2. [20 POINTS = 10 + 10] A SAMPLE OF 16 CAR POLICIES WAS DRAWN FOR THE ANALYSIS OF CAR PREMIUM DISTR

2.
[20 POINTS = 10 + 10] A SAMPLE OF 16
CAR POLICIES WAS DRAWN FOR THE ANALYSIS OF CAR PREMIUM DISTRIBUTION. ASSUME
THAT AN INDIVIDUAL PREMIUM IS NORMALLY DISTRIBUTED WITH UNKNOWN PARAMETERS.
SAMPLE SUMMARIES ARE: (SAMPLE MEAN) =
$96 AND (SAMPLE STANDARD DEVIATION) = $20. (A) A T THE 5% SIGNIFICANCE LEVEL,
IS THERE SUFFICIENT EVIDENCE THAT THE
POPULATION VARIANCE FOR A CAR POLICIY PREMIUM WAS BELOW 960? CIRCLE APPROPRIATE ANSWER: YES! OR NO!
SHOW THE TEST STATISTIC VALUE, THE CRITICAL
VALUE(S) NEEDED FOR YOUR DECISION AND
FORMULATE THE REJECTION RULE.
TEST STATISTIC VALUE =
C RITICAL VALUE(S):
R EJECTION RULE STATES… (B) A T THE 5% SIGNIFICANCE LEVEL, IS THERE
SUFFICIENT EVIDENCE THAT THE POPULATION
VARIANCE FOR A CAR POLICIY PREMIUM WAS ABOVE 250? CIRCLE APPROPRIATE ANSWER:
YES! OR NO!
S HOW THE CRITICAL VALUE(S) NEEDED FOR YOUR
DECISION AND FORMULATE THE REJECTION
RULE. TEST STATISTIC VALUE =
C RITICAL VALUE(S):
R EJECTION RULE STATES…

3.
[30 POINTS = 10 + 10 + 10]
BUREAU OF LABOR STATISTICS CONDUCTED
RESEARCH AIMED AT DRAWING CONCLUSIONS ABOUT THE AVERAGE SALARY CHANGE AMONG
SOME CATEGORIES OF WORKERS. A SAMPLE OF 25 TECHICIANS WAS SELECTED AT RANDOM
AND ANNUAL SALARY RECORDS WERE COLLECTED FOR TWO CONSECUTIVE YEARS (2012 AND
2013). THE SUMMARIES WERE FOUND AS FOLLOWS:
(2012 SAMPLE MEAN) = $41,350 AND (2013 SAMPLE MEAN) = $45,358. ALSO THE SAMPLE STANDARD DEVIATION FOR INDIVIDUAL
DIFFERENCES, D = (2013 SALARY) – (2012
SALARY), WAS FOUND AS $8,000. (A) A T
THE 1% SIGNIFICANCE LEVEL, DO RESEARCHERS HAVE EVIDENCE THAT THE POPULATION AVERAGE SALARY HAS
INCREASED? CIRCLE APPROPRIATE ANSWER:
YES! NO!
SPECIFY THE TEST STATISTIC VALUE AND CRITICAL
VALUE(S). THEN FORMULATE THE REJECTION
RULE.
TEST STATISTIC VALUE =
C RITICAL VALUE(S):
R EJECTION RULE STATES…
(B) DO RESEARCHERS HAVE EVIDENCE THAT THE
POPULATION AVERAGE SALARY HAS
CHANGED? CIRCLE APPROPRIATE ANSWER: YES! NO!
SPECIFY THE TEST STATISTIC VALUE AND CRITICAL
VALUE(S). THEN FORMULATE THE REJECTION
RULE.
TEST STATISTIC VALUE =
C RITICAL VALUE(S):
R EJECTION RULE STATES…
(C)
ESTIMATE THE POPULATION AVERAGE DIFFERENCE WITH 95% CONFIDENCE S HOW THE MID-POINT AND MARGIN OF
ERROR. A LSO SPECIFY THE CRITICAL
VALUES NEEDED FOR THIS PROCEDURE. C
RITICAL VALUE =
MID-POINT =

M ARGIN OF ERROR =
UPPER CONFIDENCE LIMIT =
LOWER CONFIDENCE LIMIT =
3
4.
[30 POINTS = 10 + 10 + 10] A STUDY AIMED
AT DRAWING CONCLUSIONS ABOUT THE PROPORTION OF CAR COLLISIONS CAUSED BY
DRIVER’S TEXTING WAS CONDUCTED. A SAMPLE OF 1,200 CASES WAS ANALYZED. IT TURNED
OUT THAT 336 COLLISIONS OCCURRED DUE TO TEXTING WHILE DRIVING. [A] IF THE HYPOTHETICAL PROPORTION
OF COLLISIONS CAUSED BY TEXTING IS 25%,
FIND THE PARAMETERS OF THE APPROXIMATE NORMAL DISTRIBUTION FOR A SAMPLE PROPORTION.
[B] AT THE SIGNIFICANCE LEVEL OF
1%, CAN YOU SAY THAT THERE IS SUFFICIENT
EVIDENCE THAT THE POPULATION PROPORTION EXCEEDS 25%?
SPECIFY THE TEST STATISTIC VALUE AND CRITICAL
VALUE(S). THEN FORMULATE THE REJECTION
RULE.
TEST STATISTIC VALUE =
C RITICAL VALUE(S):
R EJECTION RULE STATES…
[C] AT THE SIGNIFICANCE LEVEL OF
1%, CAN YOU SAY THAT THERE IS SUFFICIENT
EVIDENCE THAT THE POPULATION PROPORTION DIFFERS FROM 25%? SPECIFY THE TEST STATISTIC VALUE AND
CRITICAL VALUE(S). THEN FORMULATE THE
REJECTION RULE.
TEST STATISTIC VALUE =
C RITICAL VALUE(S):
R EJECTION RULE STATES..