STAT 3360 – EXAM 3 – PRACTICE VERSION | Complete Solution

STAT 3360 – EXAM 3 – PRACTICE VERSION
REFERENCES TO SECTIONS AND TABLES WILL NOT BE PROVIDED!
TABLES 3, 4, AND 5 WILL BE INCLUDED INTO THE EXAM PAPER!
GET READY BY AUGUST 5, 2015!
1. [SECTION 13.1 + TABLE 4]

TWO ANALYSTS (HILLARY AND BILL) INTEND TO COMPARE THE POPULATION AVERAGE TIME PER CUSTOMER SPENT BY SERVICE PERSONNEL IN TWO BANKS (CHASE AND BANK OF AMERICA).WITH THIS PURPOSE THEY COLLECTED INDEPENDENT SAMPLES OF SIZE 12 EACH, FROM BOTH SOURCES. INDIVIDUAL TIMES (IN MINUTES) ARE NORMALLY DISTRIBUTED WITH UNKNOWN PARAMETERS. TABLE BELOW PRESENTS THE SAMPLE SUMMARIES DERIVED FROM TWO INDEPENDENTLY COLLECTED SAMPLES.
SAMPLE SUMMARIES
SOURCE
SIZE
MEAN
VARIANCE = s2
v = s2 / n
CHASE
n1 = 12
122.40
900.00
75.0
BANK OF AMERICA
n2 = 12
106.74
72.00
6.0
BILL BELIEVES THAT TWO POPULATIONS SHARE THE SAME UNKNOWN COMMON VARIANCE.
A. WHAT NUMBER OF DEGREES OF FREEDOM WILL BE ASSESSED TO THE T-STATISTIC BY BILL? (DEGREES OF FREEDOM BY BILL) =
B. WHAT STANDARD ERROR WILL BE ASSESSED TO THE DIFFERENCE BETWEEN TWO SAMPLE MEANS BY BILL? PLEASE WRITE DOWN THE FORMULA! (SE FOR BILL) =
C. EVALUATE THE TEST STATISTIC. (TEST STATISTIC VALUE FOR BILL) =
D. WHAT CRITICAL VALUES WILL BILL USE? (CRITICAL VALUE FOR BILL) =
E. AT THE 5% SIGNIFICANCE LEVEL, DOES BILL HAVE EVIDENCE THAT THE POPULATION AVERAGE TIME FOR CHASE WAS HIGHER THAN THAT FOR BANK OF AMERICA? YES OR NO? BILL’S DECISION IS <REJECT / DO NOT REJECT>
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STAT 3360 – EXAM 3 – PRACTICE VERSION
2. [SECTION 13.1 + TABLE 4]
TWO ANALYSTS (HILLARY AND BILL) INTEND TO COMPARE THE POPULATION AVERAGE TIME PER CUSTOMER SPENT BY SERVICE PERSONNEL IN TWO BANKS (CHASE AND BANK OF AMERICA).WITH THIS PURPOSE THEY COLLECTED INDEPENDENT SAMPLES OF SIZE 12 EACH, FROM BOTH SOURCES. INDIVIDUAL TIMES (IN MINUTES) ARE NORMALLY DISTRIBUTED WITH UNKNOWN PARAMETERS. TABLE BELOW PRESENTS THE SAMPLE SUMMARIES DERIVED FROM TWO INDEPENDENTLY COLLECTED SAMPLES.
HILLARY BELIEVES THAT ASSUMING ENTIRELY UNKNOWN VARIANCES WOULD BE MORE APPROPRIATE.
SAMPLE SUMMARIES
SOURCE
SIZE
MEAN
VARIANCE = s2
v = s2 / n
CHASE
n1 = 12
122.40
900.00
75.0
BANK OF AMERICA
n2 = 12
106.74
72.00
6.0
A. WHAT NUMBER OF DEGREES OF FREEDOM WILL BE ASSESSED TO THE T-STATISTIC BY HILLARY? (DEGREES OF FREEDOM BY HILLARY) =
B. WHAT STANDARD ERROR WILL BE ASSESSED TO THE DIFFERENCE BETWEEN TWO SAMPLE MEANS BY HILLARY? PLEASE WRITE DOWN THE FORMULA IN EITHER CASE! (SE FOR HILLARY) =
C. EVALUATE THE TEST STATISTIC. (TEST STATISTIC VALUE FOR HILLARY) =
D. WHAT CRITICAL VALUES WILL HILLARY USE? (CRITICAL VALUE FOR HILLARY) =
E. AT THE 5% SIGNIFICANCE LEVEL, DOES HILLARY HAVE EVIDENCE THAT THE POPULATION AVERAGE TIME FOR CHASE WAS HIGHER THAN THAT FOR BANK OF AMERICA? YES OR NO? HILLARY’S DECISION IS <REJECT / DO NOT REJECT>
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STAT 3360 – EXAM 3 – PRACTICE VERSION
3. [SECTION 13.1 + TABLE 4]
THIS PROBLEM CONTINUES THE PREVIOUS ONE.
