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E 45Johan Brink, IIE24 November

Quantitative data analysisLecture 6

23-04-19

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Agenda

Chapter 14Univariate analysisBivariate analysisMultivariate analysis

Contingency Pearson’s correlationt-testChi square

Factor & cluster analysis

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Univariate analysis

One variable at a time• Frequency tables – Bar charts• Grouping of ration & interval

variables: 20-29, 30-39… - Histograms

Arithmetic mean= Sum of all values/ # Values

=33.6Median= Mid point of distribution of

valuesMode=the most frequent value in the

distribution

0

5

10

15

20

25

30

35

40

-20 21-30 31-40 41-50 51-

Ages of gymvisitors

Interval/ Ratio – Scale & distance

Ordinal -ranked

Nominal –can’t be ranked

Mean Yes No -used anyhow

No

Median

Yes Yes No

Mode Yes Yes Yes

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DispersionRange: Min to Max (7 & 3)

Variance =standard deviation²

s²= Σ (x-M)²/ (n-1) = 1,5->Standard deviation =1,22

Point (x-M)²

1 5 0

2 5 0

3 6 10

4 5 0

5 4 1

6 4 1

7 7 4

8 3 4

9 6 1

Sum 45 12

Variance = 12/(9-1) =1,5

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Spurious correlationRelationship between two variables are caused by a thirds, underlying factor

Intervening variableChain of relationships

Moderating variableThe relationship between A & B only exist if C is percent

Multivariate analysis

A B

C

A B C

A B

C

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Analyzing data

Correlation & relationshipBetween variables (questions, groups, items)Does the answers on question 1 correlate

with answers on question 2?- Different questions/items for the same

constructor does it capture a relationship

Test – significant differencesBetween variables

-Questions, groups items -Across time/treatments

Is the mean different enough given the standard deviations?

t-test (Chi-squared (nominal scales)

Differences from the expected value?

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Constructs & ItemsVariable 1

Variable 2

Variable 3

Variable 4

Variable 5

Variable 6

• Cronbach α is a measure of how well variables measures the same underlying phenomena

Construct 1 Construct 2

Item Question 1

Item Question 2

Item Question 3

Item Question 4

Item Question 5

Item Question 6

Org. Culture Performance

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Hypothesis testing

H0There is no difference between group A and group B

H1There is a difference between group A and B

H0There is no connection between variable X and Y

H1There is a connection between variable X and Y

Real relationship (unknown)

H0 true H0 false

Result of statistical test

H0 rejected

Type 1 error

Correct

H0 not rejected

Correct Type 2 error

In order to reduce the risk of type 1 error, by increasing the level of significance from 5% to 1%,the risk of committing type 2 error increases!

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Bivariate analysis

Nominal Ordinal Interval/ration

Dichotomus

Nominal – Can’t be ranked

Contingency table, Chi square, Cramer’s V

Contingency table, Chi square, Cramer’s V

Contingency table, Chi square, Cramer’s V

Contingency table, Chi square, Cramer’s V

Ordinal -Ranked

Contingency table, Chi square, Cramer’s V

Spearman’s rho

Spearman’s rho

Spearman’s rho

Interval/ratio –scale & distance

Contingency table, Chi square, Cramer’s V

Spearman’s rho

Pearson’s r Spearman’s rho

Dichotomus –Yes/No

Contingency table, Chi square, Cramer’s V

Spearman’s rho

Spearman’s rho

Phi

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Contingency tables

Reasons

Male Female

# % # %

Relaxation

3 7 6 13

Fitness 15 36 16 33

Lose weight

8 19 25 52

Build strength

16 38 1 2

Total 42 100 48 100

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Pearson's correlation

For interval/ratio variablesMeasure of the strength of

association between two variables

r = Between -1 and +1, 0= no correlation & 1= perfect correlation

r²*100% = Variation caused /explained

X Y

1 2

2 3

3 4

4 5

5 5

5 6

7 7

7 8

8 9

9 8

R=0,969

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Pearson's correlation

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Pearson's correlation

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Pearson's correlation

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t-test: A statistical test to see if there is a difference between two samples

n1=25 n2=24 Mean1=64 Mean2=56S1=10 S2=8Df=n1+n2-2=47

t= (Mean1-Mean2)/√[((n1-1)s1²+ (n2-1)s2²)/(n1+n2-2))*(1/n1+1/n2))]

t= (641-56)/√[((25-1)10²+ (24-1)8²)/(25+24-2))*(1/25+1/24))] =3,08

Statistic table t(df=47, 0,05)=2,0123,08>2,012 thus reject H0, there is a

difference!

Hypothesis: H0, the true means is equal

Alternative, H1, there is a difference

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Chi square

Is there a difference between age groups (young, middle and old) and preference for A or B?

• Nominal scales!• 40 respondents

Chi²= Σ (observed-expected)²/expected

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Chi square

Young Middle age

Old Σ

A 20 5 0 25

B 5 5 5 15

Σ 25 10 5 40

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Chi square

Young Middle age

Old Σ

A 20 (15,625)

5 (6,25) 0 (3,125)

25

B 5 (9,375)

5(3,75) 5 (1,875)

15

Σ 25 10 5 40

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Chi square

Chi²= (20-15,625)²/15,625+ (5-6,25)²/6,25+ (0-3,125)²/3,125+ (5-9,375)²/9,375+ (5-3,75)²/3,75+ (5-1,875)²/1,875=12,27

Df= (r-1)*(c-1)= 2-1*3-1=2

Chi²0,05, df=2 =>5,99

12,27> 5,99

Thus reject H0, there is a difference between A and B and age group.

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Factor analysis and Cluster analysis• Reduce the data• Variables which measures the same thing• Underlying factors