Partial preview of the text
Download Statistics and Probability: Hypothesis Testing Exercises and more Cheat Sheet Mathematics in PDF only on Docsity!
ANGEL RACOMA MA’AM GINA CATAPANG
ACAD D HUMSS WEEK 6 STATISTIC AND
PROBABILITY
TASK 1: SOLVING PROBLEMS INVOLVING TEST OF HYPOTHESIS.
(ACTIVITY 2, PP.10-11)
ACTIVITY 2: LET ME READ AND UNDERSTAND!
DIRECTIONS: CAREFULLY READ THE PROBLEM AND ANSWER THE QUESTIONS
THAT FOLLOW.
PROBLEM 1. A BANANA COMPANY CLAIMS THAT THE MEAN WEIGHT OF THEIR
BANANA IS 150 GRAMS WITH A STANDARD DEVIATION OF 18 GRAMS. DATA
GENERATED FROM A SAMPLE OF 49 BANANAS RANDOMLY SELECTED
INDICATED A MEAN WEIGHT OF 153.5 GRAMS PER BANANA. IS THERE
SUFFICIENT EVIDENCE TO REJECT THE COMPANY'S CLAIM? USE A = 0.05.
WHAT ARE THE HYPOTHESES?
U=
U≠
IS IT TWO-TAILED OR ONE-TAILED TEST?
TWO TAILED TEST
WHAT IS THE LEVEL OF SIGNIFICANCE?
A=0.
IS THE POPULATION STANDARD DEVIATION KNOWN? 5
YES
WHAT APPROPRIATE TEST STATISTIC (Z-TEST OR T-TEST) CAN YOU USE?
Z-TEST
BASED ON THE LEVEL OF SIGNIFICANCE, HYPOTHESIS TEST, AND TEST
STATISTIC, WHAT IS THE CRITICAL VALUE?
Z +1.
DRAW THE REJECTION REGION.
PROBLEM 2. THE MANUFACTURER OF AN AIRPORT BAGGAGE SCANNING
MACHINE CLAIMS IT CAN HANDLE AN AVERAGE OF 530 BAGS PER HOUR. AT
A = 0.05 IN A LEFT TAILED TEST, WOULD A SAMPLE OF 16 RANDOMLY
CHOSEN HOURS WITH A MEAN OF 510 AND STANDARD DEVIATION OF 50
INDICATE THAT THE MANUFACTURER'S CLAIM IS AN OVERSTATEMENT?
WHAT ARE THE HYPOTHESES?
U=
U< 530
IS IT TWO-TAILED TEST OR ONE-TAILED TEST?
ONE TAILED TEST
3. WHAT IS THE LEVEL OF SIGNIFICANCE?
A=0.
IS THE POPULATION STANDARD DEVIATION KNOWN OR UNKNOWN?
UNKNOWN
WHAT APPROPRIATE TEST STATISTIC (Z-TEST OR T-TEST) CAN YOU USE?
T-TEST
BASED ON THE LEVEL OF SIGNIFICANCE, HYPOTHESIS TEST, AND TEST
STATISTIC, WHAT IS THE CRITICAL VALUE?
T=1.
DRAW THE REJECTION REGION.
ACTIVITY 2. REJECT IT!
3. H:U = 15 HA:U <
THE SAMPLE MEAN IS 10, THE
SAMPLE STANDARD DEVIATION IS
6.1, AND THE SAMPLE SIZE IS 9. USE
A = 0.05.
T-TEST ;-2.
8. HO: = 25
HA: Μ <
THE SAMPLE MEAN IS 23.75, THE
SAMPLE STANDARD DEVIATION IS
4.5, AND THE SAMPLE SIZE IS 12.
USE A = 0.05.
T-TEST;1.
4. H:U 116.12 HA:U> 116.
THE POPULATION FOLLOWS A
NORMAL DISTRIBUTION WITH
STANDARD DEVIATION OF 7.18,
SAMPLE MEAN OF 118.7, AND
SAMPLE SIZE OF 21. USE A = 0.10.
T-TEST;1.
9. H:U 106.22 HA:> 106.
THE POPULATION FOLLOWS A
NORMAL DISTRIBUTION WITH
STANDARD DEVIATION OF 4.08,
SAMPLE MEAN OF 108.5 AND SAMPLE
SIZE OF 17. USE A = 0.
T-TEST;2.
5. HO: U= 215 HA: Μ ≠ 215
THE POPULATION IS APPROXIMATELY
NORMAL. THE SAMPLE MEAN IS
219.3, THE SAMPLE STANDARD
DEVIATION IS 13.12, AND THE
SAMPLE SIZE IS 22. USE A = 0.05.
T-TEST;1.
10. HO: Μ = 25.5 HA: Μ> 25. 5
THE SAMPLE MEAN IS 23.8 AND THE
SAMPLE SIZE IS 42. THE POPULATION
FOLLOWS A NORMAL DISTRIBUTION
WITH STANDARD DEVIATION 2.27.
