Download Lecture Notes 22

Lecture 22
Dustin Lueker


Similar to testing one proportion
Hypotheses are set up like two sample mean
test
◦ H0:p1-p2=0
 Same as H0: p1=p2

Test Statistic
z
( pˆ 1  pˆ 2 )  ( p1  p2 )
pˆ 1 (1  pˆ 1 ) pˆ 2 (1  pˆ 2 )

n1
n2
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
Hypothesis involves 2 parameters from 2
populations
◦ Test statistic is different
 Involves 2 large samples (both samples at least 30)
 One from each population

H0: μ1-μ2=0
◦ Same as H0: μ1=μ2
◦ Test statistic
z
( x1  x2 )  ( 1   2 )
2
1
2
2
s
s

n1 n2
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
Used when comparing means of two samples
where at least one of them is less than 30
◦ Normal population distribution is assumed for both
samples

Equal Variances
◦ Both groups have the same variability

Unequal Variances
 
2
1
2
2
◦ Both groups may not have the same variability
 
2
1
2
2
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
Test Statistic
t
( x1  x2 )  ( 1   2 )
(n1  1) s  (n2  1) s
n1  n2  2
2
1
2
2
1 1
  
 n1 n2 
◦ Degrees of freedom
 n1+n2-2
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( x1  x2 )  t / 2,n1  n2 2
(n1  1) s  (n2  1) s
n1  n2  2
2
1
2
2
1 1
  
 n1 n2 
◦ Degrees of freedom
 n1+n2-2
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
Test statistic
t
( x1  x2 )  ( 1   2 )
2
1
2
2
s
s

n1 n2

Degrees of freedom
2
s
s 
  
n1 n2 

df 
2 2
2 2
 s1   s2 
   
 n1    n2 
n1  1 n2  1
2
1
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2
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2
1
2
2
s
s

n1 n2
( x1  x2 )  t / 2,df
2
s
s 
  
n1 n2 

df 
2 2
2 2
 s1   s2 
   
 n1    n2 
n1  1 n2  1
2
1
2
2
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
How to choose between Method 1 and Method
2?
◦ Method 2 is always safer to use
◦ Definitely use Method 2
 If one standard deviation is at least twice the other
 If the standard deviation is larger for the sample with the
smaller sample size
◦ Usually, both methods yield similar conclusions
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
Comparing dependent means
◦ Example
 Special exam preparation for STA 291 students
 Choose n=10 pairs of students such that the students
matched in any given pair are very similar given
previous exam/quiz results
 For each pair, one of the students is randomly selected
for the special preparation (group 1)
 The other student in the pair receives normal
instruction (group 2)
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
“Matches Pairs” plan
◦ Each sample (group 1 and group 2) has the same
number of observations
◦ Each observation in one sample ‘pairs’ with an
observation in the other sample
◦ For the ith pair, let
Di = Score of student receiving special preparation
– score of student receiving normal instruction
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
The sample mean of the difference scores is
an estimator for the difference between the
population means
xD



n
D
i 1 i
n
D1  D2    Dn

n
We can now use exactly the same methods as
for one sample
◦ Replace Xi by Di
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
Small sample confidence interval
xD  tn 1
Note:
sD 

n
i 1
sD
n
( Di  xD )
2
n 1
◦ When n is large (greater than 30), we can use the zscores instead of the t-scores
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
Small sample test statistic for testing
difference in the population means
xD   D
t
sD n
◦ For small n, use the t-distribution with df=n-1
◦ For large n, use the normal distribution instead (z
value)
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
Ten college freshman take a math aptitude
test both before and after undergoing an
intensive training course
Then the scores for each student are paired,
as in the following table
Student
1
2
3
4
5
6
7
Before
60 73 42 88 66 77 90 63 55 96
After
70 80 40 94 79 86 93 71 70 97
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9
10
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Student
1
2
3
4
5
6
7
8
9
10
Before
60 73 42 88 66 77 90 63 55 96
After
70 80 40 94 79 86 93 71 70 97
Compare the mean scores after and before the
training course by
◦ Finding the difference of the sample means
◦ Find the mean of the difference scores
◦ Compare


Calculate and interpret the p-value for testing
whether the mean change equals 0
Compare the mean scores before and after the
training course by constructing and interpreting a
90% confidence interval for the population mean
difference
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Output from Statistical Software Package SAS
N
Mean
Std Deviation
10
7
5.24933858
Tests for Location: Mu0=0
Test
-Statistic-
-----p Value------
Student's t
Sign
Signed Rank
t
M
S
Pr > |t|
Pr >= |M|
Pr >= |S|
4.216901
4
25.5
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0.0022
0.0215
0.0059
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
Variability in the difference scores may be
less than the variability in the original scores
◦ This happens when the scores in the two samples
are strongly associated
◦ Subjects who score high before the intensive
training also dent to score high after the intensive
training
 Thus these high scores aren’t raising the variability for
each individual sample
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