Question

Use SPSS for this Application Exercise:
A staff psychologist at a police precinct believes that the new week-long training courses worsens on the job sensitivity of police officers. The psychologist designs a study where some policemen randomly get and complete the course. A month later the psychologist records the number of domestic disputes the police officers successfully resolved from their police reports. What can the psychologist conclude with α = 0.05. The success data are below.

no course	course
11.2 12.5 10.6 12.7 8.3 15.6 12.1	14.8 16.3 14.3 17.4 11.2 16.5 15.4

a) What is the appropriate test statistic?
---Select--- na z-test One-Sample t-test Independent-Samples t-test Related-Samples t-test

b)
Condition 1:
---Select--- job sensitivity no course completing the course police precinct domestic disputes
Condition 2:
---Select--- job sensitivity no course completing the course police precinct domestic disputes

c) Compute the appropriate test statistic(s) to make a decision about H₀.
(Hint: Make sure to write down the null and alternative hypotheses to help solve the problem.)
p-value =  ; Decision:  ---Select--- Reject H0 Fail to reject H0

d) Using the SPSS results, compute the corresponding effect size(s) and indicate magnitude(s).
If not appropriate, input and/or select "na" below.
d =  ;   ---Select--- na trivial effect small effect medium effect large effect
r² =  ;   ---Select--- na trivial effect small effect medium effect large effect

e) Make an interpretation based on the results.

Participants that received training had significantly less resolved domestic disputes than those that did not receive training.

Participants that received training had significantly more resolved domestic disputes than those that did not receive training.    

Participants that received training did not differ significantly on resolved domestic disputes than those that did not receive training.

Answer 1

Given that,
mean(x)=11.8571
standard deviation , s.d1=2.2307
number(n1)=7
y(mean)=15.1286
standard deviation, s.d2 =2.0295
number(n2)=7
null, Ho: u1 = u2
alternate, H1: u1 < u2
level of significance, α = 0.05
from standard normal table,left tailed t α/2 =1.94
since our test is left-tailed
reject Ho, if to < -1.94
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =11.8571-15.1286/sqrt((4.97602/7)+(4.11887/7))
to =-2.87
| to | =2.87
critical value
the value of |t α| with min (n1-1, n2-1) i.e 6 d.f is 1.94
we got |to| = 2.8701 & | t α | = 1.94
make decision
hence value of | to | > | t α| and here we reject Ho
p-value:left tail - Ha : ( p < -2.8701 ) = 0.01421
hence value of p0.05 > 0.01421,here we reject Ho
ANSWERS
---------------
a.
t test independent samples
b.
job sensitivity no course completing the course police precinct domestic disputes
c.
null, Ho: u1 = u2
alternate, H1: u1 < u2
test statistic: -2.87
critical value: -1.94
decision: reject Ho
p-value: 0.01421
d.
Effective size = modulus of (mean 1 - mean 2)/S.D pooled
standard deviation (S.D)pooled = sqrt(((n1-1)S.D1^2 + (n2-1)S.D2^2)/(n1+n2-2))
S.D pooled = sqrt(((7-1)*2.2307^2 + (7-1)*2.0295^2)/(7+7-2))
S.D pooled= 2.1324
Effective size = modulus of ( 11.8571-15.1286)/2.1324
Effective size = modulus of -1.5341
Effective size= 1.5341
large effect
e.
we have enough evidence to support the claim that job sensitivity no course completing is less the course police precinct domestic disputes