Transcript Hypothesis Testing – Two Samples

Hypothesis Testing – Two
Samples
Chapters 12 & 13
Chapter Topics
 Comparing Two Independent Samples
 Independent samples Z test for the difference in two
means
 Pooled-variance t test for the difference in two means
 F Test for the Difference in Two Variances
 Comparing Two Related Samples
 Paired-sample Z test for the mean difference
 Paired-sample t test for the mean difference
 Two-sample Z test for population proportions
 Independent and Dependent Samples
Comparing Two Independent
Samples
 Different Data Sources
Unrelated
Independent
Sample selected from one population has no effect or
bearing on the sample selected from the other
population
 Use the Difference between 2 Sample Means
 Use Z Test or t Test for Independent Samples
Independent Sample Z Test
(Variances Known)
Assumptions
Samples are randomly and independently
drawn from normal distributions
Population variances are known
Test Statistic

Z
( X 1  X 2 )  ( 1    )

2

n1


2
n2
t Test for Independent Samples
(Variances Unknown)
Assumptions
Both populations are normally distributed
Samples are randomly and independently
drawn
Population variances are unknown but
assumed equal
If both populations are not normal, need large
sample sizes
Developing the t Test for
Independent Samples
Setting Up the Hypotheses
H0:  1 =  2
H1:  1   2
OR
H0:  1 - 2 = 0
H1:  1 -  2  0
Two
Tail
H0:  1   2
H1:  1 >  2
OR
H0:  1 -  2  0
H1:  1 -  2 > 0
Right
Tail
H0:  1   2
H1:  1 <  2
OR
H0:  1 -  2  
H1:  1 -  2 < 0
Left
Tail
Developing the t Test for
Independent Samples
(continued)
Calculate the Pooled Sample Variance as
an Estimate of the Common Population
Variance
(n1  1) S  (n2  1) S
S 
(n1  1)  (n2  1)
2
p
2
1
2
2
S p2 : Pooled sample variance
n1 : Size of sample 1
S12 : Variance of sample 1
n2 : Size of sample 2
S22 : Variance of sample 2
Developing the t Test for
Independent Samples
(continued)
Compute the Sample Statistic
X

t
df  n1  n2  2
S
2
p
1
 X 2    1   2 
1 1
S   
 n1 n2 
2
p
n1  1 S   n2  1 S


 n1  1   n2  1
2
1
Hypothesized
difference
2
2
t Test for Independent Samples:
Example
You’re a financial analyst for Charles Schwab. Is there a
difference in average dividend yield between stocks
listed on the NYSE & NASDAQ? You collect the
following data:
NYSE
NASDAQ
Number
21
25
Sample Mean
3.27
2.53
Sample Std Dev 1.30
1.16
Assuming equal variances, is
there a difference in average
yield (a = 0.05)?
© 1984-1994 T/Maker Co.
Calculating the Test Statistic
X

t
1
 X 2    1  2 
1 1
S   
 n1 n2 

2
p
 3.27  2.53  0
1 
 1
1.510   
 21 25 
2
2
n

1
S

n

1
S




1
1
2
2
2
Sp 
 n1  1   n2  1
21  11.302   25  11.162


 1.502
 21  1   25  1
 2.03
Solution
H0: 1 - 2 = 0 i.e. (1 = 2) Test Statistic:
3.27  2.53
H1: 1 - 2  0 i.e. (1  2) t 
 2.03
a = 0.05
1
1
1.502   
df = 21 + 25 - 2 = 44
 21 25 
Critical Value(s):
Decision:
Reject H0
Reject H0
Reject at a = 0.05.
.025
.025
Conclusion:
There is evidence of a
-2.0154 0 2.0154 t
difference in means.
2.03
p -Value Solution
(p-Value is between .02 and .05) < (a = 0.05)
Reject.
p-Value
is between .01 and .025
2
Reject
Reject
a

-2.0154
0
2.0154
2.03
=.025
Z
Test Statistic 2.03 is in the Reject Region
Example
You’re a financial analyst for Charles Schwab. You
collect the following data:
NYSE
NASDAQ
Number
21
25
Sample Mean
3.27
2.53
Sample Std Dev 1.30
1.16
You want to construct a 95%
confidence interval for the
difference in population average
yields of the stocks listed on
NYSE and NASDAQ.
© 1984-1994 T/Maker Co.
Example: Solution
2
2
n

1
S

n

1
S




1
2
2
S p2  1
 n1  1   n2  1
21  11.302   25  11.162


 1.502
 21  1   25  1
X
1
 X 2   ta / 2,n1  n2  2
1 1
S   
 n1 n2 
2
p
1 1
 3.27  2.53  2.0154 1.502   
 21 25 
0.0088  1  2  1.4712
Independent Sample (Two Sample) tTest in JMP
Independent Sample t Test with Variances
Known
Analyze | Fit Y by X | Measurement in Y box
(Continuous) | Grouping Variable in X box
(Nominal) |  | Means/Anova/Pooled t
Comparing Two Related Samples
Test the Means of Two Related Samples
Paired or matched
Repeated measures (before and after)
Use difference between pairs
Di  X1i  X 2i
Eliminates Variation between Subjects
Z Test for Mean Difference (Variance
Known)
Assumptions
Both populations are normally distributed
Observations are paired or matched
Variance known
Test Statistic

