1. Researchers studying long term memories had 54 students study 30 “normal” words. After a 10min delay period the students tried to recall as many of the 30 words as they could. One week later the same 54 students had to study 30 “high imagery” words and again after a 10min delay period they tried to recall as many of the “high imagery” words as they could. What statistic should the researchers use to compare the mean number of normal and high imagery words that were recalled?
a. 
z for a sample mean 

b. 
single sample t 

c. 
repeated measures t 
2. A clinical psychologist wants to determine if anxiety scores are lower after 10 group therapy sessions than before the sessions. He has a group of 10 clients with anxiety take a standardized test that measures their anxiety before the 10 sessions and then again after the 10 group sessions. The data are below. Compute the effect size. Compute the D as Before–After.
Before Sessions 
After Sessions 
15 
10 
13 
12 
12 
12 
9 
8 
15 
14 
11 
10 
14 
13 
10 
9 
9 
9 
11 
11 
a. 
.33 

b. 
.58 

c. 
.25 

d. 
.76 
3.. A clinical psychologist wants to determine if anxiety scores are lower after 10 group therapy sessions than before the sessions. He has a group of 10 clients with anxiety take a standardized test that measures their anxiety before the 10 sessions and then again after the 10 group sessions. The data are below. Compute the 95% CI for the change in anxiety score in the population. Compute the D as Before–After.
Before Sessions 
After Sessions 
15 
10 
13 
12 
12 
12 
9 
8 
15 
14 
11 
10 
14 
13 
10 
9 
9 
9 
11 
11 
a. 
LB = .77, UB = 1.88 

b. 
LB = .06, UB = 2.14 

c. 
LB = .26, UB = 1.94 

d. 
LB = .48, UB = 2.68 
Ans a ) since the data is taken on same observation so we will use repeted measure t
Ans b ) using minitab>stat>baisc stat>paired t
we have
Paired TTest and CI: before, after
Paired T for before  after
N Mean StDev SE Mean
before 10 11.900 2.283 0.722
after 10 10.800 1.932 0.611
Difference 10 1.100 1.449 0.458
95% CI for mean difference: (0.063, 2.137)
TTest of mean difference = 0 (vs ≠ 0): TValue = 2.40 PValue =
0.040
effect size = mean difference / St dev =0.76
option d is true
ANs 3 ) 95% CI for mean difference: (0.063, 2.137
option b is true
. 
LB = .06, UB = 2.14 
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