exploring statistics

7/28/2019 Exploring Statistics

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STATISTICS

Help decision makers extract

maximum usefulness from

limited information

• FACILITATE WISE DECISION IN THE FACE OF UNCERTAINTY



DESCRIPTIVE STATISTICS

Techniques for the collection

and effective presentation of numerical information

• GIVE INFORMATION OR DESCRIBE THE SAMPLE

MEAN

VARIANCE

STANDARD DEVIATION



INFERENTIAL STATISTICS

Techniques for analyzing

numerical information

• INFERENCES OF THE POPULATION FROM MEAN AND VARIANCE

t – test; F – test

ANOVA

CHI-SQUARE TEST



STATISTICAL ANALYSIS

Compare the means relative tothe degree of variation among

scores in each group todetermine the probability that thecalculated differences betweenthe means reflect real differences

between subject groups and notchance occurrences

• TOOLS FOR DATA INTERPRETATION



STATISTICAL ANALYSIS

Difference between two

means is significant at 0.05level of significance implies a

probability of less than 5 out

of 100 that the difference isdue to chance

• TOOLS FOR DATA INTERPRETATION



DESCRIPTIVE STATISTICS• WHAT IS BEING DESCRIBED

Level of

measurement

Central

Tendency

Dispersion/

Variability

Relationship

NOMINAL Mode - Contingency

coefficientORDINAL Median

Mode

Range Rank order

correlation coef.

INTERVAL/

RATIO

Mean

MedianMode

Standard Dev.

VarianceRange

Pearson product

moment

correlation



MEASURES OF CENTRAL TENDENCY

Average, computed by adding a list of scores and then dividing by the number of scores.

∑X

X = --------

N

X – mean∑X – sum of the individual scores

N – number of scores

MEAN



MEASURES OF DISPERSION

Measure of the spread of a set of scores

N∑X2 – (∑X)2

s = --------------------

N(N – 1)

s – standard deviation∑X – sum of the individual scores

N – number of scores

STANDARD DEVIATION



INFERENTIAL STATISTICS• DEPEND ON HYPOTHESES AND SCALE OF VARIABLE

STATISTICAL TEST HYPOTHESIS TESTED

t - test About a single mean

Difference between 2 means

One-way ANOVA Two or more population means

are equal – single independent

variableTwo-way ANOVA Two or more population means

are equal – two independent

variables

PARAMETRIC TESTS



PARAMETRIC TEST

Normal distribution

Homogeneity of variance

Continuous equal interval

measures

• EMPLOY INTERVAL MEASUREMENT OF THE DEPENDENT VARIABLE



INFERENTIAL STATISTICS• DEPEND ON HYPOTHESES AND SCALE OF VARIABLE

STATISTICAL TEST HYPOTHESIS TESTED

Chi-square test Two variables are independent

Medians of two or more population

are equal

Mann-Whitney U-test No difference in the scores from

two population

Fisher’s Z-

transformation test

Two population correlation

coefficients are equal

NON-PARAMETRIC TESTS



NON-PARAMETRIC TEST

Require neither normaldistributions or equal group

variancesUseful for large samples thatdon’t satisfy assumptions for parametric techniques

For very small samples andfor studies involving ordinalmeasuring devices

• BASED ON NOMINAL OR ORDINAL MEASUREMENT TECHNIQUES



SELECTING A STATISTICAL TEST

TYPE AND NUMBER OF INDEPENDENT VARIABLES

INTERVAL ORDINAL NOMINAL

1 More Than 1 1 More Than 1 1 More Than 1

TypeAnd

Number

Of

DependentVariables

INTERVAL 0 - Factor Analysis

Transform ordinal variable intonominal and use C-1, or; Transform the interval variable

into ordinal and use B-2, or; Transform both variables intonominal and use C-3.

- - ROW1

1 Correlation Multiple

Correlation

Analysis of

Variance, or

t-Test

Analysis of

Variance

More

Than 1

Multiple

Correlation

- - -

ORDINAL 0 Transform ordinal variable into

nominal and use C-1, or; Transform the interval variable

into ordinal and use B-2, or; Transform the interval variableinto nominal and use C-2.

- Coefficient of

Concordance(W)

- - ROW

2

1 Spearman

Correlation,Kendall’s T

- Sign Test,

Median Test,U-Test,

Kruskal-Wallis

Friedman’s

Two-Way Analysis of

Variance

More

Than 1 - - - -

NOMINAL 0 - - - - - Chi-Square ROW

3

1 Analysis of Variance

- Sign Test,Median Test,

U-Test,

Kruskal-Wallis

- Phi Coeff. Ø, Fisher Exact

Test,

Chi-Square

-

More

Than 1 Analysis of

Variance- Friedman’s

Two-Way

Analysis of

Variance

- - -

COLUMN A COLUMN B COLUMN C



COMMON STATISTICAL TESTS• EVALUATE THE MATCH BETWEEN DATA FROM TWO OR MORE SAMPLES

t - test Nominal independent variable andordinal dependent variable with two

conditions or levels

Analysis of variance More than two conditions or more

than one independent variableChi-square statistic Two nominal variables

Spearman rank-order

correlation

Two ordinal variables

Pearson product

moment correlation

Two interval variables

Mann-Whitney U-test Nominal independent variable and

ordinal dependent variable



t – Test Statistic

Determine the probability that the

difference between the two means

reflects a real difference between the

groups of subjects rather than a

chance variation in data

X1 – X2

t = -----------------------------------------------

(N1 – 1)S12 + (N2 – 1)S2

2 x N1 + N2

N1 + N2 – 1 N1N2

• TEST WHETHER THE MEANS OF THE TWO GROUPS ARE SIMILAR



t – Test WorksheetCompute group variances

N1∑X1

2

– (∑X1)

