6.28 In a study examining the effects of individualized care of youths with seve

Question

6.28 In a study examining the effects of individualized care of youths with severe emotional prob- lems, Burchard and Schaefer (1990, personal communication) proposed to have care- givers rate the presence or absence of specific behaviors for each of 40 adolescents on a given day. To check for rater reliability, they asked two raters to rate each adolescent. The following hypothetical data represent reasonable results for the behavior of "extreme verbal abuse." Rater A Rater B Presence Absence Absence Total 14 26 40 Presence 12 25 27 13 © Cengage l earning 2013 a. What is the percentage of agreement for these raters? b. What is Cohen's kappa? c. Why is kappa noticeably less than the percentage of agreement? d. Modify the raw data, keeping N at 40, so that the two statistics move even farther apart. How did you do this?

Explanation / Answer

  

Given table data is as below col1 col2 row 1 12 2 14 row 2 1 25 26 TOTALS 13 27 N = 40 ------------------------------------------------------------------

calculation formula for E table matrix E-TABLE col1 col2 row 1 row1*col1/N row1*col2/N row 2 row2*col1/N row2*col2/N E-TABLE col1 col2 row 1 4.55 9.45 row 2 8.45 17.55 ------------------------------------------------------------------

expected frequecies calculated by applying E - table matrix formulae Oi Ei Oi-Ei (Oi-Ei)^2 (Oi-Ei)^2/Ei 12 4.55 7.45 55.5025 12.1984 2 9.45 -7.45 55.5025 5.8733 1 8.45 -7.45 55.5025 6.5683 25 17.55 7.45 55.5025 3.1625 ^2 o = 27.8025 ------------------------------------------------------------------

calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above

------------------------------------------------------------------

set up null vs alternative as
null, Ho: no relation b/w X and Y OR X and Y are independent
alternative, H1: exists a relation b/w X and Y OR X and Y are dependent
level of significance, = 0.05
from standard normal table, chi square value at right tailed, ^2 /2 =3.8415
since our test is right tailed,reject Ho when ^2 o > 3.8415
we use test statistic ^2 o = (Oi-Ei)^2/Ei
from the table , ^2 o = 27.8025
critical value
the value of |^2 | at los 0.05 with d.f (r-1)(c-1)= ( 2 -1 ) * ( 2 - 1 ) = 1 * 1 = 1 is 3.8415
we got | ^2| =27.8025 & | ^2 | =3.8415
make decision
hence value of | ^2 o | > | ^2 | and here we reject Ho
^2 p_value =0


ANSWERS
---------------
null, Ho: no relation b/w X and Y OR X and Y are independent
alternative, H1: exists a relation b/w X and Y OR X and Y are dependent
test statistic: 27.8025
critical value: 3.8415
p-value:0
decision: reject Ho

Related Questions

Navigate

• Browse (All)
• Browse 6
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.