Lecture 10 tests for proportions
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LECTURE 10 TESTS FOR PROPORTIONS. EPSY 640 Texas A&M University. TESTS FOR PROPORTIONS. Proportions involve Nominal scale data Univariate case 1 sample k samples Bivariate case association agreement. 1 Sample case. Proportion = k/n where k= # in category of interest, n=sample size

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LECTURE 10 TESTS FOR PROPORTIONS

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Lecture 10 tests for proportions

LECTURE 10TESTS FOR PROPORTIONS

EPSY 640

Texas A&M University


Tests for proportions

TESTS FOR PROPORTIONS

  • Proportions involve Nominal scale data

  • Univariate case

    • 1 sample

    • k samples

  • Bivariate case

    • association

    • agreement


1 sample case

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, andq=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


1 sample p confidence interval

1 Sample p Confidence Interval

  • The confidence interval due to Ghosh is

  • p  [n/(n+z2){ p + z2/2n  z pq/n + z2/rn2}

  • 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)


Lecture 10 tests for proportions

Excel formulas for confidence interval around a single proportion


Testing k independent proportions

Testing k independent proportions

  • For k categories with pi = proportion in category i and p1+p2+...pk=1, we can test 1. All proportions are the same2. Each pi = i based on theory or previous data


Testing k independent proportions1

Testing k independent proportions

  • For k categories with pi = proportion in category i and p1+p2+...pk=1

  • All proportions are the same p.=1/k:

    X2 = N[(p1 - p.)2/p.+ (p2-p.)2/p.+...(pk-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


Testing k independent theoretical proportions

Testing k independent theoretical proportions

  • For k categories with pi = proportion in category i and p1+p2+...pk=1, each proportion tested against a theoretical value i

    X2 = N[(p1 - 1)2/ 1+(p2- 2)2/ 2+...(pk- 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


Bivariate association

Bivariate Association

  • Two nominal variables are measured for N persons (eg. Gender and aggression status)

  • An R x C table of proportions is computedprc = proportion for row r, column c

  • Then, the sum of a row of proportions is pr. 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


Chi square association

Chi Square Association

  • For an R x C table of proportions with the sum of all cell proportions equal to 1.0,

    X2 = N[(p11 - p1.p.1)2/ p1.p.1 +(p12- p1.p.2 )2/ p1.p.2 +...(prc- pr.p.c )2/ pr.p.c]

    with (R-1)(C-1) degrees of freedom


Chi square association1

Chi Square Association

  • For an R x C table of proportions with the sum of all cell proportions equal to 1.0,

    X2 = SUM [ (Oij – Eij ]2 / Eij

    Eij = expected cell count based on row and column averages, = N*p.j*pi.

16

12

22

50

11

23

10

44

27 35 32


Lecture 10 tests for proportions

OBSERVED CELLS

DEM REP IND

M

F

X2(2,.05)=5.99


Chi square distribution

Chi Square Distribution

X2 = zi2, where zi= z-score for score I


Lecture 10 tests for proportions

TAAS PASS %

ETHNIC %

TAAS PASS N

TAAS PASS PROPORTIONS

Question: Is school passing rate related to ethnicity?


Lecture 10 tests for proportions

TAAS % PASSING

N=840

Expected %

(TAAS% - Expected)2 /Expected

Chi Square statistic

significant p< .001

Conclusion: school is associate with ethnic TAAS pass rate


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