Z
D  D
D
n
n
D
D
i 1
n
i
t Test for Mean Difference (Variance
Unknown)
Assumptions
Both populations are normally distributed
Observations are matched or paired
Variance unknown
If population not normal, need large samples
Test Statistic
D  D
t
SD
n
n
D
 Di
i 1
n
n
SD 
 ( D  D)
i 1
i
n 1
2
Dependent-Sample t Test: Example
Assume you work in the finance department. Is the
new financial package faster (a=0.05 level)? You
collect the following processing times:
User Existing System (1)
C.B.
9.98 Seconds
T.F.
9.88
M.H.
9.84
R.K.
9.99
M.O.
9.94
D.S.
9.84
S.S.
9.86
C.T.
10.12
K.T.
9.90
S.Z.
9.91
New Software (2)
9.88 Seconds
9.86
9.75
9.80
9.87
9.84
9.87
9.98
9.83
9.86
Difference Di
.10
.02
.09
.19
.07
.00
- .01
.14
.07
.05
D

D
i
n
SD 
 .072
D  D
i
n 1
 .06215
2
Dependent-Sample t Test: Example
Solution
Is the new financial package faster (0.05 level)?
H0: D  
H1: D > 
Reject
a .5 D = .072
Critical Value=1.8331
df = n - 1 = 9
Test Statistic
D  D
.072  0
t

 3.66
SD / n .06215/ 10
a .5
1.8331
Decision: Reject H0
t
3.66
t Stat. in the rejection
zone.
Conclusion: The new
software package is faster.
Confidence Interval Estimate for  D
of Two Dependent Samples
Assumptions
Both populations are normally distributed
Observations are matched or paired
Variance is unknown
100 1  a  % Confidence Interval Estimate:
D  ta / 2,n 1
SD
n
Example
Assume you work in the finance department. You want
to construct a 95% confidence interval for the mean
difference in data entry time. You collect the following
processing times:
User Existing System (1)
C.B.
9.98 Seconds
T.F.
9.88
M.H.
9.84
R.K.
9.99
M.O.
9.94
D.S.
9.84
S.S.
9.86
C.T.
10.12
K.T.
9.90
S.Z.
9.91
New Software (2)
9.88 Seconds
9.86
9.75
9.80
9.87
9.84
9.87
9.98
9.83
9.86
Difference Di
.10
.02
.09
.19
.07
.00
- .01
.14
.07
.05
Solution:
D

D
i
n
 .072
SD 
ta / 2,n 1  t0.025,9
D
i
 D
n 1
 2.2622
SD
D  ta / 2,n 1
n
 .06215 
.072  2.2622 

 10 
0.0275 <  D < 0.1165
2
 .06215
F Test for Difference in Two
Population Variances
Test for the Difference in 2 Independent
Populations
Parametric Test Procedure
Assumptions

Both populations are normally distributed
Test is not robust to this violation
Samples are randomly and independently
drawn
The F Test Statistic
2
1
2
2
S
F
S
2
1 = Variance of Sample 1
S
n1 - 1 = degrees of freedom
S
2
2= Variance of Sample 2
n2 - 1 = degrees of freedom
0
F
Developing the F Test
Hypotheses

H0 :  1 2 =  2 2

H1 :  1 2   2 2
Reject H0
Reject H0
a/2
Test Statistic


0
F = S12 /S22
Two Sets of Degrees of Freedom
Do Not
Reject
FL
a/2
FU
F
 df1 = n1 - 1; df2 = n2 - 1

Critical Values: FL( n1 -1, n2 -1) and FU( n1 -1 , n2 -1)
FL = 1/FU*
(*degrees of freedom switched)
Developing the F Test
Easier Way

Reject H0
Put the largest in the num.
Do Not
Reject
Test Statistic

F = S12 /S22
0
a
F
F
F Test: An Example
Assume you are a financial analyst for Charles Schwab. You
want to compare dividend yields between stocks listed on the
NYSE & NASDAQ. You collect the following data:
NYSE
NASDAQ
Number
21
25
Mean
3.27
2.53
Std Dev
1.30
1.16
Is there a difference in the
variances between the NYSE
& NASDAQ at the a  0.05 level?
© 1984-1994 T/Maker Co.
F Test: Example Solution
 Finding the Critical
Values for a = .05

df1  n1  1  21  1  20
df 2  n2  1  25  1  24

F.05,20,24  2.03
F Test: Example Solution
Test Statistic:
H0 :  1 2 =  2 2
2
1
2
2
S
1.30
F

