Download 1 Students will be able to interpret and compute relative frequency

1
Chapter 13
Objectives:
1. The student will be able to differentiate between the
population regression equation and the sample regression
equation as well as a population and sample correlation
analysis.
2. The student will be able to compute(using Excel) and
interpret the correlation coefficient.
3. The student will be able to compute and interpret the
intercept and slope of a regression equation.
4. The student will be able to compute and interpret the:
standard error of the estimate, coefficient of determination
and the hypothesis test for a regression equation.
5. The student will be able to use regression as a forecasting
tool.
6. The student will be able to interpret a confidence interval
for the estimates of the dependent variable.
Scatter Diagram Pg 12 and 463
PLOTTING DATA IS INFORMATIVE Excel use chart wizard/scatter.
The variable in the farthest left column is placed on the X
axis.
CORRELATION ANALYSIS DESCRIBES THE STRENGTH OF A LINEAR
RELATIONSHIP BETWEEN TWO VARIABLES
CORRELATION COEFFICIENT -1X+1
Excel>tools/data analysis/correlation
-1 implies the strongest possible negative (inverse)
relationship.
+1 implies the strongest possible positive (direct) relationship
0 implies no relationship at all with assumptions
13-1
𝑟=
∑(𝑋 − 𝑋)(𝑌 − 𝑌)
(𝑛 − 1)𝑆𝑦 𝑆𝑥
Coefficient of determination r2
Correlation/Regression
1. State a hypothesis
Your hypothesis should include a statement regarding
causation.
2/27/07
“Is an Economist Qualified To Solve Puzzle of Autism?” WSJ A1
2
police and violent crime
Vietnam war and future earnings
TV and Autism
Instrumental variables
Election year
Draft lottery
Rain or percent subscribe to
cable
http://www.correlated.org/
2. Gather data
3. Graph data
4. Statistical analysis
5. Hypothesis Test
Spurious Correlations Pg 465 Stats-Basic Statistics-correlation
(SEE HELP SHEET)
7/13/05 “Drink More, Earn More (&Give More)” WSJ
UWP AODA Spring survey on excel
1/12/10
Watching TV Linked To Higher Risk of Death WSJ D1
Steps of Hypothesis Testing Pg 332-338
1. Formulate the null hypothesis Ho.
Formulate the alternative hypothesis Ha in statistical terms.
2. Set the level of significance  and the sample size n.
3. Select the appropriate test statistic and rejection rule.
4. Collect the data and calculate the test statistic. 13-2
5. If the calculated value of the test statistic falls in the
rejection region, then reject Ho. If the calculated value of
the test statistic does not fall in the rejection region, then
do not reject Ho.
Student t-table is the minimum number of standard deviations
needed to reject Ho
t-Statistic (13-2) is the actual number of standard deviations
from Ho
If the actual number of standard deviations is greater than the
minimum, reject Ho.
Regression analysis-Stats-Regression
Regression describes a linear relationship by using the
mathematical equation below.
Population Regression
𝑌 = 𝐴 + 𝐵𝑥
Sample Regression
𝑦 = 𝑎 + 𝑏𝑥
a= intercept b = slope
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13-4 and 13-5
𝑏=𝑟
𝑆𝑦
13 − 4
𝑆𝑥
𝑎 = 𝑌̅ − 𝑏𝑋̅ 13 − 5
[2. Give me a graphical example of a relationship between to
variables that would not be represented well by a straight
line.]
DEPENDENT VARIABLE
SALES
CONSUMPTION
OUTPUT/PER ACRE
GRADES
HOUSING
INDEPENDENT VARIABLE
ADVERTISING BUDGET
INCOME
FERTILIZER
STUDY TIME
MORTGAGE RATES
OTHER
VARIABLES
CONSUMPTION
illustrate GRAPHICALLY
REGRESSION
∑(𝑦 − 𝑦)2 = ∑(𝑦′ − 𝑦)2 + ∑(𝑦 − 𝑦′)2
Total Variation (SST)
= Explained(SSR) + Unexplained(SSE)
POPULATION Yi = Bo + BiX + Ei ERROR EXISTS BECAUSE:
1. THERE IS NOT A PERFECT LINEAR RELATIONSHIP
2. THERE ARE OTHER VARIABLES THAT INFLUENCE THE DEPENDENT
VARIABLE
STANDARD ERROR OF THE ESTIMATE 13-6
∑(𝑦 − 𝑦′)2
𝑆𝑦|𝑥 = √
𝑛−2
PREDICTING THE DEPENDENT VARIABLE - POINT ESTIMATE
INTERVAL ESTIMATE
Stats-Regression-options
ASSUMPTIONS OF REGRESSION: Pg 480
1. For each value of X, there is a group of Y values, and
these Y values are normally distributed.
