inferential analysis spss

Dr Suhazeli AbdullahFMS KK MarangFMS KK MarangSempena Kursus SPSS dalam HSR

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Inferential Analysis Statistic with SPSS

“When you are willing to make sacrifices for a great cause, you will never be alone.”

1Kursus SPSS dalam HSR JKNT 2012

Which Test?Characterization of Variables

to be testedAppropriate Test of

SignificancegFirst Variable Second Variable

Continuous Dichotomous Student’s t-test / T-(Unpaired) Independent Test

Continuous Dichotomous (Paired) Paired t-test

Continuous Nominal (Qualitative) ANOVA

Continuous Continuous Pearson correlation coefficient Linear regression

O di l Di h t M Whit U t tOrdinal Dichotomous (Unpaired)

Mann Whitney U test

Ordinal Dichotomous (Paired) Wilcoxon sign rank

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( ) gKruska-Wallis test

Which Test?Characterization of Variables Appropriate Test of Significance

Which Test?to be tested

First Variable Second Variable

Ordinal Ordinal Spearman correlation coefficientOrdinal Ordinal Spearman correlation coefficient (rho) Kendall correlation coefficient (ta )(tau)

Ordinal Continuous Group the continuous variable and calculate rho or tau or chi-square

Dichotomous Dichotomous (Unpaired)

Chi-square or Fisher exact test( p )

Dichotomous Dichotomous (Paired) McNemar chi-square test

Dichotomous Nominal Chi-square test

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Nominal Nominal Chi-square test

• One sample t test(H 6 l/L?)(Ho:μ=6 mmol/L?)

• Comparing 2 means – Independent t test(Ho:μ1=μ2)(Ho:μ1 μ2)

• Comparing >2 means – One way ANOVA(Ho:μ1=μ2=μ3)

• Comparing Pre-& Post- means – Paired t test(Ho:μdifference=0)

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• Correlation(Ho:‘Rho’=0)

• Simple linear regressionC i ti ( i ti ) • Comparing proportion (association) – Chi-square test(Ho:P1=P2=….)

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File name:File name:Step 1Use “explore” if normality should be checked (n<30)(n<30)Step 2Analyze Compare Means One-sample T Test..Move the desired numerical variable(s) ( )into “Test Variable (s)” box Set desired “Test value” “Ok”

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How to How to analyze….analyze….

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Output 1:p

• p = 0.000 (<0.001)• There is significant different of mean age ofThere is significant different of mean age of

respondents from 50 years old.• The mean age of respondent was lower than 50

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years old.• We are 95% confidence that the different of mean

i th l ti i b t 2 84 t 1 3

• p = 0 000 (<0 001) p 0.000 (<0.001) • There is significant different of mean age of respondents from

50 years old. • The mean age of respondent was lower than 50 years old. • We are 95% confidence that the different of mean age in the

population is between -2.84 to -1.3.

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Step 1Step 1Analyze Descriptive Statistics Crosstabs.. Move each categorical variable to row and column box Click “ statistics” button and check the “Chi-square” option, and “Continue” Perhaps you may click “cells” button and check desired options, and “Continue” “OK”

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13• To know ass between passive smoking and SGA

Smoking * SGA Crosstabulation

41 20 6167.2% 32.8% 100.0%

Count% within Smoking

NoSmokingNormal SGA

SGATotal

67 89 15642.9% 57.1% 100.0%

108 109 21749.8% 50.2% 100.0%

Count% within SmokingCount% within Smoking

Passive

Total9 8% 50 % 00 0%g

Chi-Square Tests

10.328b 1 .0019 380 1 002

Pearson Chi-SquareContinuity Correctiona

Value dfAsymp. Sig.

(2-sided)Exact Sig.(2-sided)

Exact Sig.(1-sided)

9.380 1 .00210.488 1 .001

.002 .001

10.280 1 .001

Continuity Correctiona

Likelihood RatioFisher's Exact TestLinear-by-LinearA i ti 10.280 1 .001

217AssociationN of Valid Cases

Computed only for a 2x2 tablea.

0 cells ( 0%) have expected count less than 5 The minimum expected count is 30b

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0 cells (.0%) have expected count less than 5. The minimum expected count is 30.36.

b.

