AMS 572 Lecture Notes
Oct. 1, 2013.
Review: Inference on two population means
1. Two normal pops,  12 &  22 are known, independent samples. → exact Z
P.Q. Z 
X  Y  ( 1   2 )
 12
n1
 22

~ N 0,1
n2
2. At least one population is not normal, independent samples, both are large
( n1  30, n2  30 ) → approximate Z
(1) P.Q. (if the variances are known, by the central limit theorem)
Z
X  Y  ( 1   2 )

2
1
n1


2
2
~ N 0,1
n2
(2) P.Q. (if the variances are unknown, by the central limit theorem, and the
Slutsky’s theorem: http://en.wikipedia.org/wiki/Slutsky's_theorem)
Z
X  Y  ( 1   2 )
S12 S 22

n1 n2
~ N 0,1
3. Two normal pops,  12 &  22 are unknown but equal (  12   22   2 ),
independent samples.
Pooled variance t (exact)
P.Q. T 
X  Y  ( 1   2 )
Sp
1
1

n1 n2
~ t df
Exact d. f .  n1  n2  2 (SAS)
4. Two normal pops,  12 &  22 are unknown,  12   22 approximate t
1
P.Q. T 
X  Y  ( 1   2 )
S12 S 22

n1 n2
~ t df
More accurate d.f. ⇒ Satterthwaite method (SAS)
Quick & less accurate ⇒ d . f .  min( n1  1, n2  1) in-class exam
5. other situations ⇒ nonparametric method
Mann-Whitney U-test = Wilcoxon Rank-Sum Test (SAS)
6. Modern nonparametric method Bootstrap Resampling method
7. Transformation to Normal distribution Box-Cox transformation
e.g. X & Y are not normal, but ln(X) & ln(Y) are normal.
Inference on two population variances
* Both pop’s are normal, two independent samples
i .i.d .
Sample 1 : X 1 , X 2 ,  , X n1 ~ N (1 ,  12 )
i.i.d .
Sample 2 : Y1 , Y2 ,  , Yn2 ~ N ( 2 ,  22 )
⇒
(n1  1) S12
⇒
(n2  1) S 22
 12

2
2
~  n21 1
~  n22 1
 12
(*** Parameter of interest : 2 )
2
1. Point estimator : ˆ12  S12
ˆ 22  S 22
Def. F-distribution Let W1 ~  k21 , W2 ~  k22 , W1 ,W2 are independent.
Then, F 
W1 k1
~ Fk1 ,k2
W2 k 2
2
  P( F  Fk ,k
1
 P(
2 , ,lower
)
1
1

)
F Fk1 ,k2 , ,lower
1
~ Fk 2 ,k1
F
Fk1 ,k2 , ,lower 
1
Fk2 ,k1 , ,upper
(n1  1) S12
F
 12
(n2  1) S 22
 22
(n1  1)

(n2  1)
2. 100(1   )% CI for
1    P( F
2
,lower

S12
S 22
 12
 22
~ Fn1 1,n2 1
 12
 22
S12
S 22
 12
 22
 F
2
,upper
)
3
 P(
 P(
1
F
2

,lower
S12
S 22
F
2

,upper
 12
 22
S
S
2
1
2
2

1
F
 12

 22 F
2
,upper
S12
S 22
2
)
)
,lower
3. Test
H0 :
 12
1
 22
 12
Ha : 2  1
2
Test Statistic F0 
S12 H 0
~ Fn1 1,n2 1
S 22
At the significance level  , we reject H 0 if F0 is too large or too small.
F0  c1 , F0  c 2
* conventional boundries / thresholds
c1  Fn 1,n 1,
1
2
c2  Fn 1,n 1,
1
2
2
,upper
2
,lower
SAS program for test on 2 pop means
1. paired samples
sample 1
10 23 16 18 … 33
sample 2
15 28 21 29 … 58
data paired;
input IQ1 IQ2;
diff=IQ1-IQ2;
datalines;
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10 15
23 28
…
33 58
;
run;
proc univariate data=paired normal;
var diff;
run;
2. independent Samples
sample 1
10 23 16 18 … 33 (group 1)
sample 2
15 28 21 29 … 58 (group 2)
data indept;
input group IQ;
datalines;
1 10
1 23
1 16
…
1 33
2 15
2 28
…
2 58
;
proc sort data= indept;
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by group;
run;
proc univariate data= indept normal;
var IQ;
by group;
run;
/*use by group if we have already sorted by group– otherwise use class*/
/* If both are nomal */
proc ttest data= indept;
var IQ;
class group;
run;
/* If at least 1 pop is NOT normal */
proc NPAR1WAY data= indept;
var IQ;
class group;
Wilcoxon Rank-sum test
run;
Inference on two population means, independent samples:
Power and Sample Size calculation
– Exact or Large Sample Z-test
1. Based on the maximum error / or the length of the CI.
Suppose we are using the exact or the large sample approximate z-test ;
n1  n2  n
Suppose the maximum error is E with probability (1   )
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P(| ( X  Y )  ( 1   2 ) | E )  1  
X  Y  ( 1   2 )
 12
n1

