pearsoncorrelation (recovered 1) (recovered)

7/26/2019 PearsonCorrelation (Recovered 1) (Recovered)

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bas statistics for Social Science slide 1

Pearson

Product-MomentCorrelation


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Pearson Product-

Moment Correlation

Purpose determine relationship between twometric variables

Requirement:

DV - Interval/Ratio

IV - Interval/Ratio

berarti data nya parametrik: berjenis interval dan ratio

dependentvariable

independentvariable


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What to Expect

Strength

DescribeDirection

Calculate r

)()( SSYSSX

SXYr=

Hypothesis Test

kekuatan magnitude/ hub


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Components ofPearson ranalysis

DirectionPearson corr.

coefficient, rStrength/

MagnitudeDescriptive

Inferential

Hypothesis Test:

HO: = 0HA: 0

> 0

< 0


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1Descriptive

Calculate r using the formula

=

=

22

22 )()(

)()(

)()(

n

YY

n

XX

n

YXXY

SSYSSX

SXYr +/-

Describe the nature of relationship between the two

variables in terms of:

Direction

Strength or magnitude using Guildfords Rule

of Thumb

+ +


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Scatter Plot/Scatter gram

X

X

X

Y

Y

Positive

Relationship

Negative

Relationship

No

Relationship

Y


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Strength/MagnitudeCorrelation coefficient, r

r-1 1


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Guildford Rule of Thumb

r Strength of Relationship

< .2 Negligible Relationship

.2 - .4 Low relationship

.4 - .7 Moderate relationship

.7 - .9 High relationship

> .9 Very high relationship

tidak ada hub0.00


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2 InferentialSteps in Hypothesis Testing

1. State the null and alternative hypotheses

HO: = 0

HA: 0

< 0

> 0

2. Calculate the test statistic

2

2

1

=

n

r

rt


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3. Determine critical valuedf = n 2

One-tailed - t, df

Two-tailed - t/2, df

4. Make your decision

Criteria Decision

|tcal| > |tcritical| Reject HO

|tcal| |tcritical| Fail to reject HO

Manual

Criteria Decision

Sig-t< Reject HOSig-t Fail to reject H

SPSS

Reject HO:

Significant relationship between the two

variables

Fail to reject HO:No significant relationship between the two

variables

5. Make conclusion

O


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Example 1:

Data were collected from a randomly selected sample to determine

relationship between average assignment and test scores in statistics.

Distribution for the data is presented in the table below. Assuming

the data are normally distributed,

Data set:

Assign Test1. Calculate an appropriate correlation

coefficient,

2. Describe the nature of relationship

between the two variables, and3. Test the hypothesis on the

relationship at .01 level of

significance.

8.5 88

6 669 94

10 98

8 87

7 72

5 456 63

7.5 85

5 77

Data: 5950 Pearson 1 Class


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Example 1 Pearson correlation

1. Calculate the correlation coefficient ID X Y1 8.5 88

2 6 66

3 9 94

4 10 985 8 87

6 7 72

7 5 45

8 6 639 7.5 85

10 5 77

Summary stat:

n 10

X 72Y 775

X2 544.5

Y2 62,441

XY 5,795.5

( )( )

865.

1563.249

5.215

5.23781.26

5.215

10)775(441,62

10)72(5.544

10

)775()72(5.795,5

)()(

)()(

22

22

22

=

==

=

=

n

Y

Yn

X

X

n

YXXY

r

2. Based on r =.865, there is apositive and high relationship

between average assignment and test scores.


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3. Test the hypothesis on the relationship between the two

variables at .01 level of significance.


HO: = 0

HA: 0

2. Calculate test statistic

876.4

1774.

865.

210

865.1

0865.

21

2

2

=

=

=

=

nr

rt


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3. Determine critical value

Critical t.005, 8 = 3.355

4. Decision

Since |tcal| (4.876) is bigger than |tcritical| (3.355)

Reject the null hypothesis

5. ConclusionConclude there is a significant relationship between

average assignment and test scores at .01 level of

significance


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: Pearson Product-

Moment Correlation

Correlations

1 .865**

. .001

10 10

.865** 1

.001 .

10 10

Pearson Correlation

Sig. (2-tailed)

N

Pearson Correlation

Sig. (2-tailed)

N

Average assignment

scores

Test scores

Average

assignment

scores Test scores

Correlation is significant at the 0.01 level (2-tailed).**.


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Example 2:

Reviews of literature show that anxiety is correlated negatively with

team cohesiveness. To ascertain this relationship, Dr Irman collected

data from a randomly selected sample. Summary of the data follow:

1. Calculate an appropriate correlation

coefficient,

2. Describe the nature of relationship

between the two variables, and3. Test the hypothesis on the

relationship at .05 level of

significance.

Summary Statistics:

n 21

X 417

Y 446

X2 8,697

Y2 9,802

XY 8,566

Data: 5950 Pearson 1 Class


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1. Calculate the correlation coefficient Summary Statistics:n 21

X 417

Y 446

X2 8,697Y2 9,802

XY 8,566

( )( )

783.

6605.370

2857.290

8095.3295714.416

2857.290

21)446(802,9

21)417(697,8

21

)446()417(566,8

)()(

)()(

22

22

22

=

=

=

=

=

n

Y

Yn

X

X

n

YXY

r

2. Based on r = -.783, there is a negative and high relationship

between anxiety and team cohesiveness.


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3. Test the hypothesis on the relationship at .05 level of

significance.


HO: = 0

HA: < 0

2. Calculate test statistic

487.5

1427.

783.

221

)783.(1

0783.

21

2

2

=

=

=

=

nr

rt


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3. Determine critical value

Critical t. 05, 19 = 1.729

4. Decision

Since |tcal| (-5.487) is bigger than |tcritical| (1.729)

Reject the null hypothesis

5. ConclusionConclude there is a significant relationship between

anxiety and team cohesiveness at .05 level of

significance


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SPSS: Pearson Product-

Moment Correlation

Correlations

1 -.783**

. .000

21 21

-.783** 1

.000 .

21 21

Pearson Correlation

Sig. (1-tailed)

N

Pearson Correlation

Sig. (1-tailed)

N

Anxiety

Team cohesiveness

Anxiety

Team

cohesiveness

Correlation is significant at the 0.01 level (1-tailed).**.


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pearsoncorrelation (recovered 1) (recovered)

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