLE
X~N(, 2)
H0:  = 0
H1:   0
Derive GLRT of size .

30. ASYMPTOTIC DISTRIBUTION OF −2ln
GLRT: Reject H0 if 0
GLRT: Reject H0 if -2ln>-2ln0=c
where k is the number of parameters to be tested.
Reject H0 if -2ln>

31. TWO SAMPLE TESTS Derive GLRT of size , where X and Y are
independent; p0, p1 and p2 are unknown.