Question

QUESTION 12

1. A therapist would like to know if therapy for anxiety can be treated as effectively through teleconferencing as it can through traditional face-to-face therapy or whether one type of therapy would be more effective. After 3 months of therapy, samples of 35 clients from each of the two groups completed the Generalized Anxiety Disorder 7-Item Scale. Responses range from 0 to 21, with higher scores indicating higher anxiety (Spitzer, Kroenke, Williams, & Lowe, 2006). The Teleconferencing group of n₁ = 35 clients had a mean of M₁ = 6.35 and a standard deviation of SD₁ = 1.75. The Traditional face-to-face therapy group of n₂ = 35 clients had a mean of M₂ = 6.15 and a standard deviation SD₂ = 1.81. Identify the null hypothesis to test whether the mean anxiety scores differed between the two unrelated groups.

   	a.	µ1 ≤ µ2
   	b.	µ1 ≥ µ2
   	c.	µ1 ≠ µ2
   	d.	µ1 = µ2

QUESTION 18

For the above study, should you reject the null hypothesis?

	a.	Yes, because the t obtained value is inside the critical region.
	b.	No, because the t obtained value is not in the critical region.
	c.	Yes, because the t obtained value is not in the critical region.
	d.	No, because the t obtained value is inside the critical region.

QUESTION 19

For the above study, compute and interpret the effect side (d).

		d = 1.43, large effect
		d = 0.11, small effect
		d = 0.67, large effect
		d = 0.06, small effect

QUESTION 20

What conclusion can you make about the difference between the two types of therapies?

	a.	The results suggest that people in teleconferencing therapy had lower mean anxiety scores than people in traditional face-to-face therapy.
	b.	The results suggest that people in teleconferencing therapy had higher mean anxiety scores than people in traditional face-to-face therapy.
	c.	The results suggest that people in teleconferencing therapy had similar mean anxiety scores to people in traditional face-to-face therapy.

Answer 1

12.

Given that,
mean(x)=6.35
standard deviation , s.d1=1.75
number(n1)=35
y(mean)=6.15
standard deviation, s.d2 =1.81
number(n2)=35
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.032
since our test is two-tailed
reject Ho, if to < -2.032 OR if to > 2.032
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =6.35-6.15/sqrt((3.0625/35)+(3.2761/35))
to =0.47
| to | =0.47
critical value
the value of |t α| with min (n1-1, n2-1) i.e 34 d.f is 2.032
we got |to| = 0.46997 & | t α | = 2.032
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 0.47 ) = 0.641
hence value of p0.05 < 0.641,here we do not reject Ho
ANSWERS
---------------
option:D
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 0.47
critical value: -2.032 , 2.032
decision: do not reject Ho
p-value: 0.641
we do not have enough evidence to support the claim that whether the mean anxiety scores differed between the two unrelated groups
18.
No, because the t obtained value is inside the critical region
19.
Effective size = (mean difference)/S.D pooled
S.Dpooled = sqrt((SD1^2 + SD2^2) ⁄ 2)
S.D pooled = sqrt((1.75^2 + 1.81^2)/2)
S.D pooled = 1.780
Effective size = (6.35-6.15)/1.780
Effective size = 0.11
small effect
20.
option:C
The results suggest that people in teleconferencing therapy had similar
mean anxiety scores to people in traditional face-to-face therapy.