Lecture 6 - MGCR 271 Business
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Unformatted text preview: MGCR 271 Business
Sta+s+cs
 Make
Inference
 Ramnath
Vaidyanathan
 Two Way Tables Chapter 9 Response Variable Explanatory Variable 1 sample None 2 samples Independent >2 samples Pooled 2-prop. z-test, Pooled Two Way Tables Chi-Square Test 7 Unpooled 2-prop. z-test, Unpooled 6 1-prop. z-test 5 Categ. Categ. 8 11
 Quant. Logistic Regression Motivation Amy is a wine and music enthusiast. She felt that the wine people purchased by a customer would depend on the music that he/she was being exposed to at the time of the purchase. She wants to test her claim and hires you. What advice would you provide her? What kind of data should Amy collect? How would you go about testing her claim? Is
the
type
of
wine
purchased
in
a
 supermarket
driven
by
the
music
played?
 Raw Data Two Way Table Subject 1 2 3 ... 147 Music French None French ... None Wine Other French French ... Other Categorical Var. 1
 Wine Wine French French French French Other Other 39 39 35 35 Music Music None Total None Total 43 30 30 69 78 147 Total Total 74 74 73 43 73 69 78 147 Categorical Var. 2
 Wine French Total Wine
 French
 How
can
you
test
her
Claim?
 Other p ˆ 74 p1 ˆ 30
 French French None 39Total 30 None 69Oth 147 Fre 43 78 39 Other 30 35 69 Wine Music 73 Wine French None 35 Total 43 74 78French 147 3 ... ... .. 73 p ˆ Music
 French 39 147Other Other Total 35 74 Total 30 43 73 3 7 p1 ˆ 