 22
~ N (0,1)
n2
P (  E  X  Y  ( 1   2 )  E )  1  
E
P(
 12
n1

 22
E

2
1
n1
n
n


 22
X  Y  ( 1   2 )
 12
n2
n1
 Z

 22

n2
E
 12
n1

 22
)  1
n2
2
n2
Z  2  12   22
E
( Z  2 ) 2 ( 12   22 )
E2
100(1   ) % CI for (1   2 )
L  2 E
X  Y  Z 2
L  2  Z 2
n
 12
n
 12
n


 22
n
 22
n
4( Z  2 ) 2 ( 12   22 )
L2
2. Based on the power of the test
H 0 : 1  2  1
H a : 1  2   2  (or , )1
1    power  P(reject H 0 | H a )
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 P ( Z 0  Z | 1  2   2 )
 P(
X  Y  1
 12
n
 P(
 12
 2  1
 12
n
n
n
n
Z
 Z | 1   2   2 )
n
X  Y  2
n
Z 

 22

 22

 22
 Z 
n
 2  1
 12
n

| 1  2   2 )
 22
n
 Z 
n
 Z   ( 12   22 )
2

(1-sided test, exact Z)
 2  1 2
( Z  Z  ) 2 ( 12   22 )
  2  1 
2
(1-sided test, approximate Z)
( Z 2  Z  ) 2 ( 12   22 )
  2  1 
2
(2-sided test, exact or approximate Z)
Inference on two population means, independent samples:
Power and Sample Size Determination
– Pooled Variance T-test
1. Sample size calculation in a C.I. scenario (Maximum error)
iid
X1 ,
, X n1 ~ N ( 1 ,  2 ) ,
P.Q: T 
iid
Y1 ,
, Yn2 ~ N ( 2 ,  2 )
( X  Y )  ( 1  2 )
~ t2 n  2
2
Sp
n
100(1-α)% CI for 1  2 is ( X  Y )  t
 The length of the CI : L= 2  t
 n  8(
t2 n  2, 2  S p
L
2 n  2,

2
2 n  2,
 Sp

 Sp
2
2
n
2
n
)2
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2. Inference on the test situation
 H 0 : 1  2  0

 H a : 1  2    0
Data: Two independent samples
iid
X1 ,
iid
, X n1 ~ N ( 1 ,  2 ) ,
Y1 ,
, Yn2 ~ N ( 2 ,  2 )
Here  12   2 2   2 and n1  n2  n .
For a given α (e.g. 0.05 or 0.01) and a power = (1-β) (e.g. 85%),
calculate the sample size.

(e.g. Eff =1)

(X Y)  0
( X  Y ) H0

~ t2 n  2
T.S : T0 =
1 1
2
Sp

Sp
n1 n2
n
Def.: Effect size =
At α = 0.05, reject H 0 in favor of H a iff T0  t2 n  2,
Power = 1-β = P(reject H 0 | H a )= P(T0  t2 n 2, | H a : 1  2    0)
= P(
= P(
(X Y)
 t2 n  2, | H a : 1  2  )
2
Sp
n
(X Y)  