USE A = 0.01.
Z-TEST;=-4.
TASK 1: DETERMINING THE CRITICAL VALUE AND IDENTIFYING THE
REJECTION REGION UNDER THE NORMAL CURVE. (ACTIVITY 4.2,
PP.13-14)
ACTIVITY 4.2: BORDERLINE
DIRECTIONS: CAREFULLY READ ANALYZE THE FOLLOWING SITUATIONS.
IDENTIFY THE INFORMATION BEING ASKED. THEN, DETERMINE THE CRITICAL
VALUE AND SHADE THE AREA OF THE REJECTION REGION UNDER THE
NORMAL CURVE.
1. SUPPOSE THAT IN THE PAST, 40% OF ALL ADULTS FAVORED CAPITAL
PUNISHMENT. DO WE HAVE REASON TO BELIEVE THAT THIS
PROPORTION HAS INCREASED IF IN A RANDOM SAMPLE OF 150
ADULTS, 80 FAVORED CAPITAL PUNISHMENT? USE A 0.05 LEVEL OF
SIGNIFICANCE.
2. PROFESSORS FROM A PROFESSIONAL ORGANIZATION FOR PRIVATE
COLLEGES AND UNIVERSITIES REPORTED THAT MORE THAN 16% OF
PROFESSORS ATTENDED A NATIONAL CONVENTION IN THE PAST YEAR.
TO TEST THIS CLAIM, A RESEARCHER SURVEYED 200 PROFESSORS
AND FOUND THAT 50 ATTENDED A NATIONAL CONVENTION IN THE
PAST YEAR. AT A = 0.10, TEST IF THE FIGURE IN THE CLAIM IS
CORRECT.
3. MALAKAS MADE A CLAIM THAT AT LEAST 5% OF COLLEGE MALE
STUDENTS IN THEIR SCHOOL JOIN TRIATHLON. HIS FRIEND, MAYUMI,
FINDS THIS HARD TO BELIEVE AND DECIDED TO CHECK THE VALIDITY
OF SUCH CLAIM, SO SHE TOOK A RANDOM SAMPLE. AT 0.01, DOES
MAYUMI PROVIDE ENOUGH INDICATION TO REJECT THE CLAIM OF
MALAKAS IF THERE WERE 60 RACERS IN HER SAMPLE OF 300
EVIDENCES
4. A PHARMACEUTICAL COMPANY IS ON THE FIRST PHASE OF TESTING A
VACCINE FOR A NEW VIRUS. THEY FORM A GROUP CONSISTS OF 100
PEOPLE EACH WHO HAVE A DISEASE AND GIVEN A VACCINE. IT IS FOUND
OUT THAT, 65 RECOVERED FROM THE DISEASE. AT SIGNIFICANCE LEVEL
OF 0.01, DETERMINE THE CRITICAL VALUE AND ILLUSTRATE THE
REJECTION REGION.
5. A SAMPLE POLL OF 300 VOTERS FROM TOWN A SHOWED THAT 56%
WERE IN FAVOR OF A GIVEN CANDIDATE. AT A SIGNIFICANCE LEVEL OF
0.10, DETERMINE THE CRITICAL VALU AND ILLUSTRATE THE REJECTION
REGION.
BE CRITICAL BORDERING
THE POPULATION IS APPROXIMATELY NORMAL. THE SAMPLE MEAN IS 219.3,
THE SAMPLE STANDARD DEVIATION IS 13.12, AND THE SAMPLE SIZE IS 22.
USE Α = 0.
ANSWER: FAIL TO REJECT THE NULL HYPOTHESIS
TASK 1: SOLVING FOR THE COMPUTED Z-VALUE. (ACTIVITY 5, PP.13-
ACTIVITY 5: TELL ME THE TAIL DIRECTIONS: IN EACH PROBLEM BELOW, GIVE
THE NULL AND ALTERNATIVE HYPOTHESES AND IDENTIFY WHETHER IT IS
RIGHT-TAILED, LEFT-TAILED OR TWO-TAILED TEST.
1. A SAMPLE OF 800 ITEMS PRODUCED ON A NEW MACHINE SHOWED
THAT 48 OF THEM ARE DEFECTIVE. THE FACTORY WILL GET RID OF
THE MACHINE IF THE DATA INDICATES THAT THE PROPORTION OF
DEFECTIVE ITEMS IS SIGNIFICANTLY MORE THAN 5%. AT A
SIGNIFICANCE LEVEL OF 10% DOES THE FACTORY GET RID OF THE
MACHINE OR NOT?
HA:P= 0.05 HA:P > 0.05 RIGHT -TAILED
2. A DRUG MANUFACTURER CLAIMS THAT FEWER THAN 10% OF PATIENTS
WHO TAKE ITS NEW DRUG FOR TREATING CERTAIN PNEUMONIA WILL
EXPERIENCE NAUSEA. IN A RANDOM SAMPLE OF 250 PATIENTS, 23
EXPERIENCED NAUSEA. PERFORM A SIGNIFICANCE TEST AT THE 5%
SIGNIFICANCE LEVEL TO TEST THIS CLAIM.
