bg statistical package for social sciences spss 5 1889

Compare Means • This method is to calculate the average value in small groups and provide statistical data related to the dependent variable within the group of one or more independent v

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CHAPTER 5 QUANTITATIVE DATA ANALYSIS - TESTING HYPOTHESES ABOUT THE

MEAN VALUE

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• In the case of the mean should be compared to a

certain study criteria between two objects or

many objects we use hypothesis testing on the

mean value To perform this verification, we need

to have two variables: one or more quantitative variables to calculate the average value, a

qualitative variable used to divide the

comparison group

CHAPTER 5

QUANTITATIVE DATA ANALYSIS - TESTING

HYPOTHESES ABOUT THE MEAN VALUE

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1 Compare Means

• This method is to calculate the average value

in small groups and provide statistical data related to the dependent variable within the group of one or more independent variables

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1 Compare Means

• Performed with SPSS

Analyze\ Compare Means\ Means…

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1 Compare Means

Quantitative variables

Quatitative variables

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1 Compare Means

Check:

Anova table and eta

or Test for linearity

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1 Compare Means

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• Add a qualitative variable (grouping variables)

Add a qualitative variable

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1 Compare Means

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1 Compare Means

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2 Compare Means\ One – Sample T - Test

• Implementation of external control, 5 samples of blood sugar (have real value is 100 mg%) were

sent to a laboratory X as follows:

100, 101, 102, 103, 104 Ask that the quality of the laboratory?

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Solution

Hypothesis H0: m = ; H1: m  

1.58 SD

2.5 1

n

104 102

103 102

102 102

101 102

100 102

SD

102 5

104 103

102 101

100 m

2 2

58 1

100 102

n SD

SD

μ m

T    

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Regional accept hypothesis H0

Regional dismissed

The value T calculated is

less than tipping point

The value T calculated

is greater than tipping

point

Look up table "Student", with degrees of freedom: n = 4, t = 2.776 Thus T = 2.82 > t -> Reject H0 hypothesis, there is a difference between blood samples sent to the comparison with the results of a laboratory, concluded a laboratory

quality not achieved

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Analyze\ Compare Means\ One – Sample T test

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Analyze\ Compare Means\ One – Sample T test

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Results

Conclusion: t = 2,828, df = 4, P = 0.047: the difference

between statistical significance test results and real value of blood samples, so the X Lab results have not yet reached

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3 Compare Means\ Independent-Samples T Test…

• Compare the value of one or more quantitative

variables with a classification variable

• If there are 2 classification variable group then

have a pair to compare If there are 3 classification variable group, there are three pairs of comparison (1-2.1-3.2-3) If the variable type is 4 group we can make 6 pairs of

comparison (1-2.1-3.1-3.2-4.2-4.3-4)

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• For example, research on the number of

products consumed two customer groups, customer groups viewed the advertising and consumer groups are not considered advertising program

o The group considered:

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Hypothesis H0: m 1 = m 2 ; H 1 : m 1  m 2

124

3 164.44 204.44

80 - 140 T

164.44 n

SD SE

204.44 n

SD SE

1644 9

80 80

80 100 SD

2044 9

140 90

140 150

SD

80 10

80 20 130 120

30 40 70 80 130 100

m

140 10

90 120 140

100 240

90 160 170

140 150

m

SE SE

m m

T

2

2 2 2

2

1

2 1 2

1

2 2

2 1 2

1

2 2

2 1

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Analyze\ Compare Means\ Independent-Samples T Test…

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Results

• Testing the hypothesis "assumed Equal variances": sig =

1,000 > 0.05, should accept the hypothesis of equal

variance Therefore, use the t test results in the first series, sig = 0.006> 0.05, so reject the null hypothesis H0

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Results

• As such, the amount of products consumed between 2

groups of customers can view the ads and not see an ad

with a difference and statistical significance Group

advertising customers see more products sold clientele not see ads

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4 Compare Means\ Paired-Samples T Test…

• This is a test for the two variables are related to each other, the data used is the distance scale or scale ratio It calculates the offset values of the

two variables for each case and test the average difference is zero or not?

