STATISTICAL INFERENCES CHAPTERS 12 AND 13 HYPOTHESIS TESTING FOR PROPORTIONS AND MEANS

TESTING HYPOTHESES ABOUT PROPORTIONS PROBLEM SUPPOSE WE TOSSED A COIN 100 TIMES AND WE OBTAINED 38 HEADS AND 62 TAILS. IS THE COIN BIASED? THERE IS NO WAY TO SAY YES OR NO WITH 100% CERTAINTY. BUT WE MAY EVALUATE THE STRENGTH OF SUPPORT TO THE HYPOTHESIS THAT “THE COIN IS BIASED.”

TESTING HYPOTHESES NULL HYPOTHESIS ESTABLISHED FACT; A STATEMENT THAT WE EXPECT DATA TO CONTRADICT; NO CHANGE OF PARAMETERS. ALTERNATIVE HYPOTHESIS NEW CONJECTURE; YOUR CLAIM; A STATEMENT THAT NEEDS A STRONG SUPPORT FROM DATA TO CLAIM IT; CHANGE OF PARAMETERS

IN OUR PROBLEM

EXAMPLE WRITE THE NULL AND ALTERNATIVE HYPOTHESES YOU WOULD USE TO TEST EACH OF THE FOLLOWING SITUATIONS. (A) IN THE 1950s ONLY ABOUT 40% OF HIGH SCHOOL GRADUATES WENT ON TO COLLEGE. HAS THE PERCENTAGE CHANGED? (B) 20% OF CARS OF A CERTAIN MODEL HAVE NEEDED COSTLY TRANSMISSION WORK AFTER BEING DRIVEN BETWEEN 50,000 AND 100,000 MILES. THE MANUFACTURER HOPES THAT REDESIGN OF A TRANSMISSION COMPONENT HAS SOLVED THIS PROBLEM. (C) WE FIELD TEST A NEW FLAVOR SOFT DRINK, PLANNING TO MARKET IT ONLY IF WE ARE SURE THAT OVER 60% OF THE PEOPLE LIKE THE FLAVOR.

ATTITUDE ASSUME THAT THE NULL HYPOTHESIS IS TRUE AND UPHOLD IT, UNLESS DATA STRONGLY SPEAKS AGAINST IT.

TEST MECHANIC FROM DATA, COMPUTE THE VALUE OF A PROPER TEST STATISTICS, THAT IS, THE Z-STATISTICS. IF IT IS FAR FROM WHAT IS EXPECTED UNDER THE NULL HYPOTHESIS ASSUMPTION, THEN WE REJECT THE NULL HYPOTHESIS.

COMPUTATION OF THE Z – STATISTICS OR PROPER TEST STATISTICS

CONSIDERING THE EXAMPLE AT THE BEGINNING:

THE P – VALUE AND ITS COMPUTATION THE PROBABILITY THAT IF THE NULL HYPOTHESIS IS CORRECT, THE TEST STATISTIC TAKES THE OBSERVED OR MORE EXTREME VALUE. P – VALUE MEASURES THE STRENGTH OF EVIDENCE AGAINST THE NULL HYPOTHESIS. THE SMALLER THE P – VALUE, THE STRONGER THE EVIDENCE AGAINST THE NULL HYPOTHESIS.

THE WAY THE ALTERNATIVE HYPOTHESIS IS WRITTEN IS HELPFUL IN COMPUTING THE P - VALUE NORMAL CURVE

IN OUR EXAMPLE, P – VALUE = P( z < - 2.4) = 0.0082 INTERPRETATION: IF THE COIN IS FAIR, THEN THE PROBABILITY OF OBSERVING 38 OR FEWER HEADS IN 100 TOSSES IS 0.0082

CONCLUSION: GIVEN SIGNIFICANCE LEVEL = 0.05 WE REJECT THE NULL HYPOTHESIS IF THE P – VALUE IS LESS THAN THE SIGNIFICANCE LEVEL OR ALPHA LEVEL. WE FAIL TO REJECT THE NULL HYPOTHESIS (I.E. WE RETAIN THE NULL HYPOTHESIS) IF THE P – VALUE IS GREATER THAN THE SIGNIFICANCE LEVEL OR ALPHA LEVEL.

