Transcript Pooled Variance t Test - College of Business and Economics

Pooled Variance
t Test
•
•
•
Tests means of 2 independent populations having
equal variances
Parametric test procedure
Assumptions
–
–
–
Both populations are normally distributed
If not normal, can be approximated by normal distribution
(n1  30 & n2  30 )
Population variances are unknown but assumed equal
Two Independent Populations
Examples
•
•
An economist wishes to determine whether
there is a difference in mean family income
for households in 2 socioeconomic groups.
An admissions officer of a small liberal arts
college wants to compare the mean SAT
scores of applicants educated in rural high
schools & in urban high schools.
Pooled Variance t Test Example
You’re a financial analyst for Charles Schwab. You want
to see if there a difference in dividend yield between stocks
listed on the NYSE & NASDAQ.
NYSE
NASDAQ
Number
21
25
Mean
3.27
2.53
Std Dev
1.30
1.16
Assuming equal variances, is
there a difference in average
yield (a = .05)?
© 1984-1994 T/Maker Co.
Pooled Variance t Test
Solution
H0: m1 - m2 = 0 (m1 = m2)
H1: m1 - m2  0 (m1  m2)
a = .05
df = 21 + 25 - 2 = 44
Critical Value(s):
R
R eject
eject H
H00
R
R eject
eject H
H00
.025
.025
-2
-2.01
.0154
54 00 2.01
2.0154
54
t
Test Statistic:
t=
3.27  2.53
1 
 1
1.510   

 21 25 
=  2.03
Decision:
Reject at a = .05
Conclusion:
There is evidence of a
difference in means
Test Statistic
Solution
X  X h a
m  m f a
c
3 .2 7  2 .5 3 f  af
0
t =
=
=  2 .0 3
1
1 I
F
F
1
1 I
1.5 1 0 

S
G 
J
H2 1 2 5 K
Hn n K
11
22
11
22
22
P
P
11
S PP
22
n
a
=
22
f  an  1f S
an  1f  an  1f
2 1  1f  a
1.3 0 f  a
2 5  1f  a
1.1 6 f
a
=
a2 1  1f  a2 5  1f
11
1
22
 S 11
22
22
11
22
22
22
22
= 1.5 1 0
Pooled Variance t Test
Thinking Challenge
You’re a research analyst for General Motors. Assuming
equal variances, is there a difference in the average miles
per gallon (mpg) of two car models (a = .05)?
You collect the following:
Sedan
Van
Number
15
11
Mean
22.00
20.27
Std Dev
4.77
3.64
Alone
Group Class
Test Statistic
Solution*
X  X h a
m  m f a
c
2 2 .0 0  2 0 .2 7 f  af
0
t =
=
=  1.0 0
1
1I
F
F
1
1 I
1 8 .7 9 3 