=
0.52
 39
 (39/74)
 p2 ˆ p2 ˆ 







=
0.41
 69
 (30/73)
 p ˆ p1




=
47%
 ˆ p2 ˆ (69/147)
 p ˆ H 0 : p1 − p 2 = 0 p1 ˆ p2 ˆ Other
 35
 43
 78
 Ha : p1 − p2 = 0 French
 None
 Total
 74
 73
 147
 7 Two-Proportion pz-Test, Pooled p −ˆ ˆ 1 2 z= q p1 (1−p1 ) ˆ ˆ p (1−p ) ˆ ˆ + 2n 2 n1 2 (p1 −p2 )−d0 ˆˆ Fail to Reject H0 0.3 population H0 : µ1 − µ2 = d0 0.2 Ha : µ1 − µ2 = d0 p 2 0.1 x1 − x2 ¯ ¯ α 2 !3 0.0 !! !! ! !!! !! !!! !! !! !!! !! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !! ! !! ! !! ! !! !! ! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! ! !! ! !! !! !! !! !!! !!! !!! !!! !!! !!! !!! !!! !!! !!!! !!!! !!!! !!! !!! !!! !!! !!!!! !! !!! !!!!!!! !!! !!!!!!! !!! !!!!!!! !!!!!! !! !!!!!!! ! !!!!!!! ! !!!!!!!! !!!!!!!! N (0, 1) Fail to RejectFail to Reject H0 H0 ( ˆ1 − H0 µ1(1− ˆˆ2 = −0 µ p d dˆ z = :q pp− pp2 )−(10 p) ˆ ) ˆ H 0 : p1 − p 2 = d 0 Ha : p1 − p2 = d 0 α 2 Ha :nµp1−nµp2= d0 1 1 ˆ + 22 p = n1 +n2 ˆ ˆ x1 − x2 ¯ ¯ ! ! group group a b b c c ! n1 + n2 z∗ p 2 −z !2 ∗ !1 0 1 x H 0 : p1 − p 2 = d 0 H :p −p =d ( ˆ −p 0 ˆ ∗zz = q (p1 −pˆ2 )−d0 =3 q p1 (1p1p1ˆ2 )−2d(1−p2 ) z ˆ ) ˆ p(1−p)+ pp(1−ˆ ) ˆ −ˆ ˆ ˆ p ˆ n1 + n2 n1 n2 ∗ 2 −z p1 − pn1 p1 +n2 p2 ˆ ˆ2 ˆ p = n1 +n2 ˆ ˆ area under the standard normal curve to the left of z. Moore-212007 pbs November 20, 2007 13:52 z TABLE A Standard normal probabilities z −3.4 −3.3 −3.2 −3.1 −3.0 −2.9 −2.8 −2.7 −2.6 −2.5 −2.4 −2.3 −2.2 −2.1 −2.0 −1.9 −1.8 −1.7 −1.6 −1.5 −1.4 −1.3 −1.2 −1.1 −1.0 −0.9 −0.8 −0.7 −0.6 −0.5 −0.4 −0.3 −0.2 −0.1 −0.0 .00 .0003 .0005 .0007 .0010 .0013 .0019 .0026 .0035 .0047 .0062 .0082 .0107 .0139 .0179 .0228 .0287 .0359 .0446 .0548 .0668 .0808 .0968 .1151 .1357 .1587 .1841 .2119 .2420 .2743 .3085 .3446 .3821 .4207 .4602 .5000 .01 .0003 .0005 .0007 .0009 .0013 .0018 .0025 .0034 .0045 .0060 .0080 .0104 .0136 .0174 .0222 .0281 .0351 .0436 .0537 .0655 .0793 .0951 .1131 .1335 .1562 .1814 .2090 .2389 .2709 .3050 .3409 .3783 .4168 .4562 .4960 .02 .0003 .0005 .0006 .0009 .0013 .0018 .0024 .0033 .0044 .0059 .0078 .0102 .0132 .0170 .0217 .0274 .0344 .0427 .0526 .0643 .0778 .0934 .1112 .1314 .1539 .1788 .2061 .2358 .2676 .3015 .3372 .3745 .4129 .4522 .4920 .03 .0003 .0004 .0006 .0009 .0012 .0017 .0023 .0032 .0043 .0057 .0075 .0099 .0129 .0166 .0212 .0268 .0336 .0418 .0516 .0630 .0764 .0918 .1093 .1292 .1515 .1762 .2033 .2327 .2643 .2981 .3336 .3707 .4090 .4483 .4880 T-2 TABLES ............................................................................................................................... ........................................ .04 .0003 .0004 .0006 .0008 .0012 .0016 .0023 .0031 .0041 .0055 .0073 .0096 .0125 .0162 .0207 .0262 .0329 .0409 .0505 .0618 .0749 .0901 .1075 .1271 .1492 .1736 .2005 .2296 .2611 .2946 .3300 .3669 .4052 .4443 .4840 .05 .0003 .0004 .0006 .0008 .0011 .0016 .0022 .0030 .0040 .0054 .0071 .0094 .0122 .0158 .0202 .0256 .0322 .0401 .0495 .0606 .0735 .0885 .1056 .1251 .1469 .1711 .1977 .2266 .2578 .2912 .3264 .3632 .4013 .4404 .4801 .06 .0003 .0004 .0006 .0008 .0011 .0015 .0021 .0029 .0039 .0052 .0069 .0091 .0119 .0154 .0197 .0250 .0314 .0392 .0485 .0594 .0721 .0869 .1038 .1230 .1446 .1685 .1949 .2236 .2546 .2877 .3228 .3594 .3974 .4364 .4761 .07 .0003 .0004 .0005 .0008 .0011 .0015 .0021 .0028 .0038 .0051 .0068 .0089 .0116 .0150 .0192 .0244 .0307 .0384 .0475 .0582 .0708 .0853 .1020 .1210 .1423 .1660 .1922 .2206 .2514 .2843 .3192 .3557 .3936 .4325 .4721 .08 .09 ............................................................................................................................... ........................................ .0003 .0002 .0004 .0003 .0005 .0005 Probability .0007 .0007 .0010 .0010 Table .0014 .0014 entry for z is the area .0020 under the .0019 standard normal curve .0027 .0026 to the left of z. .0037 .0036 z .0049 .0048 .0066 .0064 TABLE A Standard normal probabilities ............................................................................................................................... ........................................ .0087 .0084 z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 .0113 .0110 −3.4 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0002 .0146 .0143 −3.3 .0005 .0005 .0005 .0004 .0004 .0004 .0004 .0004 .0004 .0003 .0188 .0183 −3.2 .0007 .0007 .0006 .0006 .0006 .0006 .0006 .0005 .0005 .0005 .0239 .0233 −3.1 .0010 .0009 .0009 .0009 .0008 .0008 .0008 .0008 .0007 .0007 .0301 .0294 −3.0 .0013 .0013 .0013 .0012 .0012 .0011 .0011 .0011 .0010 .0010 .0375 .0367 −2.9 .0019 .0018 .0018 .0017 .0016 .0016 .0015 .0015 .0014 .0014 .0465 .0455 −2.8 .0026 .0025 .0024 .0023 .0023 .0022 .0021 .0021 .0020 .0019 .0571 .0559 −2.7 .0035 .0034 .0033 .0032 .0031 .0030 .0029 .0028 .0027 .0026 .0694 .0681 −2.6 .0047 .0045 .0044 .0043 .0041 .0040 .0039 .0038 .0037 .0036 −2.5 .0062 .0060 .0059 .0057 .0055 .0054 .0052 .0051 .0049 .0048 .0838 .0823 −2.4 .0985 .0082 .0080 .0078 .0075 .0073 .0071 .0069 .0068 .0066 .0064 .1003 −2.3 .1170 .0107 .0104 .0102 .0099 .0096 .0094 .0091 .0089 .0087 .0084 .1190 −2.2 .0139 .0136 .0132 .0129 .0125 .0122 .0119 .0116 .0113 .0110 .1401 .1379 −2.1 .0179 .0174 .0170 .0166 .0162 .0158 .0154 .0150 .0146 .0143 .1635 .1611 −2.0 .0228 .0222 .0217 .0212 .0207 .0202 .0197 .0192 .0188 .0183 .1894 .1867 −1.9 .0287 .0281 .0274 .0268 .0262 .0256 .0250 .0244 .0239 .0233 .2177 .2148 −1.8 .0359 .0351 .0344 .0336 .0329 .0322 .0314 .0307 .0301 .0294 .2483 .2451 −1.7 .0446 .0436 .0427 .0418 .0409 .0401 .0392 .0384 .0375 .0367 −1.6 .0548 .0537 .0526 .0516 .0505 .0495 .0485 .0475 .0465 .0455 .2810 .2776 −1.5 .0668 .0655 .0643 .0630 .0618 .0606 .0594 .0582 .0571 .0559 .3156 .3121 −1.4 .3483 .0808 .0793 .0778 .0764 .0749 .0735 .0721 .0708 .0694 .0681 .3520 −1.3 .0968 .0951 .0934 .0918 .0901 .0885 .0869 .0853 .0838 .0823 .3897 .3859 −1.2 .1151 .1131 .1112 .1093 .1075 .1056 .1038 .1020 .1003 .0985 .4286 .4247 −1.1 .1357 .1335 .1314 .1292 .1271 .1251 .1230 .1210 .1190 .1170 .4681 .4641 −1.0 .1587 .1562 .1539 .1515 .1492 .1469 .1446 .1423 .1401 .1379 −0.9 −0.8 −0.7 −0.6 .1841 .2119 .2420 .2743 .1814 .2090 .2389 .2709 .1788 .2061 .2358 .2676 .1762 .2033 .2327 .2643 .1736 .2005 .2296 .2611 .1711 .1977 .2266 .2578 .1685 .1949 .2236 .2546 .1660 .1922 .2206 .2514 .1635 .1894 .2177 .2483 .1611 .1867 .2148 .2451 How
would
you
test
if
there
were
 three
types
of
music
played?
 What
if
there
were
three
types
of
wine
 and
three
types
of
music?
 Music
 41%
x
75
 Wine
 French
 Italian
 Other
 39
 41%
x
84
 41%
x
84
 % of People Buying each Wine Type 30.5
 9.6
 34.9
 30
 34.2
 10.7
 39.1
 30
 34.2
 10.7
 39.1
 99
 41%
 (99/243)
 13%
 
