Question

A researcher would like to determine if relaxation training will increase positive mood of cancer patients. A sample of n = 16 cancer patents was asked to assess their mood before and after the relaxation training. The data showed that the average change in patients' mood is MD = 3.9 with the variance s2 = 81.

Are the data sufficient to conclude that there is a significant improvement in patients' mood after the relaxation training? Use the repeated-measures t-test with α = .05 1-tail to answer this question.

Follow the steps of hypothesis testing and insert your answers below. In your calculations, round all numbers to two decimal places to avoid rounding errors.

ANSWER H0:

H1:

Estimated standard SM:

Computed t statistic:

df for decision about H0:

Critical t-value used for decision about H0:

Decision about H0 (i.e., reject or fail to reject):

Estimated Cohen's d (if applicable):

Conclusion (i.e., is there a significant effect of test color on anxiety or not? If the effect is significant, which color produced higher anxiety? Make sure to follow the APA reporting format in your conclusion

Answer 1

Given that,
Assume,population mean(u)=3 because not given in the data
sample mean, x =3.9
standard deviation, s =9
number (n)=16
null, Ho: μ=3
alternate, H1: μ>3
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.753
since our test is right-tailed
reject Ho, if to > 1.753
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =3.9-3/(9/sqrt(16))
to =0.4
| to | =0.4
critical value
the value of |t α| with n-1 = 15 d.f is 1.753
we got |to| =0.4 & | t α | =1.753
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :right tail - Ha : ( p > 0.4 ) = 0.3474
hence value of p0.05 < 0.3474,here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=3
alternate, H1: μ>3
test statistic: 0.4
critical value: 1.753
decision: do not reject Ho
p-value: 0.3474
we do not have enough evidence to support the claim that if relaxation training will increase positive mood of cancer patients.
cohen'd size = mean difference /standard deviation
cohen'd size = (3.9-3)/9 = 0.1
low effect.