important role in many other problems where information is...
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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
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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.
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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.
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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.
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Luzon Visayas Mindanao
Unemployment 57 53 44
Undecided 72 40 48
Inflation 71 57 58
TOTAL 200 150 150
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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.
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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).
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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
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5. Calculation of Test Statistics:
500150150200Total
186585771Inflation
160484072Undecided
154445357Unemployment
TotalMindanaoVisayasLuzon
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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
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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:
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400118163119Total
70372310Above Average
174567342Average
156256467Below Average
TotalGoodFairPoor
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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
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5. Calculation of Test Statistics:
400118163119Total
70272310Above Average174567642Average
156256467Below Average
TotalGoodFairPoorPERFORMANCE
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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
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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
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Two-Fold Test ( 2 x2 contingency table)
11
22
crdf
DBCADCBABCADNx
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Problem 1: Is there any significance relationship b/n passing the board exam and success in career?
1005545Total
401525Unsuccessful
604020Successful
TotalPassFail
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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
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5. Calculation of Test Statistics
1005545Total
401525Unsuccessful
604020Successful
TotalPassFail
)55)(45)(40)(60()]4025()1520[(100 2
2 xxX
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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.
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Problem 2: Is there any significant relationship between sex and effectiveness in management?
1507278Total
753540Not Effective
753738Effective
TotalFemaleMaleEffectiveness
Sex
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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
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5. Calculation of Test Statistics
1507278Total
753540Not Effective
753738Effective
TotalFEMALEMALEEffectiveness
SEX
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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.