• Ini menunjukkan bahawa dikalangan perokok pasif, peratus Ini menunjukkan bahawa dikalangan perokok pasif, peratus SGA lebih tinggi iaitu 57.1% berbanding dengan yang tidak iaitu 32.8%.

• Dari jadual seterusnya, nilai chi square ialah 10.328 dan nilai p ialah 0.001.

• Maka terbukti ada hubungan antara perokok pasif dan • Maka terbukti ada hubungan antara perokok pasif dan kejadian SGA

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• Jadual 1: Jadual kontigensi menunjukkan hubungan antara Jadual 1: Jadual kontigensi menunjukkan hubungan antara risiko rokok dengan kejadian SGA.

KumpulanNormal SGA Jumlah

Tidak merokok

41 20 61

Perokok pasif 67 89 156Jumlah 108 109 217

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X2 = 10.328, p = 0.001

• Sekiranya dalam jadual 2x2, ada nilai sel jangkaan yang Se a ya da a jadua , ada a se ja g aa ya g kurang dari 5, maka nilai p dan nilai X2 yang dibaca ialah nilai p di baris CONTINUITY CORRECTION.

• Ianya adalah serupa seperti kiraan Yates Correction Tetapi • Ianya adalah serupa seperti kiraan Yates Correction. Tetapi sekiranya saiz sampel lebih kecil iaitu kurang dari 40, maka nilai yang dibaca ialah nilai p dan nilai 2 pada baris Fishers' Exact Test Test.

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Chi S T tChi-Square Tests

Value dfAsymp. Sig.

(2-sided)Exact Sig.(2-sided)

Exact Sig.(1-sided)

10.328b 1 .0019.380 1 .002

10 488 1 001

Pearson Chi-SquareContinuity Correctiona

Likelihood Ratio

( ) ( ) ( )

10.488 1 .001.002 .001

10.280 1 .001

Likelihood RatioFisher's Exact TestLinear-by-LinearAssociation

217AssociationN of Valid Cases

Computed only for a 2x2 tablea.

0 cells (.0%) have expected count less than 5. The minimum expected count is 30.36.

b. 1

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File name: sga.savFile name: sga.savStep 1Use “explore” if normality should be checked (n<30)Use explore if normality should be checked (n<30)

Step 2Analyze Compare Means Independent Samples T T t M th d i d i l i bl ( ) i t T Test.. Move the desired numerical variable(s) into “Test Variable (s)” box Move the categorical variable (which describes the 2 groups) into ( g p )“Grouping Variable” box Click “Define groups” button to define 2 groups then “continue” and “Ok”

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Ok

H t lHow to analyze….

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How to analyzeHow to analyze….

0= No

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SGA1= SGA

Group Statistics

108 58.666 11.2302 1.0806109 50 657 10 3512 9915

SGANormalSGA

Weight at first ANCN Mean Std. Deviation

Std. ErrorMean

Independent Samples Test

L ' T t f

109 50.657 10.3512 .9915SGA

Levene's Test forEquality of Variances

Mean Std. Error

95% ConfidenceInterval of the

Difference

t-test for Equality of Means

.787 .376 5.463 215 .000 8.0089 1.4660 5.1193 10.8984

5.461 213.251 .000 8.0089 1.4665 5.1181 10.8996

Equal variancesassumedEqual variances

t d

Weight at first ANF Sig. t df Sig. (2-tailed)

MeanDifference

Std. ErrorDifference Lower Upper

5.461 213.251 .000 8.0089 1.4665 5.1181 10.8996not assumed

Levene’s test result: As it’s P value is more than 0.05, variances of 2 groups are different. So that we can use “equal variance assumed” line result.

If p<0 05 please use “equal variance no assumed” line result

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If p<0.05 please use equal variance no assumed line result.

T li d l lTulis dalam laporan

Kumpulan N Min Ujian pp j p

Normal 108 58.67 + 11.32Ujian TUjian T

t = 5.463 <0.0005SGA 109 50.66 + 10.35

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File name: sga-pair.savStep 1Step 1If we need to check normality (n<30),Step 2Analyze Compare Means Paired-Analyze Compare Means Paired-Samples T Test..Move the desired 2 numerical variable(s) into “Paired Variables:” numerical variable(s) into Paired Variables: box then “Ok”

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How to analyze….How to analyze….How to analyze….How to analyze….

