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2017/7/28
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contents
1. introduction
2. Hypothesis testing
2.1 One sample t test
2.2 two independent-samples t test
2.3 Paired-samples t test
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Hypothesis testing
Aim: Is the average English score of students from 2
schools different or same?
Methods1. Compute and compare two population
mean directly
1
1
2
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

2
Hypothesis testing
Aim: Is the average English score of students from 2
schools different or same?
Methods 2 Do a sampling study and then do
hypothesis testing
n1  100,
X 1  82, S1  9.5
1
n2  150,
X 2  81, S 2  8.3
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

2
The reason that 81  82

True difference between two population means.

Chance (sampling error)
So the hypothesis task is to differentiate that
the difference between two samples is from the
true difference between two population means
or from chance.
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SectionⅠ
Introduction
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The purpose of hypothesis testing is to aid the
clinician, researcher, or administrator in reaching a
conclusion concerning a population by examining a
sample from the population.

“Is the effect of the new drug significant than the
old drug?”,

“which one is better between the two operations? ”

“Did the large amounts of advertising describe the
benefits of new drugs? ”
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1 What does statistic test do?
[EXAMPLE1] General the average height of 7 years old
children increases 4cm in one year. Some researcher let
100 of 7 years old children get a bread appended lysine
in everyday. After one year the average height of 100
children increases 5cm, and the standard deviation is
2cm.
Basing on the data can we think: lysine benefits growth of
stature of 7 years old children?
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Suppose
＝0
Compute T.S
Find P-value
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STEPS
2 Steps of hypothesis testing
The statisticians have made a set of steps
as fixed as legal procedure, and made some
formulas to calculate test statistic (T.S).
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Set up hypothesis and confirm α
STEPS
compute test statistic
Find P value
P>α
P≤α
Make conclusion
Reject H0, the difference
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Don’t reject H0, the
difference is not significant
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Put forward null hypothesis and
alternative hypothesis
 What is null hypothesis?
1.
The test is designed to assess the strength of
the evidence against Ho.
2. It is denoted by H0
H0：  0
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Put forward null hypothesis and
alternative hypothesis
 What is alternative hypothesis?
(1) It is contradictory to null hypothesis
(2) It is denoted by H1
H1： < 0 ，   0 or   0
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Confirm significant level 
 What is significant level？

a probability of rejecting a true null hypothesis

denoted by α (alpha) Generally, 0.05.

determined by the investigator in advance.
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Determine the appropriate T.S
The selection of test statistics is related with
many factors, such as the type of variable,
research aims and conditions proffered by the
sample.
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Find P-value and draw conclusion
The
mathematician
have
calculated
probability corresponding to every T.S, and
listed in some tables. This is the probability that
the test statistic would weigh against Ho at least
as strongly as it does for these data.
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Find P-value and draw conclusion
If P≤α, we reject Ho in favor of H1 at significant level α,
We may think that the two populations are different;
If P>α, we don`t reject Ho at significant level α. We
may think that two populations are same.
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TypeⅠerror versus typeⅡ error in
hypothesis testing
Because the predictions in H0 and H1 are written so that
they are mutually exclusive and all inclusive, we have a
situation where one is true and the other is automatically
false.
when H0 is true ,then H1 is false.

If we don’t reject H0,we have done the right thing.

If we reject H0 ,we have made a mistake.
Type Ⅰ error: Reject H0 when it is true. The probability of
type Ⅰ error is 
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TypeⅠerror versus typeⅡ error in
hypothesis testing
when H0 is false ,then H1 is true.

If we don’t reject H0 , we have made a mistake.

If we reject H0 , we have done the right thing.
TypeⅡ error : Don’t reject when it is false. The probability
of type Ⅱ error is .
 is more difficult to assess because it depends on
several factors.
1-  is called the power of the test.
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State of nature
Decision
H0 is real
H0 is false
Don’t reject
H0
Correct decision
1-α
type Ⅱ error
(β)
Reject H0
type Ⅰ error
(α)
Correct decision
Test power (1-β)
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Tradeoff between  and 
For fixed n, the lower ,
the higher . And the
higher , the lower 

