1. chapter-7 hypothesis testing for quantitative variable
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2. contents introduction Hypothesis testing 2.1 One sample t test
2.2 two independent-samples t test
2.3 Paired-samples t test
As a result of this class, you will be able to ...
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3. Methods1. Compute and compare two population mean directly
Hypothesis testing
Aim: Is the average English score of students from 2 schools different or same?
Methods1. Compute and compare two population mean directly
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4. Methods 2 Do a sampling study and then do 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
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5. True difference between two population means. Chance (sampling error)
The reason that
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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6. SectionⅠ Introduction
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7. “Is the effect of the new drug significant than the old drug?”,
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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8. 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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9. Suppose
＝0
Compute T.S
Find P-value
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10. 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).
STEPS
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11. STEPS P>α P≤α Set up hypothesis and confirm α
compute test statistic
Find P value
P>α
P≤α
Make conclusion
Reject H0, the difference
is significant.
Don’t reject H0, the
difference is not significant
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12. 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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13. 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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14. 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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15. 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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16. 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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17. 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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18. 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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19. 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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20. State of nature H0 is real H0 is false
Decision
State of nature
H0 is real
H0 is false
Don’t reject H0
Correct decision
1 - α
type Ⅱ error
(β)
Reject H0
type Ⅰ error
(α)
Test power (1-β)
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21. Tradeoff between  and 
You can not reduce two types error at the same time when n is fixed
For fixed n, the lower , the higher . And the higher , the lower   
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22. Two-sided test and one-sided test
1 Two-sided test：Interest in whether m  m0
2 One-sided test：Interest in whether m  m0 , or m  m0
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23. Comparison of sample mean
and population mean μo
hypothesis
basing on study aim
Two-sided
One-sided H0
m = m0 H1
m ≠m0
m ≤ m0
m > m0 or
m ≥m0
m < m0
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24. H0 H1 Comparison of two sample means m1 = m2 m1 ≠ m2 m 1 ≥ m2 m1 ≤ m2
hypothesis
basing on study aim
Two-sided
One-sided H0
m1 = m2 H1
m1 ≠ m2
m 1 ≥ m2
m1 < m2
m1 ≤ m2
m1 > m2 or
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25. Two-sided test Confidence level Reject region Reject region H0 T.S
Rejection region does NOT include critical value.
1 - 
a/2
a/2
Not reject region H0
T.S
Critical value
Critical value
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26. one-sided test Confidence level Reject region a H0 Critical value T.S
Rejection region does NOT include critical value.
1 -  a
Not reject region H0
Critical value
T.S
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27. one-sided test Confidence level Reject region a H0 Critical value T.S
1 - 
Rejection region does NOT include critical value. a
Not reject region H0
Critical value
T.S
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28. 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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29. How to do one-sample hypothesis test?
yes
One sample t test no
Does the sample come from normal population?
One sample rank sum test