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Descriptive statistics of each variable

Paired Samples Statistics

variable

Paired Samples Statistics

10.247 70 .3566 .0426hb2PairMean N Std. Deviation

Std. ErrorMean

10.594 70 .9706 .1160hb31

Paired Samples Test

95% ConfidenceI t l f th

Paired Differences

-.3471 .9623 .1150 -.5766 -.1177 -3.018 69 .004hb2 - hb3Pair 1Mean Std. Deviation

Std. ErrorMean Lower Upper

Interval of theDifference

t df Sig. (2-tailed)

Mean of (hypnosis-natural state)

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score is significantly more than zero.

Kumpulan n mean Test p

Hb sebelum t

70 0.347 +0 96

Paired T T t

0.004rawatan 0.96 Test

T = Hb selepas3.018

Hb selepas rawatan

Wujud perbezaan paras hemoglobin sebelum dan selepas i i i ifi

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intervensi yang significant

File name: sga.sav

Step 1Use “explore” if normality should be checked (n<30)

Step 2Analyze Compare Means One way ANOVA..Move the desired numerical variable(s) into “Dependent list:” box Move the categorical Dependent list: box Move the categorical variable (which describes groups) into “Factor” box Click “Options” button and check “Descriptive” and “Homogeneity of variance” option “continue” and

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Homogeneity-of-variance option continue and “Ok”

Step 3Step 3If ANOVA is significant, you should do Post-Hoc tests (Click “Post Hoc…” button Select one of the post hoc tests e.g. Scheffe “Continue” “Ok”

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Descriptives

Birth weight

151 2.7801 .52623 .04282 2.6955 2.8647 1.90 4.7223 2 7643 60319 12577 2 5035 3 0252 1 60 3 96

HousewifeOffice work

N Mean Std. Deviation Std. Error Lower Bound Upper Bound

95% Confidence Interval forMean

Minimum Maximum

Between-Component

Variance

23 2.7643 .60319 .12577 2.5035 3.0252 1.60 3.9644 2.8430 .55001 .08292 2.6757 3.0102 1.90 3.79

218 2.7911 .53754 .03641 2.7193 2.8629 1.60 4.72.53938 .03653 2.7191 2.8631

.03653a 2.6339a 2.9483a -.00420

Office workField workTotal

Fixed EffectsRandom Effects

Model

W i B t t i i ti It l d b 0 0 i ti thi d ff ta Warning: Between-component variance is negative. It was replaced by 0.0 in computing this random effects measure.a.

ANOVA

Birth weight

.153 2 .077 .263 .769Between Groups

Sum ofSquares df Mean Square F Sig.

62.550 215 .29162.703 217

Within GroupsTotal

Tidak wujud perbezaan bermakna diantara jenis pekerjaan dengan berat kelahiran bayi.

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y

Jika wujud, maka kita kena lihat kumpulan mana satu melalui post-hoc LSD

P t h LSDPost-hoc LSD

Multiple Comparisons

Dependent Variable: Birth weightDependent Variable: Birth weightLSD

Mean

.01572 .12073 .897 -.2222 .2537- 06289 09240 497 - 2450 1192

(J) Type of workOffice workField work

(I) Type of workHousewife

Difference(I-J) Std. Error Sig. Lower Bound Upper Bound

95% Confidence Interval

.06289 .09240 .497 .2450 .1192-.01572 .12073 .897 -.2537 .2222-.07861 .13878 .572 -.3522 .1949.06289 .09240 .497 -.1192 .2450

Field workHousewifeField workHousewife

Office work

Field work.07861 .13878 .572 -.1949 .3522Office work

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Kumpulan Min Ujian P

Pekerja Pejabat 2.67 + 0.60

AnovaSRT 2.78 + 0.53 AnovaF=0.263

0.769

K j L 2 84 0 55Kerja Luar 2.84 + 0.55

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• Does 2 variables associated?oes a ab es assoc a ed?• Direction (-ve/+ve)?• How strong?

