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Stat6 chi square test

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Stat6 chi square test

1. 1. Chi square test
2. 2. Uses and Applications Used when you have frequency distribution of qualitative type of variables. Applications: To test goodness of fit. In the 2X2 table, it is used to test whether there is an association between the row and the column variables; ie whether the distribution of individuals among the categories of one variable is independent of their distribution among the categories of the other.
3. 3. Example, Influenza vaccination trial Influenza No Total influenzaVaccine 20 220 240 (8.3%)Placebo 80 140 220 (36.4%)Total 100 360 460 (21.7%)
4. 4. The question is: Is the difference (in the percentages of influenza) due to vaccination or occurred by chance? To answer, a Chi square test is done which compares the observed numbers in each of the four categories in the contingency table with the numbers to be expected if there was no difference in the effectiveness between the vaccine and placebo.
5. 5. The expected frequencies: Influenza No Total influenzaVaccine 52.2 187.8 240Placebo 47.8 172.2 220Total 100 360 460
6. 6. Solution, continued Ho: The proportion of influenza among the vaccine group = The proportion of influenza among the placebo group. Level of significance (alpha) = 0.05 D.f. = (No. of rows-1) (No. of columns-1). Test statistics: Chi square test. ∑ (O − E ) 2 χ = 2 E
7. 7. Solution, continued (20 − 52.2) 2 (80 − 47.8) 2 (220 − 187.8) 2 (140 − 172.2) 2χ2 = + + + = 53.09 52.2 47.8 187.8 172.2The tabulated value for X2 for 1 d.f. is 3.841The calculated value (53.09) > tabulated valueTherefore reject Ho and conclude that there is statisticallysignificant difference between the 2 proportions. Thisdifference is unlikely to be due to chance.Therefore the vaccine is effective.P < 0.05