2. Data is linearly related
3. The standard deviations of these normal distributions
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are equal.
4. The Y values are statistically independent. (Time
Series)
HYPOTHESIS TEST FOR THE SLOPE 13-2
𝑡=
𝑟√𝑛 − 2
√1 − 𝑟 2
INFERENCES ABOUT THE SLOPE OF THE REGRESSION LINE
NULL HYPOTHESIS
Ho: B = 0 Accepting the null hypothesis
implies there is likely no relationship between your independent
and dependent variable in your population with unknown risk of
error.
Null Hypothesis
Ho: B  0 Rejecting the null
hypothesis, and therefore accepting the alternative, implies
there is likely a relationship between your independent and
dependent variables in the population with a known risk of
error.
If the computed "t" value form minitab is greater
than the table value, reject the null hypothesis and
accept the alternative.
If the computed "t" value from minitab is less than
the table value, accept the null hypothesis.
Type I error – rejecting a true hypothesis
Type II error – Accepting a false hypothesis
Type I and II error in our legal system
10/17/09
Presumption of Guilt WSJ W1
Excel - Select - tools/Data Analysis/Regression
Unemployment rates on excel
Chapter 5
Objectives:
1 Students will be able to interpret and
compute relative frequency, classical, and
subjective probability.
2. Students will be able to define and compute
conditional probability.
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3. Students will be able to distinguish
independent from dependent events and
them in conditional probability.
4. Students will be able to draw probability
diagrams.
relate
tree
5. Students will be able to apply the rules of
addition and multiplication.
6. Students will be able to relate Bayes'
Theorem, conditional probability,
dependents events to the tree diagram.
7. Students will be able to apply determine
the
number of possible permutations and
combinations.
PROBABILITY A measure of the likelihood that an event in
the future will happen; it can only assume a value between 0 and
1 inclusive.
0  P (X)  1
PROBABILITY OF EVENT A
P(A) = Number of favorable outcomes in the sample space
Total number of outcomes in the sample space
OBJECTIVE - DETERMINED THROUGH EXPERIMENTATION
CLASSICAL - EVENTS WITH EQUALLY LIKELY OUTCOMES
HISTORICAL- RELATIVE FREQUENCY DISTRIBUTION
SUBJECTIVE - PERSONAL ESTIMATE OF THE LIKELIHOOD OF AN
EVENT
Relative Frequency probability that an event will be the result
of a random experiment is assigned as the proportion of times
that event occurs as the outcome of the experiment in the long
run.
INCOME/DAY
0 < 100
100 < 200
200 < 300
300 < 400
400 < 500
REL. FREQ.
.1
.2
.4
.2
.1
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SET - IS A WELL DEFINED COLLECTION OF OBJECTS
Outcome - A particular result of an experiment Pg 141
SAMPLE SPACE - IS A SET OF ALL POSSIBLE OUTCOMES
SAMPLE SPACE: COIN FLIP, DIE, DICE, DECK OF CARDS,
HAND
POKER
Table #2 Pg 179 JEP Summer 1999
COUNTING A SAMPLE SPACE ROLL A TWO OR A THREE
SAMPLE SPACE IS SIX
ROLL A TWO AND A THREE ON TWO CONSECUTIVE ROLLS
SAMPLE SPACE IS 36 = 6 X 6
EVENT - A collection of one or more outcomes of an
experiment
COMPLEMENT - ALL OUTCOMES THAT ARE NOT PART OF THE EVENT
MUTUALLY EXCLUSIVE - The occurrence of any one event means
that none of the others can occur at the same time. EVENTS THAT
CAN NOT OCCUR SIMULTANEOUSLY - Pg 142
COLLECTIVELY EXHAUSTIVE - At least one of the events must occur
when an experiment is conducted. NO OTHER EVENTS ARE POSSIBLE
P(COLLECTIVELY EXHAUSTIVE EVENTS) = 1
COUNTING TECHNIQUES
Value can only Value can be chosen
be chosen 1
multiply times
Order is
Important
COMBINATION
X
PERMUTATION
X
X
Multiplication Rule*
X
X
*Value can be chosen a multiple number of times because there
are a multiple number of groups and value can be chosen from
each group.