2

N2∑X2

2

– (∑X2)

2

S1

2 = --------------------------: S22 = ---------------------------

N1(N1 – 1) N2(N2 – 1)

Calculate t-value

(N1 – 1) S12 + (N2 – 1) S22

---------------------------------- = _______

N1 + N2 – 2

2. (N1 + N2)/N1N2 = _______

3. (Step 1 x Step 2) = _______

4. (Step 3)1/2 = _______

5. X1 – X2 = _______ 6. t = Step 5/Step 4 = ________

df = N1 + N2 – 2 = ________

Look up t-value in Table for Critical Values of t = ___;

p = 0.05

GROUP I II

N =

∑X =

∑X2 =

X =

If computed

t-value

exceeds the

tabular value,

null

hypothesis is

rejected.



t – Test Worksheet

Response

Categories

X1 f fX1 fX12 X2 f fX2 fX2

2

Strongly

Agree

5 5 25 125 5 7 35 175

Agree 4 8 32 128 4 10 40 160

Undecided 3 12 36 108 3 6 18 54

Disagree 2 3 6 12 2 3 6 12

Strongly

disagree

1 2 2 2 1 4 4 4

Total 30

N1

101

∑X1

375

∑X12

30

N2

103

∑X2

405

∑X22

GROUP IIGROUP I



t – Test WorksheetCompute group variances

N1∑X1

2

– (∑X1)

2

N2∑X2

2

– (∑X2)

2

S1

2 = --------------------------: S22 = ---------------------------

N1(N1 – 1) N2(N2 – 1)

S12 = 1.21 S2

2 = 1.77

Calculate t-value

(N1 – 1) S12 + (N2 – 1) S22

---------------------------------- = 1.49

N1 + N2 – 2

2. (N1 + N2)/N1N2 = 0.07

3. (Step 1 x Step 2) = 0.104

4. (Step 3)1/2 = 0.32

5. X1 – X2 = 0.066. t = Step 5/Step 4 = 0.1875

df = N1 + N2 – 2 = 58

Look up t-value in Table for Critical Values of t = 2.00

p = 0.05

GROUP I II

N = 30 30

∑X = 101 103

∑X2 = 375 405

X = 3.37 3.43

If computed

t-value

exceeds the

tabular value,

null

hypothesis is

rejected.



Chi-Square Statistic

Useful for nominal data but can also be

used for higher scales

Apply to cases where persons, events

or objects are grouped together intotwo or more nominal categories such

as yes-no, favor-undecided-against, or

class A, B, C, D

Calculated with actual counts

• MOST WIDELY USED NON-PARAMETRIC TEST OF SIGNIFICANCE




Applied to test of significance between

the observed distribution of data

among categories and expected

distributionExamine questions of relationship

Used in one-sample analysis, two

independent samples, or

k-independent samples

• MOST WIDELY USED NON-PARAMETRIC TEST OF SIGNIFICANCE




FORMULA:

(Oi – Ei)2

ϰ2 = ∑ -------------

Ei

O – observed frequencies

E – expected frequencies

• ONE-SAMPLE CASE ANALYSIS

k

i = 1




A survey of senior officers’ interest in pursuing the OSEC wasconducted. The researcher had interviewed 200 officers and

learned of their intentions to join the OSEC. Responses were

classified into four categories, as shown:

• ONE-SAMPLE CASE ANALYSIS

PURPOSE INTEND TOPURSUE OSEC

NO.INTERVIEWED

Increase

knowledge

16 90

Need for

promotion

13 40

Relaxation from

job

16 40

Save money 15 30

TOTAL 60 200



Chi-Square Statistic• ONE-SAMPLE CASE ANALYSIS

Purpose JoinOSEC No.Interviewed Percent(N/200)

ExpectedFrequency

(% x 60)

Knowledge 16 90 45 27

Relaxation 13 40 20 12

Promotion 16 40 20 12

Save 15 30 15 9

60 200 100 60

Null hypothesis: The proportion of thepopulation who intend to join the

OSEC is independent of the purpose.

• Calculate the expected distribution by

determining that proportion of the 200

respondents were in each group.

• Apply the proportion to the number

who intend to join the OSEC.

