Count Data PDF

8/12/2019 Count Data PDF

1/26


2/26

Co u n t D a t a . Data obtained by counting, ascontrasted to data obtained by performing

measurements on continuous scales.Count data are also referred to as

enumeration data.

Statistic for test concerning differencesamong proportions

(CHI SQUARE STATISTICS, X2)

where: 0 = observed frequency

e = expected frequency

e

eX

2

2 0


3/26

Contingency Table.(Test for independence)

The X2 statistic plays animportant role in many other problemswhere information is obtained bycounting rather than measuring. Thismethod we shall describe here appliesto two kinds of problems, which differconceptually but are analyzed thesame way.


4/26

In the first kind of problem we deal with

trials permitting more than two possibleoutcomes. For instance, the weather can get

better, remain the same or get worse; an

undergraduate can be a freshman, a

sophomore, a junior, or a senior; and a moviemay be rated G, PG, R or X.

We could say that we are dealing with

multinomial (rather than binomial) trials.


5/26

Also, in the illustration of thepreceding section, each worker mighthave been asked whetherunemployment is a more seriouseconomic problem than inflation,whether inflation is a more seriouseconomic problem thanunemployment, or whether he or sheis undecided and this might haveresulted in the following table.


6/26

Luzon Visayas Mindanao

Unemployment 57 53 44

Undecided 72 40 48

Inflation 71 57 58

TOTAL 200 150 150


7/26

We refer to this kind of table as a

3 x 3 table (where 3 x 3 is read 3 x3), because it has 3 horizontal rowsand 3 vertical columns; moregenerally, when there are r horizontal

rows and c vertical columns, we referto the table as an r x c table. Here, asin the table analyzed in the precedingsection, the column totalsrepresenting the sample sizes, are

fixed. On the other hand, the rowtotals depend on the responses of thepersons interviewed, and, hence, onchance.


8/26

A. To show how an r x c table is

analyzed, let us begin by illustratingthe calculation of an e x p e c t e d c e l lf r e q u e n c y .

The expected frequency for anycell of a contingency table may beobtained by multiplying the total ofthe row to which it belongs by the

total of the column to which it belongsand then dividing by the grand totalfor the entire table. Degrees offreedom, d f= (r-1) (c-1).


9/26

Solution to Example in Chi-Square:

1. H0 : For each alternative (unemployment,undecided, and inflation), theprobabilities are the same for the threeparts of the country.

2. HA

: For at least one alternative, theprobabilities are not the same for thethree country.

3. Test Statistics:X2 < Xc

2 : NS : Accept H0

X2

> Xc2

: S : Reject H04. Rejection Region: @ 0.01 level ofsignificance

d f= (r-1) (c-1) = (3 1) (3 1) = 4X2 = 13.277


10/26

5. Calculation of Test Statistics:

500150150200Total

186585771Inflation

160484072Undecided

154445357Unemployment

TotalMindanaoVisayasLuzon


11/26

o e o e (o e)2 (o-e)2

e

57 61.6 -4.60 21.16 0.343572 64.0 8.00 64.00 1.000071 74.4 - 3.40 11.60 0.1560

53 46.2 6.80 46.24 1.000940 48.0 - 8.00 64.00 1.333357 55.8 1.20 1.44 0.025844 46.2 -2.20 4.84 1.104848 48.0 0 0 0

58 55.8 2.20 4.84 1.0867

4.0510

051.4

0 2

2

e

eX


12/26

6. Conclusion

Since X2

= 4.051 does not exceed Xc2

=13.277, the null hypothesis is accepted; thedifference between the observed andexpected frequencies may well be due tochance.

In the second kind of problem where themethod of this section applies, the columntotals as well as the row totals depend onchance. To give an example, suppose that asociologist wants to determine whether thereis a relationship between the intelligence ofboys who have gone through a special jobtraining program and their subsequentperformance in their jobs, and that a sampleof 400 cases taken from very extensive filesyielded the following results:


13/26

400118163119Total

70372310Above Average

174567342Average

156256467Below Average

TotalGoodFairPoor


14/26

Solution

1. H0 : Intelligence and on-the-jobperformance are independent

2. HA : Intelligence and on-the-jobperformance are not independent.

3. Test Statistics:X2 < Xc

2 : NS : Accept HoX2 > Xc

2 : S : Reject Ho

4. Rejection Region: @ 0.01 Level ofSignificancedf = (r 1) ( c 1) = (3 1) (3 1) = 4Xc

2 = 13.277


15/26

5. Calculation of Test Statistics:

400118163119Total

70272310Above Average

174567642Average

156256467Below Average

TotalGoodFairPoor

PERFORMANCE


16/26

12.8353265.6916.320.737

1.061430.25-5.528.523

5.6077116.6410.820.810

0.430622.094.751.356

0.366826.015.170.976

1.854096.04-9.851.842

9.5869441.00-21.046.025

0.00250.160.463.6649.1457424.3620.646.467

( o e)2

e

(o e)2o - eeo

40.8909


17/26

6. Conclusion:

Since X2 = 40.89 which exceedsXc

2 = 13.277, the null hypothesis isrejected: we conclude that there is arelationship between IQ and on-the-

job performance.

8909.40

0 2

2

e

eX


18/26

Two-Fold Test( 2 x2 contingency table)

11

2

2

crdf

DBCADCBA

BCADNx


19/26

Problem 1: Is there any significance

relationship b/n passing the board examand success in career?

1005545Total

401525Unsuccessful

604020Successful

TotalPassFail


20/26

Solution to Two Fold Examples:

1. H0 : There is no significant relationshipbetween passing the board examination andsuccess in career.

2. HA : There is a significant relationshipbetween passing the board examination and

success in career.3. Test Statistics:

X2 < Xc2 : NS : Accept Ho

X2 > Xc2 : S : Reject Ho

4. Rejection Region: @ 0.05 Level ofSignificance

d f= (r 1) (c 1) = (2 1) (2 1) = 1

X2c = 3.841


21/26

5. Calculation of Test Statistics

1005545Total

401525Unsuccessful

604020Successful

TotalPassFail

)55)(45)(40)(60(

)]4025()1520[(100 22 xxX


22/26

6. Conclusion:

Since the computed value of chi-square is greater than the criticalvalue, therefore, there is a significantrelationship between passing theboard exam and success in career,hence, the null hypothesis is rejected.


23/26

Problem 2: Is there any significantrelationship between sex and effectiveness

in management?

1507278Total

753540Not Effective

753738Effective

TotalFemaleMaleEffectiveness

Sex


24/26

Solution to Problem No. 2:

1. H0 : There is no significant relationshipbetween sex and effectiveness inmanagement.

2. HA : There is a significant relationship

between sex and effectiveness inmanagement.

3. Test Statistics:

X2 < Xc2 : NS : Accept H0

X

2

> Xc2

: S : Reject H04. Rejection Region: @ 0.05 Level ofSignificance

df = ( r 1) (c 1) = (2 1) (2 1) = 1

X2c = 3.841


25/26

5. Calculation of Test Statistics

1507278Total

753540Not

Effective

753738Effective

TotalFEMALEMALEEffectiven

ess

SEX


26/26

1068.0

)72)(78)(75)(75(

)]4037()3538[(150 2

2

xxX

6. Conclusion:

Since X2 = 0.1068 < than Xc2 = 3.841, then

the null hypothesis is accepted, therefore is no

significant relationship between sex and

effectiveness in management.

Date post:	03-Jun-2018
Category:	Documents
Upload:	rhoda-mae-dano-jandayan
View:	229 times
Download:	0 times

Count Data PDF

Documents