Page 165-170
𝐶𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛 = 𝑛𝐶𝑟 =
𝑛!
𝑟! (𝑛 − 𝑟)!
Royal Flush
𝑃𝑒𝑟𝑚𝑢𝑡𝑎𝑡𝑖𝑜𝑛 = 𝑛𝑃𝑟 =
𝑛!
(𝑛 − 𝑟)!
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Trifecta
N = TOTAL NUMBER OF OBJECTS
X = TOTAL NUMBER USED OUT OF N OBJECTS
Multiplication Rule for the number of possibilities between
two or more groups. If there are m ways of doing one thing(in
group #1) and n ways of doing another thing(In group #2), there
are m x n ways between both groups.
[By allowing people to make multiple choices for the first time
among six racial categories(White, Black, Asian, American Indian
or Alaska Native, Native Hawaiian or Pacifi Islander, or some
other) the census offered a mosaic of 63 racial options. Someone
might check both “white” and “Asian,” for example, if they have
ancestors of both races. Are 63 racial options correct?] WSJ
3/2/01
RULES FOR PROBABILITY
ADDITION RULE P(A or B) = P(A) + P(B) - P(A and B)
1. EVENTS THAT ARE MUTUALLY EXCLUSIVE -Specific
P(A OR B) = P(A)+P(B) - Special Rule Page 158 5-2
P(HEART OR SPADE) = P(H) + P(S) = 1/4 + 1/4 = ½
PAGE 150 TEXT
2.
EVENTS THAT ARE NOT MUTUALLY EXCLUSIVE-General Rule
P(A OR B)=P(A)+P(B)-P(A&B) 5-4
P(HEART OR JACK) = 13/52 + 4/52 - 1/52
P(A)=.275 P(B)=.275 Determine the probability of A or B
Company "A" has 200 employees: 55 have accounting degrees
and 55 have business degrees while 10 individuals have both
business and accounting degrees.
1. How many employees have college degrees?
2. Determine the probability of picking someone at random
from company “A” and this person having a college
degree.
Multiplication rule for probability P(A and B)=P(A).P(B|A)
Pg 156
Joint probability is the chance that two events will occur
together or simultaneously. Savings and checking account
Key words that are used to indicate joint probability are:
and, both and neither.
Independent events - if the occurrence of one is
unrelated
to the occurrence of the other. P(A AND B) = P(A) X P(B)
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This formula applies if the events occur simultaneously or
sequentially.
If the events are not independent they are dependent or have
conditional probability. P(A AND B) = P(A) X P(B|A)
[Problem 28]
P(A|B) = P(A and B)
P(B)
P(ACE) = 4/52
B REPRESENTS THE NEW SAMPLE SPACE
P(ACE|DIAMOND) = 1/13 INDEPENDENT EVENTS
MANUFACTURING FIRM BUYS 80% OF A GIVEN INPUT USED IN PRODUCTION
FROM COMPANY K AND 4% ARE DEFECTIVE. IT ALSO BUYS 20% FROM
COMPANY L AND 6% ARE DEFECTIVE.
P(K) = .8 P(L) = .2 P(D|K) = .04 P(D}L) = .06
1. DETERMINE P(K&D) = P(K) X P(D|K) = .8 X .04 = .032
P(D&K) = P(D) X P(K|D) =
Good Parts
DEF Parts
Total Parts
INPUTS
956
44
1000
K
768
32
800
L
188
12
200
[Problem 29]
Police records reveal that 10% of the accident victims who are
Wearing seat belts sustain serious injury, while 50% of those
who are not wearing seat belts sustain serious injury. Police
estimate that 60% of the people riding in cars use seat belts.
Police are called to investigate an accident in which one person
is seriously injured. Estimate the probability that he was
wearing his seat belt at the time of the crash.
LISKA'S FIVE STEP METHOD TO SUCCESS IN SOLVING CONDITIONAL
PROBABILITY.
