1. Chi-Square

2. All the tests we’ve learned so far assume that our data is normally distributed z-test t-test We test hypotheses about parameters of these normal distributions We call these tests “Parametric Tests” Chi Square

3. What if our data is not normally distributed? Skewed distributions Nominal or ordinal data Then we use “Nonparametric Tests” Tests that do not deal with parameters Chi Square

4. Parametric vs. Nonparametric Tests Parametric Tests Test hypotheses about some parameter (e.g. about the mean) Nonparametric Tests Test hypotheses about entire distribution Chi Square

5. Characteristics of the Chi-Square Distribution The major characteristics of the chi-square distribution are: 1. It is positively skewed. 2. It is non-negative. 3. It is based on degrees of freedom. 4. When the degrees of freedom change a new distribution is created.

6. The formula for  2 is: OR, sometimes written Where f 0 is the observed frequency of each category in each cell of a table.

7. Example(1): Chi-Square Test for Independence In one large factory, 100 employees were judged to be highly successful and another 100 marginally successful. All workers were asked, “Which do you find more important to you personally, the money you are able to take home or the satisfaction you feel from doing the job?” In the first group, 49% found the money more important, but in the second group 53% responded that way. Test the null hypothesis that job performance and job motivation are independent using the.01 level of significance.

8. An Example: Chi-Square Test for Independence State the research hypothesis. Are job performance and job motivation independent? State the statistical hypotheses.

9. An Example: Set the decision rule.

10. An Example: Chi-Square Test for Independence Calculate the test statistic. High SuccessMarginal SuccessTotal Money49 (51)53 (51)102 Satisfaction51 (49)47 (49)98 Total100100200

11. An Example: Chi-Square Test for Independence Calculate the test statistic. High SuccessMarginal SuccessTotal Money49 (51)53 (51)102 Satisfaction51 (49)47 (49)98 Total100100200

12. An Example: Chi-Square Test for Independence Decide if your result is significant. Retain H 0, 0.32<6.64 Interpret your results. Job performance and job motivation are independent.

13. Q2:Is the type of crime independent of whether the criminal is a stranger? Stranger Relative 12 39 379 106 727 642 Homicide Robbery Assault Homicide :Crime of Killing, Robbery: Stealing, Assault: Attacking

14. Row Total Column Total Stranger Relative 1118 787 1905 12 39 51 379 106 485 727 642 1369 Homicide Robbery Assault Is the type of crime independent of whether the criminal is a stranger?

15. Row Total Column Total E = (row total) (column total) (grand total) Stranger Relative Homicide Robbery Assault Is the type of crime independent of whether the criminal is a stranger? 1118 787 1905 12 39 51 379 106 485 727 642 1369

16. Row Total (29.93) Column Total E = (row total) (column total) (grand total) E = (1118)(51) 1905 = 29.93 Stranger Relative Homicide Robbery Assault Is the type of crime independent of whether the criminal is a stranger? 1118 787 1905 12 39 51 379 106 485 727 642 1369

17. Row Total (29.93) (21.07) (284.64) (200.36) (803.43) (565.57) Column Total E = (row total) (column total) (grand total) E = (1118)(51) 1905 = 29.93 E = (1118)(485) 1905 = 284.64 etc. Stranger Relative Homicide Robbery Assault Is the type of crime independent of whether the criminal is a stranger? 1118 787 1905 12 39 51 379 106 485 727 642 1369

18. 12 39 379 106 727 642 Homicide Robbery Forgery (29.93) (21.07) (284.64) (200.36) (803.43) (565.57 [ 10.741 ] Stranger Relative X2 =  X2 =  (O - E ) 2 E E Upper left cell: = = 10.741 (12 -29.93) 2 29.93 (E)(E) (O - E ) 2 E Is the type of crime independent of whether the criminal is a stranger?

19. 12 39 379 106 727 642 Homicide Robbery Forgery (29.93) (21.07) [15.258] (284.64) [31.281] (200.36) [44.439] (803.43) [7.271] (565.57) [10.329] [ 10.741 ] Stranger Acquaintance or Relative X2 =  X2 =  (O - E ) 2 E E Upper left cell: = = 10.741 (12 -29.93) 2 29.93 (E)(E) (O - E ) 2 E Is the type of crime independent of whether the criminal is a stranger?

20. 12 39 379 106 727 642 Homicide Robbery Forgery (29.93) (21.07) [15.258] (284.64) [31.281] (200.36) [44.439] (803.43) [7.271] (565.57) [10.329] [ 10.741 ] Stranger Acquaintance or Relative X2 =  X2 =  (O - E ) 2 E (E)(E) E Is the type of crime independent of whether the criminal is a stranger? Test Statistic X 2 = 10.741 + 31.281 +... + 10.329 = 119.319

21. Test Statistic X 2 = 119.319 with  = 0.05 and ( r -1) ( c -1) = (2 -1) (3 -1) = 2 degrees of freedom Critical Value X 2 = 5.991

22. Critical Values of Chi 2 Significance Level df0.100.050.250.010.005 1 2.70553.84155.02396.63497.8794 2 4.60625.99157.37789.210410.5965 3 6.25147.81479.348411.344912.8381

23. Test Statistic X 2 = 119.319 with  = 0.05 and ( r -1) ( c -1) = (2 -1) (3 -1) = 2 degrees of freedom 0  = 0.05 X 2 = 5.991 Reject Independence Critical Value X 2 = 5.991 Sample data: X 2 =119.319 Fail to Reject Independence

24. Test Statistic X 2 = 119.319 with  = 0.05 and ( r -1) ( c -1) = (2 -1) (3 -1) = 2 degrees of freedom 0  = 0.05 X 2 = 5.991 Reject Independence Critical Value X 2 = 5.991 Reject independence Sample data: X 2 =119.319 Fail to Reject Independence

25. Test Statistic X 2 = 119.319 with  = 0.05 and ( r -1) ( c -1) = (2 -1) (3 -1) = 2 degrees of freedom 0  = 0.05 X 2 = 5.991 Reject Independence Critical Value X 2 = 5.991 Reject independence Sample data: X 2 =119.319 Fail to Reject Independence Claim: The type of crime and knowledge of criminal are independent H o : The type of crime and knowledge of criminal are independent H 1 : The type of crime and knowledge of criminal are dependent

26. Test Statistic X 2 = 119.319 with  = 0.05 and ( r -1) ( c -1) = (2 -1) (3 -1) = 2 degrees of freedom It appears that the type of crime and knowledge of the criminal are related. 0  = 0.05 X 2 = 5.991 Reject Independence Critical Value X 2 = 5.991 Reject independence Sample data: X 2 =119.319 Fail to Reject Independence