CHI SQUARE
NON PARAMETRICTESTCHI SQUAREKarl Pearson's famous chi-square paper appeared in the spring of 1900, an auspicious beginning to a wonderful century for the field of statistics." -B. Efron, The Statistical Century
The chi-square test is a statistical test that can be used to determine whether observed frequencies are significantly different from expected frequencies.
1.Test of Goodness-of-fit2.Test of Homogeneity3.Test of Independence
Three functions of Chi-Square
Formula:
The Chi - Square Test of Goodness-of-Fit
This is a test of difference between the observed frequencies and expected frequencies.
X2 = ( O E )2 E
Where:
X2 = the chi-square test O = the observed frequencies E = the expected frequencies
Formula:
The Chi-Square Test of HomogeneityThis test is concerned with two or more samples, with one criterion variable. This test used to determine if two or more populations are homogeneous. Its data distribution are similar with respect to a particular criterion variable.
X2 = N (ad bc )2 klmnWhere:
X2 = the chi-square test N = grand total klmn = the product of the rows and columnsFormula:
The Chi-Square Test of Independence
This test used to look into whether measures taken on two criterion variables are either independent or associated with one in a given population using such variables as level of education and income, performance in the class and IQ etc.
X2 = ( O E )2 EWhere:
X2 = the chi-square test O = the observed frequencies E = the expected frequencies = summation
The Chi-Square Test of Goodness-of-FitExample: The theory of Mendel crossing of peas is in the ratio of 9:3:3:1, meaning 9 parts are smooth yellow, 3 parts smooth green, 3 parts wrinkled yellow and 1 part wrinkled green. The researcher conducted an experiment and the result was that out of 560 peas, 310 were smooth yellow, 100 were wrinkled yellow, 110 were smooth green and 40 were wrinkled green. Is there a significant difference between the observed and the expected frequencies?
Problem: Is there a significant difference between the observed and the expected frequencies?
Hypothesis: H1 : There is a significant difference between the observed and the expected frequencies.
Level of Significance: 0.05Tabular value = 7.815
Statistics:Computation: Add the ratio 9:3:3:1 = 16 (Actual Result) (Theory) Attributes Ratio Observed ExpectedSmooth Yellow: 9 310315Wrinkled Yellow: 3 100105Smooth Green: 3 110105Wrinkled Green: 1 4035_____________________________________________________Total16 560560
Then divide 560 by 16 = 35
For expected frequencies multiply;35 x 9 = 31535 x 3 = 10535 x 3 = 10535 x 1 = 35
X2 = ( O E )2 E = ( 310 315 )2 + ( 100 105 )2 + ( 110 105 )2 + ( 40 35 )2 315 105 105 a 35 = .079 + .238 + .238 + .714 = 1.269
The Chi-Square Test of Homogeneity
Example:To illustrate this, we can evaluate the attitude of a sample of Lakas and Laban parties on the issue of peace and order in Mindanao. To carry out such study, a separate random sample of members of each party is drawn from the nationwide population of Laban and Lakas and each individual in both samples responds to the scale. Scores are the classified into Favorable or Unfavorable categories. Is there a significant difference between the attitudes of two political parties on the issue of peace and order in Mindanao?
Problem: Is there a significant difference between the attitudes of two political parties on the issue of peace and order in Mindanao?
Hypothesis: H1 : There is a significant difference between the attitudes of two political parties on the issue of peace and order in Mindanao.
Level of Significance: 0.05Tabular value : 3.841
The following frequencies are obtained: FavorableUnfavorable TotalLakas 65 35 100 k a b Laban 50 50 100 l c d _________________________________________________________Total 115 85 200m n N_________________________________________________________
Statistics: X2 = N (ad bc )2 klmn = 200 [(65)(50) - (35)(50)]2 (100)(100)(115)(85) = 200[ 3250 - 1750 ]297750000 = 200(1500)2 97750000 = 450000000 97750000 = 4.604
The Chi-Square Test of IndependenceExample: 90 individuals, male and female, were given a test in psychomotor skills and their scores were classified into high and low. Is there a significant relationship between sex and scores in psychomotor skills?
Computation: HIGH LOW Sex OE O ETotal Male 18 25.56 28 20.44 46 Female 32 24.44 12 19.56 44 _________________________________________________________ Total 50 40 90 individuals
Problem: Is there a significant relationship between sex and scores in psychomotor skills?
Hypothesis: H1 : There is a significant relationship between sex and scores in psychomotor skills.
Level of Significance: 0.05Tabular value : 3.841
For expected values: Multiply the column total to the row total and divide the product by the grand total.
50 x 46 = 25.5640 x 46 = 20.44 90 9050 x 44 = 24.4440 x 44 = 19.56 90 90Statistics: X2 = ( O E )2 E = ( 18 25.56 )2 + ( 32 24.44 )2 + ( 28 20.44 )2 + ( 12 19.56 )2 25.56 24.44 20.44 19.56 = 2.236 + 2.338 + 2.796 + 2.922 = 10.292THE END