STEP 1: Skip to the end of the problem and determine what
conditional probability the problem is asking for and
write this in probability language. Example P(a|b)
STEP 2: Use formula 5-7(5-6) in the text. The numerator of the
formula is always the joint probability of the two
events a and b. the denominator is the probability of
the given information P(b).
STEP 3: Draw the outcomes tree for this problem. The first
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branch Of the outcomes tree will start with event a. the
other branches will be the complement of event a. the
key will be how you have written the question in step
#1. If you have written the question as p(a|b), the
first branch of your tree is event a.
STEP 4: Now read the problem and identify the probabilities of
the events listed on your outcomes tree and place them
on the tree.
STEP 5: Place the appropriate joint probability of a and b in
the numerator and the probability of event b in the
denominator. The probability of event b can be found by
adding the joint probabilities of all the branches of
your outcomes tree that lead to event b.
USING THE EXAMPLE ABOVE
STEP 1: Estimate the probability that the injured person was
wearing his seat belt at the time of the crash. P(S|I)
P(S AND I)
STEP 2: P(S|I) = ----------- FORMULA 5-6 or 5-7
P(I)
OUTCOMES TREE
JOINT PROB
P(I|S)=.1
I -------- .06
STEP 3:
S
P(B)=.6
AUTO ACCIDENT
P(I'|S)=.9
I'-------I -------P(I|S')=.5
S'
P(B)=.4
.54
.20
10
P(I'|S')=.5
I'-------S
S'
I
I'
=
=
=
=
.20
SEAT BELT
NO SEAT BELT
SERIOUS INJURY
NO SERIOUS INJURY
STEP 4: SEE ABOVE OUTCOMES TREE
.O6
STEP 5: P(B|I) = -------------= .23
.26
Exponent 2/2/06, 11/6/08 and 2/12/09
9/9/09
Medicine's Dangerous Guessing Game WSJ A19
LISKA'S FIVE STEP METHOD TO SUCCESS IN SOLVING CONDITIONAL
PROBABILITY USING THE MATRIX APPROACH.
STEP 1: Skip to the end of the problem and determine what
conditional probability the problem is asking for and
write this in probability language. Example P(a|b)
STEP 2: Use formula 5-7 in the text. The numerator of the
formula is always the joint probability of the two
events a and b. the denominator is the probability of
the given information P(b).
STEP 3: Construct an outcomes tree and begin with the main
event(a). Form a probability matrix this event should be
the event you are trying to find the probability for with
the new information(the person has been injured). List
the other events that can occur in this sample space and
their prior probability. List the conditional probability
of P(b|a). List the probability of being injured given
the seat belt was on and the probability of being injured
knowing the seat belt was not on. List the joint
probability of the two events P(a and b) and P(a'and b).
Outcomes tree see above notes.
MAIN EVENT
SEAT BELT
PRIOR PROB
.6
COND. PROB
INJURY P(I|S)
.1
JOINT PROB.
.06 SEAT BELT AND
INJURY
11
NO
BELT
.4
.5
PROBABILITY OF INJURY P(I)
=
.20
.26
STEP 4: Now read the problem and identify the probabilities of
the events listed on your outcomes tree and place
them in the matrix as shown.
STEP 5: Place the appropriate joint probability of a and b in
the numerator and the probability of event b in the
denominator. The probability of event b can be found by
adding the joint probabilities.
P(S|I) =
P(S AND I)
--------- =
P(I)
.06
----- = .23
.26
ANOTHER APPROACH IS TO CONSTRUCT A PROBABILITY TABLE OF THE
EVENTS
SHOWN BELOW:*
SEAT BELT
NO
SEAT BELT
JOINT PROBABILITY
Injury
No Injury
.06
.54
Probability
.6
.4
.20
.26
* PLEASE NOTE THIS TABLE COMES FROM
BRANCHES OF THE ABOVE OUTCOMES TREE.
.20
.74
THE JOINT
PROBABILITY
Computations of joint probability are shown on the next page
P(S AND I) = P(S) X P(I|S)
= .6 X .1 = .06
P(S AND I') = P(S) X P(I'|S)
= .6 X .9 = .54
P(S' AND I) = P(S') X P(I|S')
= .4 X .5 = .20
P(S' AND I') = P(S') X P(I'|S') = .4 X .5 = .20
P(S/I) =
P(S AND I)
---------P(I)
=
.06
---.26
=
.23