You can not reduce
two types error at
the same time when
n is fixed

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Two-sided test and one-sided test
1 Two-sided test：Interest in whether   0
2 One-sided test：Interest in whether   0 , or   0
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Comparison of sample mean X
and population mean μo
basing on study aim
hypothesis
Two-sided
H0
H1
 = 0
 ≠0
One-sided
 ≥0
 < 0
 ≤ 0
or
 > 0
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Comparison of two sample means X
basing on study aim
hypothesis
Two-sided
H0
H1
1 = 2
1 ≠ 2
One-sided
 1 ≥ 2
1 < 2
or
1 ≤ 2
1 > 2
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Two-sided test
Confidence level
Reject region
Reject region
1-
/2
/2
Not reject region
Critical value
H0
T.S
Critical value
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 < 0
one-sided test
Confidence level
Reject region
1-

Not reject region
Critical value
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H0
T.S
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one-sided test
  0
Confidence level
Reject region
1-

Not reject region
H0
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Critical value
T.S
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Section Ⅱ t-test
2.1 One sample t test
2.2 Two independent-samples t test
2.3 Paired-samples t test
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How to do one-sample hypothesis test?
n>50?
yes
no
One sample t test
Does the sample come from
normal population?
yes
One sample t test
no
One sample rank sum test
29
2.1 One sample t test
Model assumptions of onesample t-test
Test statistic
(1) n≥50
(2) n<50 and the sample comes
t
X  0
S/ n
from normal population.
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EXAMPLE1
Generally the average height of 7 years old
children in city A increases 4cm in one year.
One researcher let 100 children of 7 years old
randomly drawn from the city A get a bread
appended lysine in everyday. After 1 year the
average height of 100 children increases 5cm,
and the standard deviation is 2cm. Basing on
the data can we think: lysine benefits growth of
stature of 7 years old children?
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Solution:
 Ho: μ≤μo
H1: μ>μo
=0.05
 Compute T.S
x  0
54
t

5
s / n 2 / 100
df=100-1=99
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Find P-value and draw conclusion
∵ t=5>1.660
∴ P < 0.05
Because P is smaller than α, we reject Ho at
the significant level 0.05 in favor of H1 . We
can think that lysine benefits growth of stature
of 7 years old children.
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Table 2
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one-sided test
Confidence level
  0
Reject region
1-

Not reject region
H0
1.660
5
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【exercise 1】25 adult female was
chosen randomly from Zhengzhou
city in 2010 and the systolic blood
pressure was measured by
standard methods. To test whether
the average of SBP in Zhengzhou
city is same with the average
level( 126.5mmHg) in China？
118.8
125.4
123.6
123
111.6
112.3
123.3
138.1
124.3
123.3
123.1
116.9
136.5
119.1
122.3
112.1
114.1
133.4
131.3
130.1
125.8
119
137.2
123.7
125.6
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How to do two-samples hypothesis test?
Is n larger than 50 in both groups?
yes
no
Do two samples come from normal
population？
yes
no
Are two population variances equal？
yes
Two-independent
samples t test
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Wilcoxon rank sum test
no
Correction t test
37
2.2 two independent-samples t test
assumptions
Test statistic
The data of two samples
1.
must come from normal
Two population variances
are equal.
 
2
1
2
Sc (
distribution.
2.
t
X1 X 2
2
2
1
1
 )
n1 n 2
S1（n1  1） S 2（n2  1）
2
Sc 
n1  n2  2
2
2
degree of freedom
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2.2 two independent-samples t test
 When the assumption of normal distribution is
valid while the equality of variance is violated ,
we should choose correction t test (
t ' test)
 When the assumption of normal distribution is
violated , we should choose rank sum test.
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t ' test
t' 
X1  X 2
2
1
2
2
s
s