HA:P= 0.10 HA:P < 0.10 LEFT -TAILED
3. IN A GROUP OF 375 SENIOR HIGH SCHOOL STUDENTS, 40 WERE LEFT-
HANDED. IS THIS SIGNIFICANTLY DIFFERENT FROM THE PROPORTION
OF ALL SENIOR HIGH SCHOOL STUDENTS WHO ARE LEFT-HANDED,
WHICH IS 12%?
HA:P= 0.12 HA:P ≠ 0.12 TWO-TAILED
4. IN A RANDOM SURVEY OF 1000 HOUSEHOLDS IN UNLAD PROVINCE, IT
IS FOUND THAT 29% OF THE HOUSEHOLDS HAVE AT LEAST ONE
MEMBER WITH A COLLEGE DEGREE. DOES THIS FINDING CONTRADICT
THE STATEMENT THAT THE PROPORTION OF ALL SUCH HOUSEHOLDS
IN UNLAD PROVINCE IS 35 PERCENT? TEST AT A = .05 SIGNIFICANCE
LEVEL.
HA:P= 0.35 HA:P ≠ 0.35 TWO-TAILED
5. IN A RANDOM SAMPLE OF 400 ELECTRONIC GADGETS, 14 WERE
FOUND TO BE DEFECTIVE. THE MANUFACTURER WANTS TO CLAIM
THAT LESS THAN 5% OF ALL OF THEIR GAMES ARE DEFECTIVE. TEST
THIS CLAIM AT THE 0.01 SIGNIFICANCE LEVEL.
HA:P= 0.05 HA:P < 0.05 LEFT-TAILED
WHAT I CAN DO
ACTIVITY 5. FINDING ZOOM
P= 0.
SAMPLE SIZE= 180
SAMPLE PROPORTION= 0.
P= 0.
SAMPLE SIZE=
SAMPLE PROPORTION= 29 %
N=
Q =1-P 0.
ZCOM= 3.
TASK 2: DRAWING CONCLUSION ABOUT THE POPULATION
PROPORTION. (PROBLEM 1, PAGE 13)
ACTIVITY 4: DECIDE NOW, CONCLUDE LATER!
DIRECTIONS: USING THE GIVEN HYPOTHESES, COMPUTED Z-VALUE, AND
LEVEL OF SIGNIFICANCE, MAKE YOUR OWN DECISION AND CONCLUSION.
THEN, COMPLETE THE STATEMENT BY FILLING IN THE BLANK WITH THE
APPROPRIATE WORD/S. THE FIRST ONE WAS DONE FOR YOU AS A GUIDE.
PROBLEM 1
GIVEN:
HO: THE PROPORTION OF EMPLOYEES IN A SHOE FACTORY WHO SMOKE
CIGARETTE IS
30%. (H .: P = .30)
H₁: THE PROPORTION OF EMPLOYEES IN A SHOE FACTORY WHO SMOKE
CIGARETTE HAS INCREASED TO 30%.
A = 0.01 (H .: P> .30)
COMPUTED Z-STATISTIC: ZCOM = 2.56 AND CRITICAL Z- VALUE: Z TAB =
1.282 IN
DECISION: SAVE THE COMPORTED TEST SEMESTIC ZOM = 256 FALL IN THE
REJECTION RAGING REJECT THE MALL HYPOTHERS (MAL
CONCLUSION: THEMEFORE, WE CONCLUDE SHOT OF ONE LOVE OF
SUGAPFRENCE. PLANE IS EVIDENCE TO CONCLUDE THAT THE PROPORTION
OF IMPLAYERS IN A SHEE FARTING WHO SMAKE CIGARETTE HAS
INCORESEED TO 3%
TASK 3: SOLVING PROBLEMS INVOLVING TEST OF HYPOTHESIS ON
THE SAMPLE PROPORTION. (ASSESSMENT NOS. 13-15, PP.14-15)
FOR US. 13 TO 15, REFER TO THE GIVEN PROBLEM BELOW.
THE MAYOR OF A TOWN SAW AN ARTICLE CLAIMING THAT THE NATIONAL
UNEMPLOYMENT RATE IS 8%. HE WONDERED IF THIS HOLDS TRUE IN THEIR
TOWN, SO A SAMPLE OF 200 RESIDENTS WAS TAKEN. THE SAMPLE
INCLUDED 22 UNEMPLOYED RESIDENTS AND 0.05 LEVEL OF SIGNIFICANCE
WAS USED.
13.PAIR OF HYPOTHESES.
HO: P = 0.
14 THE TEST IS TWO TAILED TEST
LEVEL OF SIGNIFICANCE(A) IN THE GIVEN PROBLEM IS 0.
