1. 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 and expected counts assuming the null hypothesis Chi-squared Tests

2. The chi-square distribution can be used to see whether or not an observed counts agree with an expected counts.Let O = observed count and E = Expected count Chi-squared Distribution

3. Testing if Observed Counts are in Agreement with Known Percentages Consider items of a population distributed over k categories in in proportions If H 0 is true then we expect E i = n, expected frequency for the ith category as opposed to O i, observed frequency.

4. ObservedExpectedFrequency H4050 T6050 sum100100 An Example Biased Coin?

6. degrees of freedom = (R –1)(C – 1) R = number of rows C = number of columns

7. Is our chi square value an extreme outcome just by chance while in fact the null hypothesis is true and sample frequencies are not significantly apart from the ideal frequencies? Note that chi-squared statistic is a positive number

8. only the right-hand side of the table is used nondirectional test the statistic has no sign

9. ObservedExpected DieFrequencyFrequency 1410 2610 31710 41610 5810 6910 sum6060

11. degrees of freedom = number of terms -1

12. 2 x 2 contingency tables Chi-squared test for independence Var A Var B a1 a2 b1b2 total Ho : The two variable are independent Ha : The two variables are associated

13. Operator Result A B def not def. total 1009001000 60440500 16013401500

14. Operato r Result A B def not def. total 1009001000 60440500 16013401500 Total number of items=1500 Total number of defective items=160 Overall defective rate =160/1500=0.1067 Now, apply this rate to the number of items produced by each operator.

15. Operato r Result A B def not def. total 1009001000 60440500 16013401500 Expected defective from Operator A = 1000 * 0.1067 = 106.7 (expected not defective=1000-106.7=893.3) Expected defective from Operator B = 500 * 0.1067 = 53.3 (expected not defective=500-53.3=446.7)

16. Operato r Expected A B def not def. total 106.7893.3 53.3446.7 Result Operato r A B def not def. total 100900 1000 60440500 16013401500

17. r x c contingency tables SAANODSD Gr 112184812 Gr2482210810 Gr3104121012

18. use when you have categorical data measure the difference between actual counts and expected counts test the independence of two variables Assumptions: data set is a random sample you have at least 5 counts in each category degrees of freedom = (categories var1 -1)(categories var2 -1)