S
G 
J
H1 5 1 1K
Hn n K
11
22
11
22
22
P
P
11
S PP
22
n
a
=
22
f  an  1f S
an  1f  an  1f
1 5  1f  a
4 .7 7 f  a
1 1  1f  a
3 .6 4 f
a
=
a1 5  1f  a1 1  1f
11
1
22
 S 11
22
22
11
22
22
22
22
= 1 8 .7 9 3
One-Way ANOVA F-Test
2 & c-Sample Tests with
Numerical Data
22 &
&C
C-S
-Saam
mpple
le
TTeessts
ts
M
Meeaann
22
PPoo
oole
ledd
VVaaria
riannccee
tt Te
Tesstt
##
SSaam
mple
pless
VVaaria
riannccee
C
C
O
Onnee-W
-Waayy
A
AN
NO
OVVA
A
M
Meeddia
iann
FF TTeesstt
(2
(2 SSaam
mpple
less))
22
W
Wilc
ilcooxxon
on
R
Raank
nk SSum
um
TTeesstt
##
SSaam
mple
pless
C
C
K
Kru
russkkaal-lW
Waallis
llis R
Raank
nk
TTeesstt
Experiment
•
Investigator controls one or more independent
variables
–
–
•
Observes effect on dependent variable
–
•
Called treatment variables or factors
Contain two or more levels (subcategories)
Response to levels of independent variable
Experimental design: Plan used to test
hypotheses
Completely Randomized Design
•
Experimental units (subjects) are assigned
randomly to treatments
–
•
One factor or independent variable
–
•
Subjects are assumed homogeneous
2 or more treatment levels or classifications
Analyzed by:
–
–
One-Way ANOVA
Kruskal-Wallis rank test
Randomized Design Example
FFaaccto
torr le
leve
vels
ls
(T
(Tre
reaatm
tm eennts
ts))
E
Exxppeerim
rim eennta
tall
uunnits
its
FFaaccto
torr (T
(Tra
rain
inin
ingg M
M eeth
thoodd))
LLeeve
LLeeve
LLeeve
vell 11
vell 22
vell 33
KKK KKK KKK
D
Deeppeennddeenntt
va
varia
riabble
le
2211 hhrs
rs..
1177 hhrs
rs..
3311 hhrs
rs..
2277 hhrs
rs..
2255 hhrs
rs..
2288 hhrs
rs..
(R
(Reessppoonnssee))
2299 hhrs
rs..
2200 hhrs
rs..
2222 hhrs
rs..
One-Way ANOVA
F-Test
•
•
Tests the equality of 2 or more (c) population
means
Variables
–
One nominal scaled independent variable
•
–
•
2 or more (c) treatment levels or classifications
One interval or ratio scaled dependent variable
Used to analyze completely randomized
experimental designs
One-Way ANOVA
F-Test Assumptions
•
Randomness & independence of errors
–
•
Normality
–
•
Independent random samples are drawn
Populations are normally distributed
Homogeneity of variance
–
Populations have equal variances
One-Way ANOVA
F-Test Hypotheses
•
H0: m1 = m2 = m3 = ... = mc
–
–
•
All population means are
equal
No treatment effect
f(X )
H1: Not all mj are equal
–
–

At least 1 population mean is
different
Treatment effect
m1  m2  ...  mc is wrong
m1 = m2 = m3
X
m1 = m2
X
f(X )
m3
One-Way ANOVA
Basic Idea
•
•
•
•
Compares 2 types of variation to test equality
of means
Ratio of variances is comparison basis
If treatment variation is significantly greater
than random variation then means are not
equal
Variation measures are obtained by
‘partitioning’ total variation
ANOVA Partitions Total
Variation
Total variation
Variation due to
treatment
Sum of squares among
 Sum of squares between
 Sum of squares model
 Among groups variation

Variation due to
random sampling
Sum of squares within
 Sum of squares error
 Within groups variation