• For example, compare the starting salary and

current salary Comparison of the level of interest

of 2 fragrances

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• For example : Treatment of 8 patients with

medication lower blood X, the results before and after the treatment as follows (mmg):

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 The average value of the difference:

 The variance of the difference:

15 8

10.35 15

n SD

d T

: có

Ta

10.35 107.2

S

107.2 7

15 20

7

15 30

15 20

15 15

15 20

15 15

SD

d

2 2

pressure before treatment and after treatment

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Analyze\ Compare Means\ Paired-Samples T Test…

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Analyze\ Compare Means\ Paired-Samples T Test…

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 The average deviation = 15, standard deviation (SD) =

10.35, t = 4.09 degrees of freedom, df = 7, significance level is 0.005 < 0.05 Reject H0 hypothesis

 To conclude, there is a difference in blood pressure after

the treatment, drug X has the effect of lowering blood pressure

Results

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5 Compare Means\ One – Way Anova

• Is a method of comparative statistics on average

between multiple independent form

• This test is based on the variability of the

observed values within the Group and between group

• Includes 2 form variable: dependent variable

(quantitative scale) grouped variables

(qualitative scale)

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The test of equal variances:

• If sig  0.05 of the variance in each group, statistical

tables "ANOVA" good use to test the hypothesis H0 If

the value of sig < 0.05, Reject H0 hypothesis, if the value of sig  0.05 H0 hypothesis is accepted Using

test "Post hoc" (with variance equal conditions) to find pairs of samples with different mean values The test

"Post hoc" is often used in this case is "Bonferroni or Tukey's"

 Bonferroni: Often used when the test a few pairs

comparisons average

 Tukey's: Often used when comparing the large

number of average pair

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The test of equal variances:

• If sig < 0.05, the variances of the groups are

not equal, statistical tables "ANOVA" no good use to test the hypothesis H0 Using test "Post hoc" (with variance conditions are not equal)

to find pairs of samples with different mean values If there is a valid sample pairs sig< 0.05, Reject H0 hypothesis The test "Post hoc" is

often used in this case is "Tamhane's T2"

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Analyze\ Compare Means\ One – Way Anova

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1

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Click

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2

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Exercises

Using data practices, test relationship of the following variables:

1) C3 & tncn 2) C3 & hocvan 3) C3 & tngd

4) Tuoi & tngd 5) C29c & hocvan 6) Sonk & hocvan 7) C36.2 & hocvan 8) C36.7 & tngd

9) C36.9 & dotuoi

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6 Test "ANOVA" many factors

• Analyze\ General linear model\ Univariate

Characteristics: analysis of the relationship

between a dependent variable and multiple variable factors (qualitative)

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• Analyze\ General linear model\ Univariate

Model for the study:

Y =  + A + B + AB + 

In which:

Y: The value of the dependent variable

: Average overall

A: Effects of fixed or random factor A to Y

B: Effects of fixed or random factor B to Y

AB: Effects of interaction between two factors AB

: residuals

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Analyze\ General linear model\ Univariate

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1

2

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For example, research on the value of the

average platelet (1000/mm3) of 3 groups of patients (fever, dengue, viral fever) in terms

of blood tests early or late Figures are as follows:

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• Analyze\ General linear model\ Multivariate

Characteristics: Analysis of the relationship

between the dependent variable and much more variable factors (qualitative)

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Analyze\ General linear model\ Multivariate

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1

2

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For example

Research on the height and age of the 18 students from urban areas (1), and 14 students from rural areas (2) as follows: The question is there any difference in height and age between students in two different areas?

• Analyze\ General linear model\ Multivariate

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id area age (month) Height (cm) area age (month) Height (cm)

Tiêu đề	Quantitative data analysis testing hypotheses about the mean value
Chuyên ngành	Statistics
Thể loại	Chapter

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