TWO ANALYSTS (HILLARY AND BILL) INTEND TO COMPARE THE POPULATION AVERAGE TIME PER CUSTOMER SPENT BY SERVICE PERSONNEL IN TWO BANKS (CHASE AND BANK OF AMERICA).WITH THIS PURPOSE THEY COLLECTED INDEPENDENT SAMPLES OF SIZE 12 EACH, FROM BOTH SOURCES. INDIVIDUAL TIMES (IN MINUTES) ARE NORMALLY DISTRIBUTED WITH UNKNOWN PARAMETERS. TABLE BELOW PRESENTS THE SAMPLE SUMMARIES DERIVED FROM TWO INDEPENDENTLY COLLECTED SAMPLES.
SAMPLE SUMMARIES
SOURCE
SIZE
MEAN
VARIANCE = s2
v = s2 / n
CHASE
n1 = 12
122.40
900.00
75.0
BANK OF AMERICA
n2 = 12
106.74
72.00
6.0
NOW THE CONFIDENCE INTERVALS ARE SUPPOSED TO BE DERIVED FOR THE DIFFERENCE BETWEEN THE TWO POPULATION MEANS.
A. ENTER RESULTS FROM PREVIOUS PROBLEM HERE. (DEGREES OF FREEDOM BY BILL) = (DEGREES OF FREEDOM BY HILLARY) =
B. ENTER RESULTS FROM PREVIOUS PROBLEM HERE. (SE BY BILL) = (SE BY HILLARY) =
C. TO ESTIMATE THE DIFFERENCE BETWEEN THE TWO POPULATION MEANS AT THE 95% CONFIDENCE, WHAT CRITICAL VALUE WILL BILL USE? (CRITICAL VALUE BY BILL) =
D. TO ESTIMATE THE DIFFERENCE BETWEEN THE TWO POPULATION MEANS AT THE 95% CONFIDENCE, WHAT CRITICAL VALUE WILL HILLARY USE? (CRITICAL VALUE BY HILLARY) =
E. PRESENT THE CONFIDENCE INTERVALS AND INDICATE THE MID-POINT AND MARGIN OF ERROR IN EACH CASE. CONFIDENCE INTERVAL FOR BILL: (MID-POINT) ± (MARGIN OF ERROR) CONFIDENCE INTERVAL FOR HILLARY: (MID-POINT) ± (MARGIN OF ERROR)
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STAT 3360 – EXAM 3 – PRACTICE VERSION
4. [SECTION 13.3 + TABLE 4].
BILL AND HILLARY COMPLAINED ABOUT THE INCREASED PROPERTY TAX. TO CONVINCE THE MUNICIPAL AUTHORITIES, THEY COLLECTED 16 MATCHED RECORDS (FROM THE SAME HOUSEHOLDS) FOR 2 CONSECUTIVE YEARS, 2011 AND 2012. THE SAMPLE SUMMARIES ARE SHOWN IN THE TABLE BELOW. FOR THE SAKE OF SIMPLICITY, USE $K = $1,000 AS 1.
SUMMARIES
YEAR
MEAN (IN $K)
SD FOR DIFFERENCES (IN $K)
2011
4.568
2012
5.648
2.400
A. ESTIMATE THE POPULATION AVERAGE FOR THE DIFFERENCE (YEAR 2012) – (YEAR 2011) WITH THE 95% CONFIDENCE. SHOW ALL CHARACTERISTICS OF THE CONFIDENCE INTERVAL BELOW. CRITICAL VALUE = STANDARD ERROR FOR DIFFERENCE BETWEEN SAMPLE MEANS = MARGIN OF ERROR = MID-POINT = UPPER CONFIDENCE LIMIT = LOWER CONFIDENCE LIMIT = WIDTH OF THE INTERVAL =
B. AT THE SIGNIFICANCE LEVEL OF 5%, DO BILL AND HILLARY HAVE ENOUGH EVIDENCE THAT THE AVERAGE PROPERTY TAX IN THE YEAR 2012 WAS HIGHER THAN IN 2011? SHOW THE FOLLOWING ELEMENTS OF YOUR DECISION. CRITICAL VALUE(S) = TEST STATISTIC = REJECTION RULE: REJECT THE HYPOTHESIS ABOUT EQUAL POPULATION MEANS IF ... DECISION: THERE IS / IS NOT ENOUGH EVIDENCE THAT THE POPULATION AVERAGE PROPERTY TAX WAS HIGHER IN 2012 COMPARED TO 2011.
