A therapist measures the relationship between a patient\'s expectations that the

Question

A therapist measures the relationship between a patient's expectations that therapy will be successful and the actual success of the therapy. The following table shows the results of this hypothetical study using a sample of 50 clients.

(a) Compute the phi correlation coefficient. (Round your answer to three decimal places.)


(b) Using a two-tailed test at a 0.05 level of significance, state the decision to retain or reject the null hypothesis. Hint: You must first convert r to 2.

Retain the null hypothesis.

Reject the null hypothesis.    

Therapy
Successful Yes No Totals Client
expects
success Yes 23 8 31 No 6 13 19 Totals 29 21

Explanation / Answer

a.

phi correlation cofficient formula = a.db.c/sqrt(e.f.g.h)

a=23,b=8,c=6,d=13 given data

Let the totals be a + b = e =31

c + d = f =19

a + c = g = 29

b + d = h = 21

= (23*13-8*6)/sqrt(31*19*29*21)= 0.4190

The interpretation of the value of phi coefficient of correlation is illustrated below:

1) If the value lies between -1.0 to -0.7, it indicates a strong-negative correlation.

2) If the value lies between -0.7 to -0.3, it concludes a weak-negative correlation.

3) If the value lies between -0.3 to 0.3, it denotes very little or no correlation.

4) If the value lies between 0.3 to 0.7, it indicates a weak-positive correlation.

5) If the value lies between 0.7 to 1.0, it represents a strong-positive correlation

b.

Given table data is as below MATRIX col1 col2 TOTALS row 1 23 8 31 row 2 6 13 19 TOTALS 29 21 N = 50 ------------------------------------------------------------------

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 17.98 13.02 row 2 11.02 7.98 ------------------------------------------------------------------

calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above Oi Ei Oi-Ei (Oi-Ei)^2 (Oi-Ei)^2/Ei 23 17.98 5.02 25.2004 1.4016 8 13.02 -5.02 25.2004 1.9355 6 11.02 -5.02 25.2004 2.2868 13 7.98 5.02 25.2004 3.1579 ^2 o = 8.7818 ------------------------------------------------------------------

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 = 8.7818
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| =8.7818 & | ^2 | =3.8415
make decision
hence value of | ^2 o | > | ^2 | and here we reject Ho
^2 p_value =0.003


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: 8.7818
critical value: 3.8415
p-value:0.003
decision: reject Ho

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