Question

An educational psychologist wants to see the impact of mnemonics on memory. To test this the psychologist designs a study where participants are randomly assigned to a mnemonic method or not. However, before the participants are assigned they are paired for vocabulary knowledge. All the participants are then asked to read a passage from a book. A few days later, the participants are asked to reproduce the passage. Below are the number of correctly recalled words when reproducing the passage. What can the psychologist conclude with an α of 0.05?

mnemonic	control
210 596 397 271 332 226 429 348 486 471	223 412 402 285 353 243 443 340 582 490

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

b)
Condition 1:
---Select--- the book OR correctly recalled words OR control OR the passage OR mnemonic
Condition 2:
---Select--- the book OR correctly recalled words OR control OR the passage OR mnemonic

c) Input the appropriate value(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 OR 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 OR trivial effect OR small effect OR medium effect OR large effect
r² = _____ ;   ---Select--- na OR trivial effect OR small effect OR medium effect OR large effect

e) Make an interpretation based on the results.

A) Participants learning the mnemonic method significantly recalled more correct words than those in the control.

B) Participants learning the mnemonic method significantly recalled less correct words than those in the control.   

C) There was no significant difference in correctly recalling words between the mnemonic method and the control.

Answer 1

a) related sample t test

b) condition1 = mnemonic

Condition 2 = Control

c)

SAMPLE 1	SAMPLE 2	difference , Di =sample1-sample2	(Di - Dbar)²
210	223	-13.000	151.290
596	412	184.000	34114.090
397	402	-5.000	18.490
271	285	-14.000	176.890
332	353	-21.000	412.090
226	243	-17.000	265.690
429	443	-14.000	176.890
348	340	8.000	75.690
486	582	-96.000	9082.090
471	490	-19.000	334.890

	sample 1	sample 2	Di	(Di - Dbar)²
sum =	3766	3773.00	-7.000	44808.100

mean of difference ,    D̅ =ΣDi / n =   -0.700                  
                          
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    70.5597                  
                          
std error , SE = Sd / √n =    70.5597   / √   10   =   22.3130      
                          
t-statistic = (D̅ - µd)/SE = (   -0.7   -   0   ) /    22.3130   =   -0.0314
                          
Degree of freedom, DF=   n - 1 =    9                  
  
p-value =        0.5122   [excel function: =t.dist.rt(t-stat,df) ]               
Decision:   p-value>α , Fail to reject null hypothesis                      

d)

cohen's d = |Dbar/ std dev| =    0.01 (small)

e)

C) There was no significant difference in correctly recalling words between the mnemonic method and the control.