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STAT 3360 – EXAM 3 – PRACTICE VERSION
5. [SECTION 12.2 + TABLE 5]
TEST SCORES IN STAT 1342 COURSE ARE ASSUMED TO BE NORMALLY DISTRIBUTED WITH UNKNOWN POPULATION SUMMARIES. A SAMPLE OF 16 STUDENTS WAS SELECTED FOR A STUDY OF VARIABILITY IN THE TEST SCORES. THE SAMPLE SUMMARIES OBTAINED FROM THE SAMPLE ARE SHOWN BELOW.
[SAMPLE MEAN] = 73.7 AND [SAMPLE STANDARD DEVIATION] = 8.0.
A. ESTIMATE THE POPULATION VARIANCE WITH THE 99% CONFIDENCE. ROUND THE CONFIDENCE LIMITS TO THE 2ND DECIMAL PLACE.
CRITICAL VALUE
UPPER CONFIDENCE LIMIT
LOWER CONFIDENCE LIMIT
B. AT THE 5% SIGNIFICANCE LEVEL, DO WE HAVE SUFFICIENT EVIDENCE THAT THE POPULATION VARIANCE EXCEEDED 30? SHOW THE FOLLOWING ELEMENTS OF YOUR DECISION. CRITICAL VALUE(S) = TEST STATISTIC = REJECTION RULE: REJECT THE NULL HYPOTHESIS IF ... DECISION: THERE IS / IS NOT ENOUGH EVIDENCE THE POPULATION VARIANCE EXCEEDED 30
C. AT THE 5% SIGNIFICANCE LEVEL, DO WE HAVE SUFFICIENT EVIDENCE THAT THE POPULATION VARIANCE WAS BELOW 128? SHOW THE FOLLOWING ELEMENTS OF YOUR DECISION. CRITICAL VALUE(S) = TEST STATISTIC = REJECTION RULE: REJECT THE NULL HYPOTHESIS IF ... DECISION: THERE IS / IS NOT ENOUGH EVIDENCE THE POPULATION VARIANCE WAS BELOW 128
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STAT 3360 – EXAM 3 – PRACTICE VERSION
6. [SECTION 12.1 + TABLE 4]
TEST SCORES IN STAT 1342 COURSE ARE ASSUMED TO BE NORMALLY DISTRIBUTED WITH UNKNOWN POPULATION SUMMARIES. A SAMPLE OF 16 STUDENTS WAS SELECTED FOR A STUDY OF POPULATION AVERAGE IN THE TEST SCORES. THE SAMPLE SUMMARIES OBTAINED FROM THE SAMPLE ARE SHOWN BELOW.
[SAMPLE MEAN] = 73.7 AND [SAMPLE STANDARD DEVIATION] = 8.0.
A. ESTIMATE THE POPULATION AVERAGE WITH THE 99% CONFIDENCE. ROUND THE CONFIDENCE LIMITS TO THE 2ND DECIMAL PLACE.
CRITICAL VALUE
UPPER CONFIDENCE LIMIT
LOWER CONFIDENCE LIMIT
B. AT THE 5% SIGNIFICANCE LEVEL, DO WE HAVE SUFFICIENT EVIDENCE THAT THE POPULATION AVERAGE EXCEEDED 70? SHOW THE FOLLOWING ELEMENTS OF YOUR DECISION. CRITICAL VALUE(S) = TEST STATISTIC = REJECTION RULE: REJECT THE NULL HYPOTHESIS IF ... DECISION: THERE IS / IS NOT ENOUGH EVIDENCE THE POPULATION AVERAGE EXCEEDED 70
C. AT THE 5% SIGNIFICANCE LEVEL, DO WE HAVE SUFFICIENT EVIDENCE THAT THE POPULATION AVERAGE WAS BELOW 78? SHOW THE FOLLOWING ELEMENTS OF YOUR DECISION. CRITICAL VALUE(S) = TEST STATISTIC = REJECTION RULE: REJECT THE NULL HYPOTHESIS IF ... DECISION: THERE IS / IS NOT ENOUGH EVIDENCE THE POPULATION AVERAGE WAS BELOW 78
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STAT 3360 – EXAM 3 – PRACTICE VERSION
7. [SECTION 13.5 + TABLE 3 + TABLE 4 (ROW = ∞)]
DEPARTMENT OF PUBLIC SAFETY COLLECTED RECORDS ABOUT DRIVERS FROM TWO CATEFORIES. THE DATA SUMMARIES SHOW THAT AMONG 200 YOUNG DRIVERS, 27 HAD SPEEDING TICKETS. IN THE MEANTIME, AMONG 200 AGED DRIVERS, THERE WERE ONLY 13 DRIVERS WITH SPEEDING TICKETS.
