1. LECTURE 10 TESTS FOR PROPORTIONS EPSY 640 Texas A&M University

2. TESTS FOR PROPORTIONS Proportions involve Nominal scale data Univariate case *1 sample *k samples Bivariate case *association *agreement

3. 1 Sample case Proportion = k/n where k= # in category of interest, n=sample size Var (p) = p(1-p)/n = pq/n, where p= proportion, n= # in sample, and q=1-p If you have a single case, then Var(p)=p(1-p) = pq ex. Single roll of two dice, getting a 7 has p=1/6, so for a single roll, Var(p)=1/6*5/6 = 5/36, SD=sqrt(5/36) =.373

4. 1 Sample p Confidence Interval The confidence interval due to Ghosh is p  [n/(n+z 2 ){ p + z 2 /2n  z  pq/n + z 2 /rn 2 } ex. The probability of admission to med school is.08 for applicants. The 95% confidence interval around a single case is (.01,.81) For the roll of a 7, the 95% CI is (.005,.846)

5. Excel formulas for confidence interval around a single proportion

6. Testing k independent proportions For k categories with p i = proportion in category i and p 1 +p 2 +...p k =1, we can test 1. All proportions are the same 2. Each p i =  i based on theory or previous data

7. Testing k independent proportions For k categories with p i = proportion in category i and p 1 +p 2 +...p k =1 All proportions are the same p.=1/k: X 2 = N[(p 1 - p.) 2 /p. + (p 2 -p.) 2 /p.+...(p k -p.) 2 /p.] with degrees of freedom k-1 ex. The reported proportions of ethnic groups in Bryan is.33 Caucasian,.42 Hispanic, and.25 African American with 8000 students. The chi square is 347.2 with 2df. This is significant at p=.0001

8. Testing k independent theoretical proportions For k categories with p i = proportion in category i and p 1 +p 2 +...p k =1, each proportion tested against a theoretical value  i X 2 = N[(p 1 -  1 ) 2 /  1 +(p 2 -  2 ) 2 /  2 +...(p k -  k ) 2 /  k ] with degrees of freedom k-1 ex. The reported proportions of ethnic groups in Bryan is.33 Caucasian,.42 Hispanic, and.25 African American with 8000 students. National proportions are.74,.14, and.12. The chi square is 7423.9 with 2df. This is significant at p=.0001

9. Bivariate Association Two nominal variables are measured for N persons (eg. Gender and aggression status) An R x C table of proportions is computed p rc = proportion for row r, column c Then, the sum of a row of proportions is p r. and for a column the sum is p.c Association is defined as departure from average expected proportion for each cell cumulated over the cells

10. Chi Square Association For an R x C table of proportions with the sum of all cell proportions equal to 1.0, X 2 = N[(p 11 - p 1. p.1 ) 2 / p 1. p.1 +(p 12 - p 1. p.2 ) 2 / p 1. p.2 +...(p rc - p r. p.c ) 2 / p r. p.c ] with (R-1)(C-1) degrees of freedom

11. Chi Square Association For an R x C table of proportions with the sum of all cell proportions equal to 1.0, X 2 = SUM [ (Oij – Eij ] 2 / Eij Eij = expected cell count based on row and column averages, = N*p. j *p i. 161222 112310 50 44 27 35 32

12. 16122250 11231044 27353294 0.1527840.1980530.181077expected 0.134450.1742870.159348cell p's 14.361718.6170217.02128expected 12.638316.3829814.97872cell n's 0.1868872.3518781.456277chi square 0.2123722.6725891.65486cell values X 2 =8.534863 prob =0.014018 OBSERVED CELLS DEM REP IND MFMF X 2 (2,.05)=5.99

13. Chi Square Distribution X 2 =  z i 2, where z i = z-score for score I

14. TAAS PASS %ETHNIC % TAAS PASS N TAAS PASS PROPORTIONS Question: Is school passing rate related to ethnicity?

15. N=840 TAAS % PASSING Expected % (TAAS% - Expected) 2 /Expected Chi Square statistic significant p<.001 Conclusion: school is associate with ethnic TAAS pass rate