(31/243)
 46%
 (113/243)
 1
 19
 11
 31
 35
 35
 43
 113
 46%
x
75
 French
 Italian
 None
 Total
 75
 84
 84
 243
 46%
x
84
 Compute
Test
Score
(Chi‐Square)
 The
chi‐square
sta+s+c
(χ2)
is
a
measure
of
how
much
the
observed
 cell
counts
in
a
two‐way
table
diverge
from
the
expected
cell
counts.
 The
formula
for
the
χ2
sta+s+c
is
(summed
over
all
r
*
c
cells
in
the
 table)
 Large
values
for
χ2
represent
strong
devia+ons
from
the
expected
distribu+on
 under
the
 H0,
providing
evidence
against
 H0.
However,
since
 χ2
is
a
sum,
how
 large
a
 χ2
is
required
for
sta+s+cal
significance
will
depend
on
the
number
of
 comparisons
made.
 
Compute
Test
Score
(Chi‐Square)
 H0:
No
relaHonship
between
music
and
wine






Ha:
Music
and
wine
are
related

 Observed
counts 
 
 






Expected
counts
 We
calculate
nine
X2
 components
and
sum
them
to
 produce
the
X2
sta+s+c:
 Iden+fy
Test
Sta+s+c:
χ2

Distribu+on The
χ2
distribu+ons
are
a
family
of
distribu+ons
that
can
take
only
posi+ve
 values,
are
skewed
to
the
right,
and
are
described
by
a
specific
degrees
of
 freedom.

 Table
F
gives
upper
cri+cal
 values
for
many
χ2
 distribu+ons.
 
Compute
p‐Value
 For
the
chi‐square
test,
H0
states
that
there
is
no
associa+on
between
the
row
 and
 column
 variables
 in
 a
 two‐way
 table.
 The
 alterna+ve
 is
 that
 these
 variables
are
related.
 If
H0
is
true,
the
chi‐square
test
has
approximately
a
χ2
distribuHon
with
(r
−
1) (c
−
1)
degrees
of
freedom.

 The
p‐value
for
the
chi‐square
test
is
 the
area
to
the
right
of
χ2
under
the
 χ2
distribu+on
with
df
(r−1)(c−1):

 P(χ2
≥
X2).
 χ2
Table
 df
=
(r−1)(c−1)
 Ex:
In
a
4x3
 table,
df
=
 3*2
=
6
 If
χ2
=
16.1,
 the
p‐value
 is
between
 0.01−0.02.
 
Compute
p‐Value
 H0:
No
rela+onship
between
music
and
wine






Ha:
Music
and
wine
are
related

 We
found
that
the
X2
sta+s+c
under
H0
is
18.28.

 The
two‐way
table
has
a
3x3
design
(3
levels
of
music
 and
3
levels
of
wine).Thus,
the
degrees
of
freedom
for
 the
X2
distribu+on
for
this
test
is:

 (r
–
1)(c
–
1)
=
(3
–
1)(3
–
1)
=
4

 
16.42
<
X2
=18.28
<
18.47
 
0.0025
>
p‐value
>
0.001


very
significant
 There
is
a
significant
rela+onship
between
the
type
of
music
played
and
wine
 purchases
in
supermarkets.
 8 Two
Way
Tables:
Chi‐Square
Test
 χ= p p ˆ p(39.5−30)2 2 2 =( χ (H0= pr χ2 (p2 .5−0r − 1)(c − 1)) df :−44− αdf c=α1)) (35−39.8) French 3 French (43 ( 1 − 1)( .5)39 α . . . 39.8 . . . ... 44.5(35−39.8)Oij −Eij )2 (Oij −r= )2 ( 2 2 χ2 Eij =r0 Eij 147 None Other χ = 2 Ha : p1r−ij 39.8 (43−44.5) E p2 c 2 44.5 Music (43 B
 2 (35.3−39)2 −44.5) p c Wine French None Total p 1 44.5 c 35 χ2 ( = − 69 Frenchdf 39 (r 30 1)(c − 1)) .3 2 1 2.5−30)2 (Oij −Eij ) Other 35 43 78 (39 α 2 (df = (r − 1)(c − 1)) χ α Eij Total 74 73 147 39.5 A
 2 None French (39.5−30) 1 French 39.5 Other 44.5 2 39.8 p44..3 35 5 χ (df = (r − 1)(c − 1)) (35−39.r 2 8)2 χ 39.8 p1 ˆ p2 ˆ H0 : A not related to B Ha : A is related to B (35.3−39) 35.3 2 r c = (Oij −Eij )2 Eij c (43−44.5)2 p 44.5 α 1 χ2 (df = (r − 1)(c − 1)) χ2 ∗ χ= 2 (Oij −Eij )2 Eij (39.5−30)2 39.5 (35−39.8)2 39.8 (43−44.5)2 44.5 χ2 (df = (r − 1)(c − 1)) χ2 = p (Oij −Eij )2 Eij Interpre+ng
Chi‐Square
Output
 41%
x
75
 Expected Counts French
 Italian
 Other
 39
 30.5
 9.6
 34.9
 30
 34.2
 10.7
 39.1
 30
 34.2
 10.7
 39.1
 99
 41%
 13%
 46%
 1
 19
 11
 31
 35
 35
 43
 113
 French
 Italian
 None
 Total
 75
 84
 84
 243
 Interpre+ng
Chi‐Square
Output
 The
 values
 summed
 to
 make
 up
 χ2
 are
 called
 the
 χ2
 components.
 When
the
test
is
sta+s+cally
significant,
the
 largest
 components
point
 to
the
condi+ons
most
different
from
the
expecta+ons
based
on
H0.
 Music
and
wine
purchase
decision
 X2
components
 Two
chi‐square
components
contribute
 most
to
the
X2
total

the
largest
effect
is
 for
sales
of
Italian
wine,
which
are
 strongly
affected

 by
Italian
and
French

 music.

 0.5209




2.3337






0.5209

 0.0075




7.6724






6.4038

 0.3971




0.0004






0.4223
 Actual
propor+ons
show
that
Italian
 music
helps
sales
of
Italian
wine,
but
 French
music
hinders
it.
 What
if
we
conducted
a
Chi‐Square
 Test
for
the
first
situa+on
with
two
 wines
and
two
types
of
music?
 Music
 47%
x
74
 Wine
 French
 39
 47%
x
73
 % of People Buying each Wine Type 34.7
 30
 34.3
 69
 47%
 (69/147)
 Other
 35
 39.3
 43
 38.7
 78
 53%
 (78/147)
 53%
x
74
 French
 None
 Total
 74
 73
 147
 53%
x
73
 Response Variable Explanatory Variable 1 sample None 2 samples Independent >2 samples Pooled 2-prop. z-test, Pooled Two Way Tables 7 Unpooled 2-prop. z-test, Unpooled 6 1-prop. z-test 5 Categ. Categ. 8 11
 Quant. Logistic Regression Chi‐Square
Test:
Summary
 The chi-square test is an overall technique for comparing any number of population proportions, testing for evidence of a relationship between two categorical variables. We can either:   Test for Independence: Take one random sample and classify the individuals in the sample according to two categorical variables (attribute or condition) observational study, historical design.   Compare Several Populations: Randomly select several random samples each from a different population (or from a population subjected to different treatments) experimental study. Both models use the chi-square test to test of the hypothesis of no relationship. Comparing
Several
Popula+ons:
χ2
Test
 Select
 independent
 SRSs
 from
 each
 of
 c
 popula+ons,
 of
 sizes
 n1,
 n2,
 .
 .
 .
 ,
 nc.
 Classify
 each
 individual
 in
 a
 sample
 according
 to
 a
 categorical
 response
 variable
 with
 r
 possible
 values.
 There
 are
 c
 different
 probability
 distribu+ons,
one
for
each
popula+on.
 The
 null
 hypothesis
 is
 that
 the
 distribu+ons
 of
 the
 response
 variable
 are
 the
 same
 in
 all
 c
 popula+ons.
 The
 alterna+ve
 hypothesis
 says
 that
 these
 c
 distribu+ons
 are
 not
all
the
same.
 When
is
it
safe
to
use
the
χ2
test?
  The samples are simple random samples (SRS).  All individual expected counts are 1 or more.  No more than 20% of expected counts are less than 5 For a 2x2 table, this implies that all four expected counts should be 5 or more. Marginal
and
Condi+onal
Distribu+ons
 Is
there
a
rela+onship
between
Age
 and
Level
of
Educa+on
aiained?
 Marginal
Distribu+ons
 Condi+onal
Distribu+ons
 Does
background
music
in
 supermarkets
influence
customer
 purchasing
decisions?
 Wine
purchased
for
each
kind
of
music
 played
(column
percents)
 Music
played
for
each
kind
of
wine
 purchased
(row
percents)

