LOS Learning Platform x C Chegg study Guided so x P Contingency Table x VA To pi
ID: 3227375 • Letter: L
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LOS Learning Platform x C Chegg study Guided so x P Contingency Table x VA To pic Eight Homework x a C www.webassign.net/web/Student/Assignment-Responses/last?dep 15352565-Q7 7. 04 points I Previous Ans Priviterastats2 17.E.021 My Note Ask Yo 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. Relapse Cravings Yes 21 No 26 28 32 60 (a) Conduct a chi-square test for independence at a 0.05 level of significance. Round your answer to two decimal places.) 3.00 Decide whether to retain or reject the null hypothesis. 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.) x v- 12 You may need to use the appropriate table in Appendix Bto answer this question, Refer to steps for computing the chi-square test for independence (Section 17.6) and effect size (Section 17,8). Additional Materials Section 17.1 9:47 AM 4x 4/26/2017 Type here to searchExplanation / Answer
Given table data is as below
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calculation formula for E table matrix
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expected frequecies calculated by applying E - table matrix formulae
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calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above
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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 = 7.1861
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| =7.1861 & | ^2 | =3.8415
make decision
hence value of | ^2 o | > | ^2 | and here we reject Ho
^2 p_value =0.0073
ANSWERS
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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: 7.1861
critical value: 3.8415
p-value:0.0073
decision: reject Ho
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