A. AT THE 1% SIGNIFICANCE LEVEL, DO YOU HAVE SUFFIICIENT EVIDENCE THAT THE THEORETICAL PROPORTIONS DIFFER? SHOW THE SUMMARIES AND SPECIFY THE FOLLOWING ELEMENTS OF YOUR DECISION MAKING PROCESS. NULL HYPOTHESIS STATES ... ALTERNATIVE STATES ... CRITICAL VALUE(S) = TEST STATISTIC FOR COMPARISON OF 2 POPULATION PROPORTIONS IS [SHOW THE FORMULA] STANDARD ERROR = DECISION IS: REJECT THE NULL HYPOTHESIS / DO NOT REJECT IT [CIRCLE ONE!]
B. AT THE SAME 1% SIGNIFICANCE LEVEL, DO YOU HAVE SUFFICIENT EVIDENCE THAT THE AGED DRIVERS ARE LESS LIKELY TO BE TICKETED? SHOW THE SUMMARIES AND SPECIFY THE FOLLOWING ELEMENTS OF YOUR DECISION MAKING PROCESS. NULL HYPOTHESIS STATES ... ALTERNATIVE STATES ... CRITICAL VALUE(S) = TEST STATISTIC FOR COMPARISON OF 2 POPULATION PROPORTIONS IS [SHOW THE FORMULA] STANDARD ERROR = DECISION IS: REJECT THE NULL HYPOTHESIS / DO NOT REJECT IT [CIRCLE ONE!]
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STAT 3360 – EXAM 3 – PRACTICE VERSION
8. [SECTION 12.3 + TABLE 3 + TABLE 4 (ROW = ∞)]
MARKET RESEARCH ANALYSTS ARE MAKING CONCLUSIONS ABOUT THE PROPORTION CUSTOMERS WILLING TO ADD EXTRA SERVICES AT THE LOW PRICE. THEY BELIEVE THAT THE THEORETICAL PROPORTION MAY BE HIGHER THAN 25%. A SAMPLE OF 1,200 RESPONDENTS WAS COLLECTED. IT SHOWS THAT 333 ARE WILLING TO ADD THE NEW SERVICES AT THE LOW PRICE.
SAMPLE SIZE
1,200
WILLING
333
SAMPLE PROPORTION
333/1,200 = 0.2775
A. DETERMINE THE PARAMETERS OF THE NORMAL APPROXIMATION FOR THE DISTRIBUTION OF THE SAMPLE PROPORTION. MEAN = STANDARD DEVIATION =
B. AT THE 1% SIGNIFICANCE LEVEL, DO ANALYSTS HAVE EVIDENCE THAT THE PROPORTION OF CUSTOMERS WILLING TO ADD NEW SERVICES EXCEEDED THE LEVEL OF 25%? (YES / NO) TEST STATISTIC = STANDARD ERROR = CRITICAL VALUE =
C. STATE THE REJECTION RULE <REJECT THE NULL HYPOTHESIS IF ...>
D. IN THE FUTURE ANALYSTS WILL DESIGN A NEW STUDY. THEY INTEND TO ESTIMATE THE POPULATION PROPORTION WITH THE 99% CONFIDENCE. THEY ALSO WANT TO MAKE SURE THAT THE MARGIN OF ERROR WILL NOT EXCEED 4%. HOW MANY RECORDS SHOULD THEY COLLECT WITHOUT KNOWING THE SAMPLE PROPORTION? SHOW THE CRITICAL VALUE FOR THIS PROBLEM AND EVALUATE THE SMALLEST SAMPLE SIZE NEEDED FOR THE STUDY. CRITICAL VALUE = THE SMALLEST SAMPLE SIZE =
(MAKE SURE THAT THIS IS AN INTEGER NUMBER!)