 A
Paradox!!
 Is
there
discrimina+on?
 How
would
you
test
this
Hypothesis?
 Is
there
discrimina+on
really….?
 Simpson’s
Paradox!!!!
 Slope = # successful / # unsuccessful = odds 33
 Slope = # successful / # unsuccessful = odds 34
 35
 36
 Simpson’s
Paradox
 An
associa+on
or
comparison
that
holds
for
all
of
several
groups
can
reverse
 direc+on
 when
 the
 data
 are
 combined
 (aggregated)
 to
 form
 a
 single
 group.
 This
reversal
is
called
Simpson’s
paradox.
 Example:
Hospital
death
 rates
 On
the
surface,
 Hospital
B
would
 seem
to
have
a
beier
 record.
 But
once
pa+ent

 condi+on
is
taken

 into
account,
we

 see
that
hospital
A

 has
in
fact
a
beier

 record
for
both
pa+ent
condi+ons
(good
and
poor).

 Here
pa+ent
condi+on
was
the
lurking
variable.
 Moore-212007 pbs November 20, 2007 13:52 T-20 Chi-Square Distribution Table TABLES df
=
(r−1)(c−1)
 Table entry for p is the ∗ critical value χ 2 with probability p lying to its right. Probability p (χ 2)* TABLE F χ 2 distribution critical values df 1 2 3 4 5 6 7 8 9 10 11 12 13 .25 1.32 2.77 4.11 5.39 6.63 7.84 9.04 10.22 11.39 12.55 13.70 14.85 15.98 .20 1.64 3.22 4.64 5.99 7.29 8.56 9.80 11.03 12.24 13.44 14.63 15.81 16.98 .15 2.07 3.79 5.32 6.74 8.12 9.45 10.75 12.03 13.29 14.53 15.77 16.99 18.20 ............................................................................................................................... ............................................................................................................................ Tail probability p .10 2.71 4.61 6.25 7.78 9.24 10.64 12.02 13.36 14.68 15.99 17.28 18.55 19.81 .05 3.84 5.99 7.81 9.49 11.07 12.59 14.07 15.51 16.92 18.31 19.68 21.03 22.36 .025 5.02 7.38 9.35 11.14 12.83 14.45 16.01 17.53 19.02 20.48 21.92 23.34 24.74 .02 5.41 7.82 9.84 11.67 13.39 15.03 16.62 18.17 19.68 21.16 22.62 24.05 25.47 .01 6.63 9.21 11.34 13.28 15.09 16.81 18.48 20.09 21.67 23.21 24.72 26.22 27.69 .005 7.88 10.60 12.84 14.86 16.75 18.55 20.28 21.95 23.59 25.19 26.76 28.30 29.82 .0025 9.14 11.98 14.32 16.42 18.39 20.25 22.04 23.77 25.46 27.11 28.73 30.32 31.88 .001 10.83 13.82 16.27 18.47 20.51 22.46 24.32 26.12 27.88 29.59 31.26 32.91 34.53 .0005 12.12 15.20 17.73 20.00 22.11 24.10 26.02 27.87 29.67 31.42 33.14 34.82 36.48 ...
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This note was uploaded on 08/31/2010 for the course MANAGEMENT MGCR 271 taught by Professor Vaidyanathan during the Summer '10 term at McGill.

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