30. 2.1 One sample t test Test statistic
Model assumptions of one-sample t-test
(1) n≥50
(2) n<50 and the sample comes from normal population.
Test statistic
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31. 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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32. Solution: Compute T.S df=100-1=99 Ho: μ≤μo H1: μ>μo =0.05
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33. 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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34. Table 2
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35. one-sided test Confidence level Reject region a H0 1.660 1 - 
Rejection region does NOT include critical value. a
Not reject region H0
1.660 5
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36. 【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？
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37. How to do two-samples hypothesis test?
Is n larger than 50 in both groups?
yes no
Do two samples come from normal population？
Are two population variances equal？
Two-independent samples t test
Correction t test
Wilcoxon rank sum test
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38. 2.2 two independent-samples t test
assumptions
The data of two samples must come from normal distribution.
Two population variances are equal.
Test statistic
degree of freedom
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39. 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 ( test)
When the assumption of normal distribution is violated , we should choose rank sum test.
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40. test
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41. 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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42. Table 5.1 potency for two samples
sample sample 2
Calculated :
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43. Solution : Ho: μ1=μ2 H1: μ1≠μ2 α= 0.05 Compute t
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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44. Table 2
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45. Two-sided test Confidence level Reject region Reject region H0 1 - 
Rejection region does NOT include critical value.
1 - 
a/2
a/2
Not reject region H0
4.24
-2.101
2.101
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46. EXAMPLE 2
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47. H0: Normal
H1: Not normal
SAMPLE1 normal
SAMPLE2 normal
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48. Result of correction t test
Result of t test
Tests of equality of variance
Result of correction t test
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49. If 12≠22 12= 22 To test equality of variances?
Ho: 12= 22 , H1: 12≠22 =0.05
Compute F
Conclusion : find F critical value in table 4. If
12≠22
12= 22
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50. , so we can think: 12= 22 . we have known: S12 =0.105, n1 =10,
Ho: 12= 22 , H1: 12≠22 ,=0.05
Compute F
Conclusion : find F critical-value in table 4.
, so we can think: 12= 22 .
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51. EXERCISE 2 Table 1 increase of concentration of Hb in two groups
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52. Analyze→ Descriptive Statistics→ Explore Dependent list→ y
Factor list→group
Plots→√Normality plots with tests
Continue OK
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53. Analyze→ Compare means→ Independent- sample T Test
Test Variable(s) → y
Grouping Variable→group
Define Groups→ Group 1: 1； Group 2: 2
Continue OK
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54. 【SPSS】
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55. 【SPSS】
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56. 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, ).
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57. How to report the result?
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58. Does the difference of paired-samples come from normal population?
How to do paired-samples hypothesis test?
n is the number of pairs
n>50?
yes
Paired-samples t test no
Does the difference of paired-samples come from normal population?
rank sum test

59. 2.3 Paired-samples t test Test statistic Model assumptions
The differences among each paired-samples must come from normal distribution population.
Test statistic
numbers of pairs
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60. d: difference between each pair; : sample mean of difference;
New concepts:
d: difference between each pair;
: sample mean of difference;
Sd: sample standard deviation of difference;
n: number of pairs of sample.
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61. 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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62. 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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63. randomization pair 20 rabbits Treat 1 10 rabbits ! “ # $ % & ‘ ( ) *
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64. 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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65. 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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66. Table 3 Repair estimates(in hundreds of dollars)
car
Garage 1
Garage 2
Difference(d) d2 1
7.6
7.3
0.3
0.09 2
10.2
9.1
1.1
1.21 3
9.5
8.4 4
1.3
1.5
-0.2
0.04 5
3.0
2.7 6
6.3
5.8
0.5
0.25 7
5.3
4.9
0.4
0.16 8
6.2
0.9
0.81 9
2.2
2.0
0.2 10
4.8
4.2
0.6
0.36 11
11.3
11.0 12
12.1 13
6.9
6.1
0.8
0.64 14
6.7 15
7.5
Totals
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67. Solution: This is comparison of paired data, so we should use paired-samples t test.
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68. Population mean of difference
1 Set up the hypothesis and confirm α
Ho: µd= 0
H1: µd > 0
α= 0.05
2 Compute t
Population mean of difference
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69. 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 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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70. Table 2
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71. EXAMPLE 3
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72. Compute difference of each pairs
Normality tests of difference
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73. Normal
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74. Analyze→ Compare means→ one sample t Test Test Variables →difference
Method1
Analyze→ Compare means→ one sample t Test
Test Variables →difference
Test value → 0 OK
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75. 2018/12/7

76. Analyze→ Compare means→ paired-samples t Test
Method 2
Analyze→ Compare means→ paired-samples t Test
Paired Variables→garage1－garage2 OK
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77. 2018/12/7

78. 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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79. 2018/12/7

80. To test homogeneity of two population variances
Ho: 12= 22 , H1: 12≠22 =0.05
Compute F
Conclusion : not significant, we can think the two population variance are same
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81. Solution : Ho: μ1≤μ2 H1: μ1>μ2 α= 0.05 Compute t
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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82. Thanks for your attention!
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