 1.25
2
S
1.16
H1 :  1 2   2 2
a  .05
df1  20 df2  24
Critical Value(s):
Decision:
Do not reject at a = 0.05.
Reject
.05
0
2.33
1.25
2
F
Conclusion:
There is insufficient
evidence to prove a
difference in variances.
F Test: One-Tail
H0: 12  22
H1: 12 < 22
or
a = .05
FL n1 1,n2 1 
Reject
H0: 12  22
H1: 12 > 22
1
FU  n2 1,n1 1
Reject
a  .05
a  .05
0
Degrees of
freedom
switched
F
FL n1 1,n2 1
0
FU  n1 1,n2 1
F
F Test: One-Tail
Easier Way

Reject H0
Put the largest in the num.
Do Not
Reject
Test Statistic

F = S12 /S22
0
a
F
F
Z Test for Differences in Two
Proportions (Independent Samples)
What is It Used For?
To determine whether there is a difference
between 2 population proportions and whether
one is larger than the other
Assumptions:
Independent samples
Population follows binomial distribution
Sample size large enough: np  5 and n(1-p)  5
for each population
Z Test Statistic
Z
 p
1
 p 2 
p Pd q Pd p Pd q Pd

n1
n2
p Pd
q Pd
Z
 p
1
n1 p 1  n2 p 2

n1  n2
n1q1  n2 q 2

n1  n2
 p 2    p1  p2 
p 1q1 p 2 q 2

n1
n2
The Hypotheses for the
Z Test
Research Questions
Hypothesi
s
No Difference Prop 1  Prop 2 Prop 1  Prop 2
Any Difference Prop 1 < Prop 2 Prop 1 > Prop 2
H0
p1 - p2  
H1
p1 - p 2  0
p1 - p2  0
p1 - p2 < 0
p1 - p2  0
p1 - p2 > 0
Z Test for Differences in Two
Proportions: Example
As personnel director, you
want to test the perception
of fairness of two methods
of performance evaluation.
63 of 78 employees rated
Method 1 as fair. 49 of 82
rated Method 2 as fair. At
the 0.01 significance level,
is there a difference in
perceptions?
63
 0.8077
78
49
p2 
 0.598
82
p 1 

np  5
n1  78
n2  82
n1  p  5
Calculating the Test Statistic
Z
 p
1
 p 2    p1  p2 
p Pd q Pd p Pd q Pd

n1
n2
 2.898
p Pd
q Pd
n1 p 1  n2 p 2

n1  n2
n1q1  n2 q 2

n1  n2
Z Test for Differences in Two
Proportions: Solution
H0: p1 - p2 = 0
H1: p1 - p2  0
a = 0.01
n1 = 78 n2 = 82
Critical Value(s):
Reject H 0
Reject H 0
.005
.005
-2.58 0 2.58 Z
Test Statistic:
Z  2.90
Decision:
Reject at a = 0.01.
Conclusion:
There is evidence of a
difference in proportions.
Confidence Interval for Differences in
Two Proportions
 The 100 1  a  % Confidence Interval for
Differences in Two Proportions
 p
1
 p 2   z
1
p 1 1  p 1 
a
2
n1

p 2 1  p 2 
n2
Confidence Interval for Differences in Two
Proportions: Example
As personnel director, you
want to find out the
perception of fairness of
two methods of
performance evaluation. 63
of 78 employees rated
Method 1 as fair. 49 of 82
rated Method 2 as fair.
Construct a 99%
confidence interval for the
difference in two
proportions.
63
 0.8077
78
49
p2 
 0.598
82
p 1 

np  5
n1  78
n2  82
n1  p  5
Confidence Interval for Differences in Two
Proportions: Solution
 p
1
 p 2   z
1
p 1 1  p 1 
a
2
n1

p 2 1  p 2 
n2
0.80771  0.8077 0.59761  0.5976
 0.8077  0.5976  2.5758

78
82
0.0294 < p1  p2 < 0.3909
We are 99% confident that the difference between two
proportions is somewhere between 0.0294 and 0.3909.
Z Test for Differences in Two
Proportions (Dependent Samples)
What is It Used For?
To determine whether there is a difference
between 2 population proportions and whether
one is larger than the other
Assumptions:
Dependent samples
Population follows binomial distribution
Sample size large enough: np  5 and n(1-p)  5
for each population
Z Test Statistic for Dependent
Samples
a d
Z
a d
Sample One
Category 1
Category 2
Sample Two
Category 1 Category 2
a
b
c
d
a+c
b+d
This Z can be used when a+d>10
If 10<a+d<20, use the correction in the text
a+b
c+d
N
Unit Summary
 Compared Two Independent Samples
Performed Z test for the differences in two
means
Performed t test for the differences in two
means
Performed Z test for differences in two
proportions
 Addressed F Test for Difference in Two Variances
Unit Summary
 Compared Two Related Samples
Performed dependent sample Z tests for the
mean difference
Performed dependent sample t tests for the
mean difference
Performed Z tests for proportions using
dependent samples