 t2 n 2, 
| H a : 1  2  )
2
2
Sp
Sp
n
n
= P(T  t2 n2,  Eff *


n
)
| H a : 1  2  ) (Effect size= 
 Sp
2
 t2 n2,   t2 n2,  Eff *
n
2
 n  2(
t2 n2,  t2 n2, 
Eff
)2
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Example A new method of making concrete blocks has been proposed. To test whether
or not the new method increases the compressive strength, 5 sample blocks are made by
each method.
New Method
14
15
13
15
16
Old Method
13
15
13
12
14
a. Get a 95% for the mean difference of the 2 methods.
b. At  = 0.05, can you conclude the new method is better? Provide p-value.
Write the SAS program for part (b).
Solution
a. Assume both populations are normal.
First, we check whether  12   22
H 0 :  12   22
H a :  12   22
S12
Test Statistic : F0  2
S2
n
Recall S 2 
 (x
i 1
i
 x) 2
n 1
 F0  1  F4, 4, 0.1,upper  4.11
It is reasonable to assume  12   22
Pooled-variance statistic (PQ)
t
( X 1  X 2 )  ( 1   2 )
Sp
1
1

n1 n2
~ t n1  n2 2
95% CI for (1   2 ) is ( X 1  X 2 )  t n1  n2 2,0.025  S p 
1
1

n1 n2
10
Sp 
(n1  1) S12  (n2  1) S 22
n1  n2  2
b. Assume both populations are normal.
First, we check whether  12   22
By part (a), we found that it is reasonable to assume  12   22
H 0 : 1   2  0
H a : 1   2  0
Test Statistic : T0 
X1  X 2  0
Sp
1
1

n1 n2
H0
~ t n1  n2 2
At  =0.05, we reject H 0 if T0  t n1  n2  2,0.05  t 8,0.05 .
But T0 ( 1.66)  t 8,0.05 ( 1.86)
 We cannot reject H 0 at  =0.05.
SAS
data block ;
input method $ strength ;
datalines ;
new 14
new 15
new 13
new 15
new 16
old 13
old 15
old 13
old 12
old 14
;
run ;
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proc univariate data=block normal plot ;
class method ;
var strength ;
run ;
proc ttest data=block ;
class method ;
var strength ;
run ;
proc npar1way data=block ;
class method ;
var strength ;
run ;
Example An experiment was done to determine the effect on dairy cattle of a
diet supplement with liquid whey. While no differences were noted in milk
production between the group with a standard diet (hay + grain + water) and the
experimental group with whey supplement (hay + grain + whey), a considerable
difference was noted in the amount of hay ingested. For a 2-tailed test with
 =0.05, determine the approximate number of cattle that should be included in
each group if we want   0.1 for 1   2  0.5 . Previews study has shown
  0.8
Solution
H 0 : 1   2  0
H a : 1   2    0
1.  12   22   2
2. either both populations are normal or both sample size are large.
12
n 
2 2 ( Z  2  Z  ) 2
2

(0.8) 2 (1.96  1.28) 2
 53.75
(0.5) 2
 n  54
Example Do fraternities help or hurt your academic progress at college? To
investigate this question, 5 students who joined fraternities in 1998 were
randomly selected. It was shown that their GPA before and after they joined the
fraternities are as follows.
Student
1
2
3
4
5
Before
3
4
3
3
2
After
2
3
3
2
1
Diff.
1
1
0
1
1
Please test the hypothesis at  =0.05
Solution
H 0 : d  0
H a : d  0
Assumption : the difference follows a normal distribution.
X d  0.8, S d  0.447, n  5,   0.05
Test statistic : T0 
X d  0 H0
~ t n 1
Sd n
T0  4.02  t 4,0.025  2.776
We reject H 0 at  =0.05 and conclude fraternities does hurt…
SAS
data frat ;
input before after ;
diff = before – after ;
datalines ;
13
32
43
33
32
21
;
run ;
proc univariate data=frat normal ;
var diff ;
run ;
G*Power:
http://www.psycho.uni-duesseldorf.de/abteilungen/aap/gpower3/
Dear students: Please install G*Power on your own computer. This is a very
effective power calculation tool. For two-sided exact T tests, for example, it
will give you the accurate power calculation – better than the approximate
hand calculation formula where we dropped a smaller probability in the
calculation.
You can follow the following website to learn how to use G*Power:
1. One population mean:
http://www.ats.ucla.edu/stat/gpower/one_sample.htm
2. Two population means (paired samples):
http://www.ats.ucla.edu/stat/gpower/pairedsample.htm
3. Two population means (independent samples):
http://www.ats.ucla.edu/stat/gpower/indepsamps.htm
You can also find many other helpful sites by typing at Google:
g power, t test (or typing the key words: g*power, t test)
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