ASSUMPTIONS AND CONDITIONS RANDOMIZATION INDEPENDENT OBSERVATIONS 10% CONDITION SUCCESS/FAILURE CONDITION

EXAMPLE 1 THE NATIONAL CENTER FOR EDUCATION STATISTICS MONITORS MANY ASPECTS OF ELEMENTARY AND SECONDARY EDUCATION NATIONWIDE. THEIR 1996 NUMBERS ARE OFTEN USED AS A BASELINE TO ASSESS CHANGES. IN 1996, 31% OF STUDENTS REPORTED THAT THEIR MOTHERS HAD GRADUATED FROM COLLEGE. IN 2000, RESPONSES FROM 8368 STUDENTS FOUND THAT THIS FIGURE HAD GROWN TO 32%. IS THIS EVIDENCE OF A CHANGE IN EDUCATION LEVEL AMONG MOTHERS?

EXAMPLE 1 CONT’D (A) WRITE APPROPRIATE HYPOTHESES. (B) CHECK THE ASSUMPTIONS AND CONDITIONS. (C) PERFORM THE TEST AND FIND THE P – VALUE. (D) STATE YOUR CONCLUSION. (E) DO YOU THINK THIS DIFFERENCE IS MEANINGFUL? EXPLAIN.

SOLUTION

EXAMPLE 2 IN THE 1980s IT WAS GENERALLY BELIEVED THAT CONGENITAL ABNORMALITIES AFFECTED ABOUT 5% OF THE NATION’S CHILDREN. SOME PEOPLE BELIEVE THAT THE INCREASE IN THE NUMBER OF CHEMICALS IN THE ENVIRONMENT HAS LED TO AN INCREASE IN THE INCIDENCE OF ABNORMALITIES. A RECENT STUDY EXAMINED 384 CHILDREN AND FOUND THAT 46 OF THEM SHOWED SIGNS OF AN ABNORMALITY. IS THIS STRONG EVIDENCE THAT THE RISK HAS INCREASED? ( WE CONSIDER A P – VALUE OF AROUND 5% TO REPRESENT STRONG EVIDENCE.)

EXAMPLE 2 CONT’D (A) WRITE APPROPRIATE HYPOTHESES. (B) CHECK THE NECESSARY ASSUMPTIONS. (C) PERFORM THE MECHANICS OF THE TEST. WHAT IS THE P – VALUE? (D) EXPLAIN CAREFULLY WHAT THE P – VALUE MEANS IN THIS CONTEXT. (E) WHAT’S YOUR CONCLUSION? (F) DO ENVIRONMENTAL CHEMICALS CAUSE CONGENITAL ABNORMALITIES?

SOLUTION

INFERENCES ABOUT MEANS TESTING HYPOTHESES ABOUT MEANS ONE – SAMPLE t – TEST FOR MEANS PROBLEM Test HO:  = 0

ASSUMPTIONS AND CONDITIONS INDEPENDENCE ASSUMPTION RANDOMIZATION CONDITION 10% CONDITION NEARLY NORMAL CONDITION OR LARGE SAMPLE

STEPS IN TESTING NULL HYPOTHESIS HO:  = 0 ALTERNATIVE HYPOTHESIS HA:  > 0 or HA:  < 0 or HA:  ≠ 0  

t HAS STUDENT’S t – DISTRIBUTION WITH n – 1 DEGREES OF FREEDOM. ATTITUDE: Assume that the null hypothesis HO is true and uphold it, unless data strongly speaks against it. STANDARD ERROR TEST STATISTICS t HAS STUDENT’S t – DISTRIBUTION WITH n – 1 DEGREES OF FREEDOM.

P-value: Let to be the observed value of the test statistic. HA P-value NORMAL DISTRIBUTION CURVE  HA:  > 0 P(t > to)   HA:  < 0 P(t <to) HA:  ≠ 0 P(t > |to|) + P(t < -|to|)

CONCLUSION  

EXAMPLES FROM PRACTICE SHEET