Question

# 1. Researchers studying long term memories had 54 students study 30 “normal” words. After a 10-min...

1. Researchers studying long term memories had 54 students study 30 “normal” words. After a 10-min 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 10-min 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. 0.33 b. 0.58 c. 0.25 d. 0.76

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 T-Test 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)
T-Test of mean difference = 0 (vs ≠ 0): T-Value = 2.40 P-Value = 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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