Total Variation
e
S S T = X 1111  X
22
j  eX
2211
 X
22
j    eX
n
ncc cc
R
R eessp
po
on
nssee,, X
X
``X
G
G ro
rou
up
p 11
G
G ro
rou
up
p 22
G
G ro
rou
up
p 33
 X
22
j
Among-Groups Variation
e
S S A = n 11 X 11  X
22
j
e
 n 22 X 22  X
22
j    n eX
cc
cc
R
R eessp
po
on
nssee,, X
X
`X3
``X
`X1
G
G ro
rou
up
p 11
`X2
G
G ro
rou
up
p 22
G
G ro
rou
up
p 33
 X
22
j
Within-Groups Variation
c
S S W = X 1111  X 11
22
h  cX
2211
 X 11
22
h   cX
n
ncc cc
R
R eessp
po
on
nssee,, X
X
`X3
`X1
G
G ro
rou
up
p 11
G
G ro
rou
up
p 22
`X2
G
G ro
rou
up
p 33
 X cc
22
h
One-Way ANOVA
Test Statistic
•
Test statistic
– F = MSA / MSW
•
•
•
MSA is Mean Square Among
MSW is Mean Square Within
Degrees of freedom
– df1 = c -1
– df2 = n - c
•
•
c = # Columns (populations, groups, or levels)
n = Total sample size
One-Way ANOVA
Summary Table
D
S o u rc e
D eeg re e s
o
o
off
off
V a ria tio n F
ree d o m
Fre
Sum of
S q u a re s
M
M eea n
S q u a re
(V a ria n c e )
F
F
A
mo n g
Am
(F
(Faac to r)
cc - 1
S
A
SA
SS
M
M SA =
A
SA
MS
S
/(c - 1 ) M S W
A/(c
SA
SS
W
in
ithin
W ith
(E
r)
rror)
(E rro
n -c
S
W
SW
SS
M
W ==
SW
MS
S
/(n -- cc)
W /(n
SW
SS
T
tal
ota
To
n -1
S
T ==
ST
SS
S
W
SW
SS
A++S
SA
SS
One-Way ANOVA
Critical Value
If means are equal,
F = MSA / MSW  1.
Only reject large F!
R
R eeje
jecctt H
H 00
D
Do
oN
No
ott
R
R eeje
jecctt H
H0
a
0
F
F
00
F UU ((aa ;; cc 11,, nn  cc ))
Always One-Tail!
© 1984-1994 T/Maker Co.
One-Way ANOVA
F-Test Example
As production manager, you
want to see if 3 filling
machines have different mean
filling times. You assign 15
similarly trained &
experienced workers,
5 per machine, to the
machines. At the .05 level, is
there a difference in mean
filling times?
Mach1Mach2Mach3
25.40 23.40 20.00
26.31 21.80 22.20
24.10 23.50 19.75
23.74 22.75 20.60
25.10 21.60 20.40
One-Way ANOVA
F-Test Solution
H0: m1 = m2 = m3
H1: Not all equal
a = .05
df1 = 2 df2 = 12
Critical Value(s):
Test Statistic:
F =
a = .05
0
3 .8 9
F
MSA
MSW
=
2 3 .5 8 2 0
.9 2 1 1
= 2 5 .6
Decision:
Reject at a = .05
Conclusion:
There is evidence pop.
means are different
Summary Table
Solution
S o u rc e o f D e g re e s o f S u m o f
V a ria tio n
F re e d o m S q u a re s
Am ong
(M a c h in e s )
3 -1 = 2
4 7 .1 6 4 0
W ith in
(E rro r)
1 5 - 3 = 1 2 1 1 .0 5 3 2
T o ta l
1 5 - 1 = 1 4 5 8 .2 1 7 2
M ean
S q u a re
(V a ria n c e )
F
2 3 .5 8 2 0
2 5 .6 0
.9 2 1 1
Summary Table
Excel Output
One-Way ANOVA Thinking
Challenge
You’re a trainer for Microsoft
Corp. Is there a difference
in mean learning times of
12 people using 4 different
training methods (a =.05)?
M1
10
9
5
M2 M3 M4
11 13 18
16
8 23
9
9 25
Alone
© 1984-1994 T/Maker Co.
Group Class
One-Way ANOVA Solution*
H0: m1 = m2 = m3 = m4
H1: Not all equal
a = .05
df1 = 3 df2 = 8
Critical Value(s):
a = .05
0
4 .0 7
F
Test Statistic:
F =
MSA
MSW
=
116
10
= 1 1.6
Decision:
Reject at a = .05
Conclusion:
There is evidence pop.
means are different
Summary Table
Solution*
S o u rc e o f
V a ria tio n
Am ong
(M e th o d s )
W ith in
(E rro r)
T o ta l
D e g re e s o f S u m o f
F re e d o m
S q u a re s
M ean
S q u a re
(V a ria n c e )
F
1 1 .6
4 -1 = 3
348
116
12 - 4 = 8
80
10
12 - 1 = 11
428