n1 n2
2
 s1 s2 
 

n

n
1
2

  2
2
s1 2
s2 2
( ) ( )
n1
n
 2
n1  1 n2  1
2
2
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Example 2
Company officials were concerned about the
length of time a particular drug product retained
its potency. A random sample, sample 1, of
n1=10 bottles of the product was drawn from the
production line and analyzed for potency. A
second sample, sample 2, of n2=10 bottles was
obtained and stored in a regulated environment
for a period of one year. Whether the two
population mean are different at 0.05 level?
Suppose the two samples come from normal
population.
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Table 5.1 potency for two samples
sample 1
sample 2
10.2 10.6
9.8 9.7
10.5 10.7
9.6 9.5
10.3 10.2
10.1 9.6
10.8 10.0
10.2 9.8
9.8 10.6
10.1 9.9
x1  10.37, s1  0.105
2
Calculated :
x2  9.83, s2  0.058
2
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Solution :
Ho: μ1=μ2 H1: μ1≠μ2
Compute t
t
α= 0.05
x1  x2
s12 (n1  1)  s22 (n2  1) 1
1
( 
)
n1  n2  2
n1 n2
 4.24
t 0.05,18=2.101, so P<. We reject Ho in favor of H1 at
level 0.05, then we can think their potencies are different.
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Table 2
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Two-sided test
Confidence level
Reject region
/2
Reject region
1-
/2
Not reject region
-2.101
H0
2.101
4.24
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EXAMPLE 2
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H0: Normal
H1: Not normal
  0.05
SAMPLE1 : TS  0.953, P  0.701
SAMPLE 2 : TS  0.935, P  0.495
SAMPLE1
normal
SAMPLE2
normal
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Result of t test
Tests of equality of variance
Result of correction t test
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To test equality of variances?
Ho: 12= 22 , H1: 12≠22
=0.05
Compute F
2
S1 (l arg er )
F 2
, 1  n1  1, 2  n2  1
S 2 ( smaller )
Conclusion : find F critical value in table 4.
If F  F ,  , P  
12≠22
If F  F ,  , P  
12= 22
1, 2
1, 2
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we have known: S12 =0.105, n1 =10,
S22 =0.058, n2 =10
Ho: 12= 22 , H1: 12≠22 ,=0.05
Compute F
0.105
F
 1.81, 1  10  1  9, 2  10  1  9
0.058
Conclusion : find F critical-value in table 4.
2=  2 .
,
so
we
can
think:

F  F0.05,9,9  4.03, P  
1
2
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EXERCISE 2
Table 1 increase of concentration of Hb in two groups
increase of concentration of Hb （g/L）
group
new drug group
30.5
21.4
25.0
34.5
33.0
32.5
29.5
25.5
24.4
23.6
routine drug group
19.5
19.0
13.0
24.7
21.5
22.0
19.0
15.5
24.5
23.4
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Analyze→ Descriptive Statistics→ Explore
Dependent list→ y
Factor list→group
Plots→√Normality plots with tests
Continue
OK
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Analyze→ Compare means→ Independentsample T Test
Test Variable(s) → y
Grouping Variable→group
Define Groups→ Group 1: 1； Group 2: 2
Continue
OK
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【SPSS】
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【SPSS】
Independent Samples Test
Levene's T est for
Equality of Variances
y
Equal variances assumed
Equal variances not
assumed
F
1.345
Sig.
.261
t-test for Equality of Means
t
4.137
4.137
18
Sig. (2-tailed)
.001
Mean
Difference
7.7800
Std. Error
Difference
1.8807
17.461
.001
7.7800
1.8807
df
95% Confidence
Interval of the
Difference
Lower
Upper
3.8288
11.7312
3.8200
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11.7400
How to report the result?
The data of two samples were adequately normally
distributed（Shapiro-Wilk test：P1=0.466；P2＝
0.482） and two population variances were equal at
the significant level 0.10（F＝1.345；P=0.261）, so
two independent samples t test was used（t=4.137；
df=18；P=0.001）. The results indicated a
statistically significant difference between effects of
two drugs at two-sided significant level 0.05 and the
average increase of concentration of Hb was higher
in patients taking the new drug, which could also be
observed from the 95% confidence interval of the
difference of two population means (3.829, 11.731).
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How to report the result?
Table 2 Average increase of concentration of Hb in two groups ( x  s )
groups
n
the increase of concentration of Hb（g/L）
new drug group
10
27.99±4.56
routine drug group
10
20.21±3.82
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How to do paired-samples hypothesis test?
n is the number of pairs
n>50?
yes
no
Paired-samples t
Does the difference of paired-samples
test
come from normal population?
yes
Paired-samples t test
no
rank sum test
58
2.3 Paired-samples t test
Model assumptions
Test statistic
The differences among each
paired-samples must come
from normal distribution
d 0
t
,  n  1
sd / n
population.
numbers of pairs
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New concepts:
d: difference between each pair;
d : sample mean of difference;
Sd: sample standard deviation of difference;
n: number of pairs of sample.
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2.3 Paired-samples t test
When the assumption of normal distribution of
difference is violated, we should make data
transformation or choose rank sum test.
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There are two forms in paired t-test:

1 The study objects are matched by certain
conditions (the same weights 、the same age or
the same sex). Then the two study objects of each
pair are assigned randomly to different groups.

