Count Data. Data obtained by counting, as contrasted to data obtained by performing measurements on continuous scales.

Count data are also referred to as enumeration data.

Statistic for test concerning differences among proportions

(CHI – SQUARE STATISTICS, X2)

where: 0 = observed frequencye = expected frequency

e

eX2

2 0

Contingency Table. (Test for independence)

The X2 statistic plays an important role in many other problems where information is obtained by counting rather than measuring. This method we shall describe here applies to two kinds of problems, which differ conceptually but are analyzed the same way.

In the first kind of problem we deal with trials permitting more than two possible outcomes. 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 movie may be rated G, PG, R or X.

We could say that we are dealing with multinomial (rather than binomial) trials.

Also, in the illustration of the preceding section, each worker might have been asked whether unemployment is a more serious economic problem than inflation, whether inflation is a more serious economic problem than unemployment, or whether he or she is undecided and this might have resulted in the following table.

Luzon Visayas Mindanao

Unemployment 57 53 44

Undecided 72 40 48

Inflation 71 57 58

TOTAL 200 150 150

We refer to this kind of table as a 3 x 3 table (where 3 x 3 is read “3 x 3”), because it has 3 horizontal rows and 3 vertical columns; more generally, when there are r horizontal rows and c vertical columns, we refer to the table as an r x c table. Here, as in the table analyzed in the preceding section, the column totals representing the sample sizes, are fixed. On the other hand, the row totals depend on the responses of the persons interviewed, and, hence, on chance.

A. To show how an r x c table is analyzed, let us begin by illustrating the calculation of an expected cell frequency.

The expected frequency for any cell of a contingency table may be obtained by multiplying the total of the row to which it belongs by the total of the column to which it belongs and then dividing by the grand total for the entire table. Degrees of freedom, df = (r-1) (c-1).

Solution to Example in Chi-Square:1. H0 : For each alternative (unemployment,

undecided, and inflation), the probabilities are the same for the three parts of the country.

2. HA : For at least one alternative, the probabilities are not the same for the three country.

3. Test Statistics:X2 < Xc

2 : NS : Accept H0X2 > Xc

2 : S : Reject H04. Rejection Region: @ 0.01 level of

significance df = (r-1) (c-1) = (3 – 1) (3 – 1) = 4X2 = 13.277

5. Calculation of Test Statistics:

500150150200Total

186585771Inflation

160484072Undecided

154445357Unemployment

TotalMindanaoVisayasLuzon

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 0.156053 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 0 058 55.8 2.20 4.84 1.0867

4.0510

051.40 22

eeX

6. Conclusion

Since X2 = 4.051 does not exceed Xc2 =

13.277, the null hypothesis is accepted; the difference between the observed and expected frequencies may well be due to chance.

In the second kind of problem where the method of this section applies, the column totals as well as the row totals depend on chance. To give an example, suppose that a sociologist wants to determine whether there is a relationship between the intelligence of boys who have gone through a special job–training program and their subsequent performance in their jobs, and that a sample of 400 cases taken from very extensive files yielded the following results:

400118163119Total

70372310Above Average

174567342Average

156256467Below Average

TotalGoodFairPoor

Solution

1. H0 : Intelligence and on-the-job performance are independent

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

3. Test Statistics:X2 < Xc

2 : NS : Accept HoX2 > Xc

2 : S : Reject Ho4. Rejection Region: @ 0.01 Level of

Significancedf = (r – 1) ( c – 1) = (3 – 1) (3 – 1) = 4 Xc

2 = 13.277

5. Calculation of Test Statistics:

400118163119Total

70272310Above Average174567642Average

156256467Below Average

TotalGoodFairPoorPERFORMANCE

12.8353265.6916.320.7371.061430.25-5.528.5235.6077116.6410.820.8100.430622.094.751.3560.366826.015.170.9761.854096.04-9.851.8429.5869441.00-21.046.0250.00250.160.463.6649.1457424.3620.646.467

( o – e)2

e(o – e)2o - eeo

40.8909

6. Conclusion:

Since X2 = 40.89 which exceeds Xc

2 = 13.277, the null hypothesis is rejected: we conclude that there is a relationship between IQ and on-the-job performance.

8909.400 22

eeX

Two-Fold Test ( 2 x2 contingency table)

11

22

crdf

DBCADCBABCADNx

Problem 1: Is there any significance relationship b/n passing the board exam and success in career?

1005545Total

401525Unsuccessful

604020Successful

TotalPassFail

Solution to Two – Fold Examples:1. H0 : There is no significant relationship

between passing the board examination and success in career.

2. HA : There is a significant relationship between passing the board examination and success in career.

3. Test Statistics:X2 < Xc

2 : NS : Accept HoX2 > Xc

2 : S : Reject Ho4. Rejection Region: @ 0.05 Level of

Significancedf = (r – 1) (c – 1) = (2 – 1) (2 – 1) = 1X2

c = 3.841

5. Calculation of Test Statistics

1005545Total

401525Unsuccessful

604020Successful

TotalPassFail

)55)(45)(40)(60()]4025()1520[(100 2

2 xxX

6. Conclusion:

Since the computed value of chi-square is greater than the critical value, therefore, there is a significant relationship between passing the board exam and success in career, hence, the null hypothesis is rejected.

Problem 2: Is there any significant relationship between sex and effectiveness in management?

1507278Total

753540Not Effective

753738Effective

TotalFemaleMaleEffectiveness

Sex

Solution to Problem No. 2:

1. H0 : There is no significant relationship between sex and effectiveness in management.

2. HA : There is a significant relationship between sex and effectiveness in management.

3. Test Statistics:X2 < Xc

2 : NS : Accept H0

X2 > Xc2 : S : Reject H0

4. Rejection Region: @ 0.05 Level of Significancedf = ( r – 1) (c – 1) = (2 – 1) (2 –1) = 1X2

c = 3.841

5. Calculation of Test Statistics

1507278Total

753540Not Effective

753738Effective

TotalFEMALEMALEEffectiveness

SEX

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.