• Calculate chi-square:

x 2 = 9.89

df = k – 1 = 4 -1 = 3

D. Obtain the critical test value for x 2

considering df = 3; p = 0.05

x 2 = 7.82

Decision: Reject hypothesis

(O - E)2

X 2 = ∑ ----------------------

E

(16 –27)2 (13-12)2 (16-12)2 (15-9)2

X 2 = ----------- + ----------- + ----------- + ----------

27 12 12 9




FORMULA:

(Oi – Ei)2

ϰ2 = ∑∑ -------------

Ei

O – observed frequencies

E – expected frequencies

• TWO-SAMPLE CASE ANALYSIS

k

i = 1

Appropriate for situation in which a test for differences

between samples is required.




An officer has become interested in whether drinking alcoholhas some deleterious effects on police behavior. He wonder

whether drinking alcohol triggers police misdemeanors.

• TWO-SAMPLE CASE ANALYSIS

ALCOHOL

DRINKER

YES NO

Heavy 12 4

Moderate 9 6

Non-drinker 13 22

TOTAL 34 32




AlcoholDrinker YES NO Row Total

Heavy 12 4 16

Moderate 9 6 15

Non-drinker 13 22 35

Column

Total

34 32 66

• TWO SAMPLE CASE ANALYSISNull hypothesis: There is no difference in

police misdemeanor occurrencesbetween alcohol drinker and non-

drinker.

• Calculate the expected observations

in each cell by multiplying the two

marginal totals common to aparticular cell and dividing this

product by the total observations.

• Calculate chi-square:

x 2 = 6.86

df = (r – 1)(c – 1) = (3-1)(2-1) = 2

D. Obtain the critical test value for x 2

considering df = 2; p = 0.05

x 2 = 5.99

Decision: Reject hypothesis

(12 –8.24)2 (9-7.73)2 (13-18.03)2 (4-7.76)2 (6-7.27)2 (22-16.97)2

X 2 = ------------ + ----------- + --------------- + ----------- + ---------- + --------------

8.24 7.73 18.03 7.76 7.27 16.97

Alcohol

Drinker

YES NO Row Total

Heavy 8.24 7.76 16

Moderate 7.73 7.27 15

Non-drinker 18.03 16.97 35

Column

Total

34 32 66



Chi-Square Statistic• SAMPLE CASE ANALYSIS

• A researcher is interested in student attitude toward compulsory attendanceto OSEC recreational games. A random sample of students is drawn from

each of the four classes of the National Police College. The students in the

sample are then asked to respond whether they strongly agree, agree or

disagree to compulsory attendance

CLASS STRONGLY

AGREE

AGREE DISAGREE TOTAL

OSEC 60 12 48 20 80

OSEC 61 7 20 33 60

OSEC 62 6 19 35 60

OSEC 63 5 3 32 40

TOTAL 30 90 120 240




CLASS SD A D TOTAL

O-60 12 48 20 80

O-61 7 20 33 60

O-62 6 19 35 60

O-63 5 3 32 40

TOTAL 30 90 120 240

• SAMPLE CASE ANALYSIS

Null Hypothesis: The four class

populations do not differ in their

attitude toward compulsory attendance

to OSEC recreational games.

• Determine the value of the expected

outcome, E = (∑k∑r)/N

• Compute chi-square value:x 2 = 40.28

df = (r – 1)(c – 1) = 6

C. Obtain the critical test value:

x 2 = 12.59

p = 0.05

Conclusion: Class and attitude toward

compulsory attendance to OSEC

recreational games are related (not

independent) in the student population.

The null hypothesis is rejected.

CLASS SD A D TOTAL

O-60 10 30 40 80

O-61 7.5 22.5 30 60

O-62 7.5 22.5 30 60

O-63 5 15 20 40

TOTAL 30 90 120 240




• To study whether a relationship may exist between level of investigative competence and the amount of education received

by the investigators.

EDUCATION HighlyCompetent

BarelyCompetent

Incompetent TOTAL

High school 75 54 12 141

College 64 106 28 198

Masteral 28 82 51 161

TOTAL 167 242 91 500




EDUC. HC BC I TOTAL

HS 75 54 12 141

BS 64 106 28 198

MS 28 82 51 161

TOTAL 167 242 91 500


• Determine the value of the

expected outcome, E = (∑k∑r)/N

• Compute chi-square value:

x 2 = 59.37

df = (r – 1)(c – 1) = 4


x 2 = 9.49

p = 0.05

EDUC. HC BC I TOTAL

HS 47 68 26 141

BS 66 96 36 198

MS 54 78 29 161

TOTAL 167 242 91 500




• Decide whether the four groups really differ in opinionconcerning the components of a program toward crime

prevention.

PROGRAM Police LGU NGO Civilian TOTAL

Foot patrol 83 67 114 95 359

Motorized

patrol

37 33 86 55 211

TOTAL 120 100 200 150 570




Program P L N C TOTAL

Foot patrol 83 67 114 95 359

Motorized

patrol

37 33 86 55 211

TOTAL 120 100 200 150 570


• Determine the value of the

expected outcome, E = (∑k∑r)/N

• Compute chi-square value:

x 2 = 5.55

df = (r – 1)(c – 1) = 4


x 2 = 7.82

p = 0.05

Program P L N C TOTAL

Foot patrol 76 63 126 94 359

Motorized

patrol

44 37 74 56 211

TOTAL 120 100 200 150 570

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