2 One study objects receive two different
disposals.The aim is to infer whether there is
difference between the effect of two disposals.
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randomization
Treat 1
10 rabbits
pair
!“#$%&‘()*
20 rabbits
!“#$%&‘()*
Treat 2
10 rabbits
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While analyzing paired data, the differences
between each paired are more important than the
raw data. The aim is to compare whether the
efficiency of two factors is different.
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Example 3
Insurance adjusters are concerned about
the high estimates they are receiving from
garage 1 for auto repairs compared to garage
2. To verify their suspicions, that is, the mean
repair estimate for garage 1 is greater than
that for garage 2, each of 15 cars recently
involved an accident was taken to both
garages for separate estimates of repair
costs. Is true their suspicions?
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Table 3
car
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Repair estimates(in hundreds of dollars)
Garage 2
1
7.6
7.3
0.3
0.09
2
10.2
9.1
1.1
1.21
3
9.5
8.4
1.1
1.21
4
1.3
1.5
-0.2
0.04
5
3.0
2.7
0.3
0.09
6
6.3
5.8
0.5
0.25
7
5.3
4.9
0.4
0.16
8
6.2
5.3
0.9
0.81
9
2.2
2.0
0.2
0.04
10
4.8
4.2
0.6
0.36
11
11.3
11.0
0.3
0.09
12
12.1
11.0
1.1
1.21
13
6.9
6.1
0.8
0.64
14
7.6
6.7
0.9
0.81
15
8.4
7.5
0.9
0.81
x2  6.23
 d  9.2
 d 2  7.82
Totals
x1  6.85
Difference(d)
d2
Garage 1
66
Solution: This is comparison of paired data,
so we should use paired-samples t test.
d
t
Sd / n
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1 Set up the hypothesis and confirm α
Ho: µd= 0
H1: µd > 0
Population mean
α= 0.05
of difference
2 Compute t
d

d
sd 

sd
 0.61
d2 (
d
t
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n
n 1

n

d )2 n
0.61

7.82  (9.2) 2 15
 0.394
15  1
 6.00
0.394 / 15
68

3 Confirm p-value and draw conclusion
df =14, t = 6.0, t(14, 0.05)=1.761
so P < 0.05
We reject Ho in favor of H1 at level 0.05. We believe
that the suspicions of the insurance adjusters are true,
that is, the mean repair estimate for garage 1 is greater
than garage 2
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Table 2
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EXAMPLE 3
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Compute difference of
each pairs
Transform→ Compute
Target Variable→ difference
Numeric Expression→garage1－garage2
OK
Normality tests of
difference
Analyze→ Descriptive Statistics→ Explore
Dependent list→difference
Plots→√Normality plots with tests
Continue
OK
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Normal
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Method1

Analyze→ Compare means→ one sample t Test

Test Variables →difference

Test value → 0

OK
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Method 2

Analyze→ Compare means→ paired-samples t Test

Paired Variables→garage1－garage2

OK
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Paired Samples Test
Paired Differences
Pair 1
garage1 - garage2
Mean
.61333
Std. Deviation
.39437
Std. Error
Mean
.10182
95% Confidence
Interval of the
Difference
Lower
Upper
.39494
.83173
t
6.023
df
14
Sig. (2-tailed)
.000
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Exercise
One doctor want to explore whether the height of adult
males is higher than that of the adult females. He chose
randomly 64 males and 49 females and measured their
heights one by one. The outcome are as follows
Question: Is the height of adult males higher than that
of the adult females.
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x1  175cm
x2  160cm
s1  4cm
s2  3.5cm
n1  64
n2  49
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To test homogeneity of two population variances
Ho: 12= 22 , H1: 12≠22
=0.05
Compute F
2
S1 (larger )
F 2
 1.31, 1  63, 2  48
S 2 ( smaller )
Conclusion : not significant, we can think the two
population variance are same
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Solution :
Ho: μ1≤μ2
Compute t
t
H1: μ1>μ2
α= 0.05
x1  x2
s12 (n1  1)  s22 (n2  1) 1
1
( 
)
n1  n2  2
n1 n2
 20.84
P<. We reject Ho in favor of H1 at level 0.05, then we
can think height of adult males higher than that of the adult
females.
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