A psychologist studying addiction tests whether cravings for cocaine and relapse

Question

A psychologist studying addiction tests whether cravings for cocaine and relapse are independent. The following table lists the observed frequencies in the small sample of people who use drugs.

Retain the null hypothesis.Reject the null hypothesis.    

(b) Compute effect size using and Cramer's V. Hint: Both should give the same estimate of effect size. (If necessary, round your intermediate steps to two or more decimal places. Round your answers to two decimal places.)

Obs. Freq. Relapse Yes No Cravings Yes 21 12 33 No 7 18 25 28 30 N = 58

Explanation / Answer

a.

b.

Effect size Phi = sqrt(chisquare/n)

= sqrt(7.24/58)

= 0.35

df = degree of freedom = min(r-1,c-1) =min(2-1,2-1) = 1*1 =1 =0.30(medium)

cramer's V = sqrt(chisquare/n*df)

= sqrt(7.24/(58*0.30)

= 0.64

Given table data is as below MATRIX col1 col2 TOTALS row 1 21 12 33 row 2 7 18 25 TOTALS 28 30 N = 58 ------------------------------------------------------------------

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

expected frequecies calculated by applying E - table matrix formulae E-TABLE col1 col2 row 1 15.93 17.07 row 2 12.07 12.93 ------------------------------------------------------------------

calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above Oi Ei Oi-Ei (Oi-Ei)^2 (Oi-Ei)^2/Ei 21 15.93 5.07 25.7 1.61 12 17.07 -5.07 25.7 1.51 7 12.07 -5.07 25.7 2.13 18 12.93 5.07 25.7 1.99 ^2 o = 7.24 ------------------------------------------------------------------

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 =3.84
since our test is right tailed,reject Ho when ^2 o > 3.84
we use test statistic ^2 o = (Oi-Ei)^2/Ei
from the table , ^2 o = 7.24
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 3.84
we got | ^2| =7.24 & | ^2 | =3.84
make decision
hence value of | ^2 o | > | ^2 | and here we reject Ho
^2 p_value =0.01


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: 7.24
critical value: 3.84
p-value:0.01
decision: reject Ho

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