# STATISTICAL INFERENCE PART VI HYPOTHESIS TESTING 1.

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STATISTICAL INFERENCE PART VI HYPOTHESIS TESTING 1

Inference About the Difference of Two Population Proportions Population 1 Population 2 PARAMETERS: p 1 PARAMETERS: p 2 Statistics: Sample size: n 1 Sample size: n 2 Independent populations

SAMPLING DISTRIBUTION OF A point estimator of p 1 -p 2 is The sampling distribution of is if n i p i  5 and n i q i  5, i=1,2.

STATISTICAL TESTS Two-tailed test H o :p 1 =p 2 H A :p 1  p 2 Reject H 0 if z z  /2. One-tailed tests H o :p 1 =p 2 H A :p 1 > p 2 Reject H 0 if z > z  H o :p 1 =p 2 H A :p 1 < p 2 Reject H 0 if z < -z 

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.

SOLUTION H 0 : p 1 = p 2 H A : p 1 < p 2 and, so pooled sample proportion is Test Statistic:

Decision Rule: Reject H 0 if z < -z 0.05 =-1.96. Conclusion: Because z = -1.4518 > -z 0.05 =- 1.96, not reject H 0 at  =0.05. Manufacturer is wrong. There is no significant difference between the proportions of union members in these two companies.

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. p 1 : proportion of all men who regularly take aspirin and suffer from heart attack. p 2 : proportion of all men who do not take aspirin and suffer from heart attack

Test of Hypothesis for p 1 -p 2 H 0 : p 1 -p 2 = 0 H A : p 1 -p 2 < 0 Test Statistic: Conclusion: Reject H 0 since p-value=P(z<-5.02)  0

Confidence Interval for p 1 -p 2 A 100(1-  C.I. for p   p  is given by:

Inference About Comparing Two Population Variances Population 1 Population 2 PARAMETERS: Statistics: Sample size: n 1 Sample size: n 2 Independent populations

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

STATISTICAL TESTS Two-tailed test H o :(or ) H A : Reject H 0 if F F 1-  /2,n 1 -1,n 2 -1. One-tailed tests H o : H A : Reject H 0 if F > F 1- , n 1 -1,n 2 -1 H o : H A : Reject H 0 if F < F ,n 1 -1,n 2 -1

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. 14

Example, cont. Appraisal values (in thousand \$)Statistics With discussion97, 101,102,95,98,103n 1 =6 s 1 ²=9.867 Without discussion118, 109, 84, 85, 100, 121, 115, 93, 91, 112 n 2 =10 s 2 ²=194.18 15 Ho: versus H 1 : Reject Ho. Group discussion reduces the variability in appraisals.

TEST OF HYPOTHESIS HOW TO DERIVE AN APPROPRIATE TEST 16 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.

MOST POWERFUL TEST (MPT) H 0 :  =  0  Simple Hypothesis H 1 :  =  1  Simple Hypothesis Reject H 0 if (x 1,x 2,…,x n )  C The Neyman-Pearson Lemma: Reject H 0 if 17 Proof: Available in text books (e.g. Bain & Engelhardt, 1992, p.g.408)

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

Solution 19

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 20

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

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 μ 1, but only on the fact that  0 >  1. So this is a UMPT of H 0 :  =  0 vs H 1 :  <  0. 22

UNIFORMLY MOST POWERFUL (UMP) TEST To find UMPT, we can also use Monotone Likelihood Ratio (MLR). If L=L(  0 )/L(  1 ) depends on (x 1,x 2,…,x n ) only through the statistic y=u(x 1,x 2,…,x n ) 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. 23

UNIFORMLY MOST POWERFUL (UMP) TEST Theorem: If a joint pdf f(x 1,x 2,…,x n ;  ) has MLR in the statistic Y, then a UMP test of size  for H 0 :  0 vs H 1 :  >  0 is to reject H 0 if Y  c where P(Y  c  0 )= . for H 0 :  0 vs H 1 :  <  0 is to reject H 0 if Y  c where P(Y  c  0 )= . 24

EXAMPLE X~Exp(  ) H 0 :  0 H 1 :  >  0 Find UMPT of size . 25

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. 26

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

28 GENERALIZED LIKELIHOOD RATIO TEST (GLRT) GLRT: Reject H 0 if  0

EXAMPLE X~N( ,  2 ) H 0 :  =  0 H 1 :    0 Derive GLRT of size . 29

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

TWO SAMPLE TESTS 31 Derive GLRT of size , where X and Y are independent; p 0, p 1 and p 2 are unknown.

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