r>0 75 very good/perfect correlationr>0.75 very good/perfect correlation

r>0.50 - <0.75 moderate/good l ticorrelation

r>0.25 - <0.50 fair correlation

r<0.25 (little or no correlation)

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File name: recall scoreStep 1Graphs Scatter Simple Move 2 Graphs Scatter… Simple Move 2 variables into Y axis (Dependent*) and X axis (Independent*) “Ok”(Independent ) Ok[*If there is a direction of relationship]

Step 1Step 1Analyze Correlate… Bivariate Move 2

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variables into “Variables:” box – Check “Pearson” or “Spearman” “Ok”

How to analyze….How to analyze…. 10How to analyze….How to analyze….

mm

ol/l

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8

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hole

stro

l lev

el in

m

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2 3

diet score reflecting the content of cholestrol

70605040302010

bloo

d c 4

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-There is a linear relationship between 2relationship between 2

variables.-Positive correlationPositive correlation

-If diet score increase, blood cholesterol level

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will increase.


12 4

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Output 5: P-value

r

- We conclude that in the population, there is a linearcorrelation between diet score and blood cholesterol levelcorrelation between diet score and blood cholesterol level

(p<0.001).- Our sample suggests that there is a good positive linear

correlation between diet score and blood cholesterol levelcorrelation between diet score and blood cholesterol level (r=0.666).

- It means that, if the diet score is increased, the blood h l t l l l ill b i d

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cholesterol level will be increased.

• We conclude that in the population, there is a linear correlation We conclude that in the population, there is a linear correlation between diet score and blood cholesterol level (p<0.001).

• Our sample suggests that there is a good positive linear correlation between diet score and blood cholesterol level (r=0.666).

• It means that if the diet score is increased the blood • It means that, if the diet score is increased, the blood cholesterol level will be increased.

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• Does 2 variables associated?• Direction ( ve/+ve)?• Direction (-ve/+ve)?

• How strong?• Strength of association

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File name: recall scoreStep 1Step 1As above

Step 1Analyze Regression Linear Move 2 variables into Analyze Regression… Linear Move 2 variables into Dependent and Independent boxes Click the “Statistics…” button Check “Confidence Intervals” Continue Ok

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How to analyze….How to analyze…. 10Double How to analyze….How to analyze….

mm

ol/l

9

8

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click on the graph

5

hole

stro

l lev

el in

m

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5

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2 3

diet score reflecting the content of cholestrol

70605040302010

bloo

d c 4

3

114

6

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- It is linear and probably both normality & equal variance assumptions are satisfied

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q p


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3

4

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Output 7:p

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• We conclude that in the population, there We conclude that in the population, there is a linear correlation between diet score and blood cholesterol level (p<0.001).

• Our sample suggests that the ‘linear’ relationship can be interpreted as………..Blood cholesterol level=3.25 + (0.074*diet score).

• It means that if the diet scores higher by 100 units, the blood cholesterol level will be i d b 7 4 itincreased by 7.4 units.

• We are 95% sure that the ‘population’ beta ill li b t 0 062 d 0 085

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will lie between 0.062 and 0.085.

• We conclude that in the population there is • We conclude that in the population, there is a linear correlation between diet score and blood cholesterol level (p<0.001). b ood c o es e o e e (p 0.00 ).

• Our sample suggests that the ‘linear’ relationship can be interpreted asrelationship can be interpreted as………..Blood cholesterol level=3.25 + (0.074*diet score)score).

• It means that if the diet scores higher by 100 units the blood cholesterol level will be units, the blood cholesterol level will be increased by 7.4 units. W 95% th t th ‘ l ti ’ b t

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• We are 95% sure that the ‘population’ beta will lie between 0.062 and 0.085.

chol(pred.)=7.30 + (0.3*age) - (.54*exercise) + (.39*diet)

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• There is a significant linear relationship between diet inventory There is a significant linear relationship between diet inventory index and cholesterol level (p<.001). Those with 1 unit more in the index, have cholesterol level higher for 0.39 mmol/L (95% CI: 0 29 0 50 mmol/L)0.29, 0.50 mmol/L).

• With the 3 significant variables the model explains 69% of • With the 3 significant variables, the model explains 69% of variation of the blood cholesterol level in the sample (R2=0.69)

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Thank youy

inferential analysis spss

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