Assistant prof. Dr. Mayasah A. Sadiq FICMS-FM The Chi-Square test Assistant prof. Dr. Mayasah A. Sadiq FICMS-FM 2

DATA QUALITATIVE QUANTITATIVE CHI SQUARE TEST T-TEST

The most obvious difference between the chi‑square tests and the other hypothesis tests we have considered (T test) is the nature of the data. For chi‑square, the data are frequencies rather than numerical scores.

For testing significance of patterns in qualitative data. Chi-squared Tests For testing significance of patterns in qualitative data. Test statistic is based on counts that represent the number of items that fall in each category Test statistics measures the agreement between actual counts(observed) and expected counts assuming the null hypothesis

Chi Square FORMULA:

Chi square distribution

Applications of Chi-square test: Goodness-of-fit The 2 x 2 chi-square test (contingency table, four fold table) The a x b chi-square test (r x c chi-square test)

Steps of CHI hypothesis testing 1. Data :counts or proportion. 2. Assumption: random sample selected from a population. 3. HO :no sign. Difference in proportion no significant association. HA: sign. Difference in proportion significant association.

4. level of sign. df 1st application=k-1(k is no. of groups) df 2nd &3rd application=(column-1)(row-1) IN 2nd application(conengency table) Df=1, tab. Chi= 3.841 always Graph is one side (only +ve)

5. apply appropriate test of significance

6. Statistical decision & 7. Conclusion Calculated chi <tabulated chi P>0.05 Accept HO,(may be true) If calculated chi> tabulated chi P<0.05 Reject HO& accept HA.

The Chi-Square Test for Goodness-of-Fit The chi-square test for goodness-of-fit uses frequency data from a sample to test hypotheses about the shape or proportions of a population. The data, called observed frequencies, simply count how many individuals from the sample are in each category.

Example Eye colour in a sample of 40 Blue 12,brown 21,green 3,others 4 Eye colour in population Brown 80% Blue 10% Green 2% Others 8% Is there any difference between proportion of sample to that of population .use α0.05

Observed counts(frequency) Figure 18.1 Distribution of eye colors for a sample of n = 40 individuals. The same frequency distribution is shown as a bar graph, as a table, and with the frequencies written in a series of boxes.

Expected counts(frequency) Expected blue10/100*40=4 Expected brown=80/100*40=32 Expexcted green=2/100*40=0.8 Expected others=8/100*40=3

1. Data Represents the eye colour of 40 person in the following distribution Brown=21 person,blue=12 person,green=3,others=4

2. Assumption Sample is randomly selected from the population.

3. Hypothesis Null hypothesis: there is no significant difference in proportion of eye colour of sample to that of the population. Alternative hypothesis: there is significant difference in proportion of eye colour of sample to that of the population.

4. Level of significance; (α =0.05); 5% Chance factor effect area 95% Influencing factor effect area d.f.(degree of freedom)=K-1; (K=Number of subgroups) =4-1=3 D.f. for 0.5=7.81

7.81 Accept Ho Influencing factor effect area 95% Reject Ho Chance factor effect area 5% 5%

Figure 18.4 For Example 18.1, the critical region begins at a chi-square value of 7.81.

5. Apply a proper test of significance

Chi Square FORMULA:

Observed counts(frequency) Figure 18.1 Distribution of eye colors for a sample of n = 40 individuals. The same frequency distribution is shown as a bar graph, as a table, and with the frequencies written in a series of boxes.

Expected counts(frequency) Expected blue=10/100*40=4 Expected brown=80/100*40=32 Expexcted green=2/100*40=0.8 Expected others=8/100*40=3

=(12-4)² (21-32)² (3-0.8)² (4-3)² ------------ +---------- +----------- + -------- 4 32 0.8 3 =(64/4) + (121/32)+(4.8/0.8)+(1/3) =16+3.78+6+0.3= Calculated chi =26.08

7.81 Accept Ho Influencing factor effect area 95% Reject Ho Chance factor effect area 5% 5%

6. Statistical decision Calculated chi> tabulated chi P<0.5

7. Conclusion We reject H0 &accept HA: there is significant difference in proportion of eye colour of sample to that of the population.