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STAT 3360 – EXAM 3 – PRACTICE VERSION
9. [SECTION 15.1 + TABLE 5]
INSTRUCTORS BELIEVE THAT SCIENCE MAJOR STUDENTS WHO TAKE MATH AND STAT COURSES SHOULD SHOW 20% OF A GRADES, 30% OF B GRADES, 25% OF C GRADES, 15% OF D GRADES, AND 10% OF F GRADES. THE TABLE BELOW SHOWS OBSERVED COUNTS FOR ONE SEMESTER.
GRADE
A
B
C
D
F
TOTAL
OBSERVED
138
198
138
72
54
600
PROPORTION
20%
30%
25%
15%
10%
100%
EXPECTED
A. FIND THE EXPECTED COUNTS (FILL THE CELLS IN THE TABLE ABOVE).
B. EVALUATE THE TEST STATISTIC AND SPECIFY THE NUMBER OF DEGREES OF FREEDOM. TEST STATISTIC = DEGREES OF FREEDOM =
C. AT THE SIGNIFICANCE LEVEL OF 5%, DO YOU HAVE EVIDENCE THAT THE GRADE DISTRIBUTION DEVIATED FROM THE HYPOTHETICAL ASSUMPTION? (YES / NO) CRITICAL VALUE = REJECTION RULE STATES: < FILL THE GAP>
D. AT THE SIGNIFICANCE LEVEL OF 1%, DO YOU HAVE EVIDENCE THAT THE GRADE DISTRIBUTION DEVIATED FROM THE HYPOTHETICAL ASSUMPTION? (YES / NO) CRITICAL VALUE = REJECTION RULE STATES: < FILL THE GAP>
10. [(SECTIONS 16.1 – 16.4) + TABLE 4]
ANALYSTS USE A SAMPLE OF 32 WEEKS TO BUILD THE LINEAR MODEL AIMED AT PREDICTING THE WEEKLY REVENUE (Y) AS A LINEAR FUNCTION OF ADVERTISEMENT COST (X).
THE MODEL WAS FOUND AS (Y-HAT) = 4,058 + 40.796 X.
MEASUREMENT ERRORS ARE ASSUMED TO BE NORMALLY DISTRIBUTED. THE T-STATISTIC FOR SIGNIFICANCE OF THE ESTIMATED SLOPE SHOWS THE VALUE OF (T = [SLOPE] / (SE) = 2.632).
A. WHAT IS THE STANDARD ERROR (SE) OF THE ESTIMATED SLOPE? STANDARD ERROR =
B. AT THE 1% SIGNIFICANCE LEVEL, DO ANALYSTS HAVE ENOUGH EVIDENCE TO CONCLUDE THAT ADVERTISEMENT COST WOULD POSITIVELY IMPACT THE WEEKLY REVENUE? (YES / NO)TEST STATISTIC = CRITICAL VALUE(S) = REJECTION RULE STATES: <FILL THE GAP, PLEASE!>
C. ESTIMATE THE POPULATION SLOPE WITH 99% CONFIDENCE AND PRESENT IT IN THE FORMAT (MID-POINT) ± (MARGIN OF ERROR). MID-POINT = MARGIN OF ERROR = CRITICAL VALUE(S) =
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STAT 3360 – EXAM 3 – PRACTICE VERSION
11. [SECTION 7.5 + TABLE 2]
STUDENTS ARRIVE AT THE GEMS CENTER WITH THE INTENSITY 0.15 PER MINUTE. A RANDOM VARIABLE Y MEASURES THE NUMBER OF STUDENTS ARRIVED AT THE CENTER DURING THE OBSERVATION PERIOD OF 30 MINUTES. ASSUMING THAT ARRIVALS FOLLOW POISSON ASSUMPTIONS FIND THE FOLLOWING PROBABILITIES.
A. WHAT IS THE CHANCE THAT THERE WILL BE AT MOST 4 STUDENTS ARRIVING WITHIN THE OBSERVATION PERIOD? P [Y ≤ 4] =
B. WHAT IS THE CHANCE THAT AT LEAST 7 STUDENTS WILL ARRIVE DURING THE OBSERVATION PERIOD? P [Y ≥ 7] =
C. WHAT IS THE CHANCE THAT 5 TO 10 STUDENTS WILL ARRIVE DURING THE THE OBSERVATION PERIOD? P [5 ≤ Y ≤ 10] =
D. WHAT IS THE CHANCE THAT THE NUMBER OF STUDENTS ARRIVING DURING THE OBSERVATION PERIOD WILL BE FEWER THAN 8 AND GREATER THAN 4? P [4 < Y < 8] =
E. WHAT IS THE CHANCE THAT 4 OR 5 STUDENTS WILL ARRIVE DURING THE OBSERVATION PERIOD? P [Y = 4 OR 5] =
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