Applications of Chi-square test: Goodness-of-fit The 2 x 2 chi-square test (contingency table, four fold table) The a x b chi-square test (r x c chi-square test)

The Chi-Square Test for Independence The second chi-square test, the chi-square test for independence, can be used and interpreted in two different ways: 1. Testing hypotheses about the relationship between two variables in a population, or(2×2) 2. Testing hypotheses about differences between proportions for two or more populations.(a×b)

The Chi-Square Test for Independence (cont.) The data, called observed frequencies, simply show how many individuals from the sample are in each cell of the matrix. The null hypothesis for this test states that there is no relationship between the two variables; that is, the two variables are independent.

2× 2 chi square (contingency table )  

The Chi-Square Test for Independence (cont.) The calculation of chi-square is the same for all chi-square tests:

Chi Square FORMULA:

2nd application

Example A total 1500 workers on 2 operators(A&B) Were classified as deaf & non-deaf according to the following table.is there association between deafness & type of operator .let α 0.05

Result not deaf. deaf total Operator A 100 900 1000 B 60 440 500 total 160 1340 1500

Operator Result A B deaf not deaf. total 100 900 1000 60 440 500 160 1340 1500 Total number of items=1500 Total number of defective items=160

 

Operator Result A B def not def. total 100 900 1000 60 440 500 160 1340 1500 Expected deaf from Operator A = 1000 * 160/1500 = 106.7 (expected not deaf=1000-106.7=893.3) Expected deaf from Operator B = 500 * 160/1500 = 53.3

Result Operator A B def not def. total 100 900 1000 60 440 500 160 1340 1500 Operator Expected A B def not def. total 106.7 893.3 53.3 446.7

1. Data Represent 1500 workers,1000 on operator A 100 of them were deaf while 500 on operator B 60 of them were deaf

2. Assumption Sample is randomly selected from the population.

3. Hypothesis HO: there is no significant association between type of operator & deafness. HA:there is significant association between type of operator & deafness.

4. Level of significance; (α = 0.05); 5% Chance factor effect area 95% Influencing factor effect area d.f.(degree of freedom)=(r-1)(c-1) =(2-1)(2-1)=1 D.f. 1 for 0.05=3.841

3.841 Accept Ho Influencing factor effect area 95% Reject Ho Chance factor effect area 5% 5%

5. Apply a proper test of significance

=(100-106.7)² ( 900-893.3)² (60-53.3)² --------------- + ---------------- + -------------- 106.7 893.3 53.3 +(440-446.7)² ---------------= 446.7 =0.42+0.05+o.84+0.10 =1.41

3.841 Accept Ho Influencing factor effect area 95% Reject Ho Chance factor effect area 5% 5%

6. Statistical decision Calculated chi< tabulated chi P>0.5

7. Conclusion We accept H0 HO may be true There is no significant association between type of operator & deafness.

Applications of Chi-square test: Goodness-of-fit The 2 x 2 chi-square test (contingency table, four fold table) The a x b chi-square test (r x c chi-square test)

aₓ b SA A NO D SD Gr 1 12 18 4 8 12 Gr2 48 22 10 8 10 Gr3 10 4 12 10 12

 

Degree of freedom The d.f depends on colunms number & rows number. or (r-1 ) (c-1) i.e. if 3 row ,4 colunms Df=(3-1)(4-1) df =6 If 3 rows,3 colunms Df=(3-1)(3-1) Df=4

Yates Correction When we apply 2x2 chi-square test and one of the expected cells was <5 Or when we apply axb chi-square test and one of the expected cells was <2, Or when the grand total is <40 we have to apply Yates' correction formula;

YATES = ∑ -------------------------------------

Note When 2x2 chi-square test have a zero cell (one of the four cells is zero) we can not apply chi-square test because we have what is called a complete dependence criteria. But for axb chi-square test and one of the cells is zero when can not apply the test unless we do proper categorization to get rid of the zero cell.

12 3    Or 5 6 7 11 12 3    5 6 7 13