Question

Based upon the work of Elizabeth Loftus, a researcher was interested in testing the most effective way to implant false memories in people. First, she had all subjects watch a 60 minute movie. Then she randomly assigned these subjects to one of four conditions, each of which used different assessment techniques to influence subjects to think they saw something that didn't actually occur. Condition A utilized falsified photos, condition B utilized falsified video, condition C utilized misleading narrative, and condition D was a control group with no intervention at all. The data below shows how many false memories subjects in each condition 'remembered'.
Condition A: {xA} = (4,8,6,6,5,9)
Condition B: {xB} = (9,10,7,9,11,7)
Condition C: {xC} = (2,1,2,2,4,3)
Condition D: {xD} = (0,2,1,1,0,3)

Question 8:
Assume that we made a priori hypothesis regarding the relationship between each of the three experimental conditions and the control group (i.e. A to D, B to D, and C to D). Is there any point in assessing the differences between the means of these groups? If no, explain why. If yes, describe what you would do to assess these differences.

Question 9:
Are there any potential differences between the means of the three experimental conditions (i.e. A to B, A to C, and B to C)? If no, explain how you know this. If yes, describe what you would do to assess these differences.

Answer 1

	Original Observations		Squared Observations
	A	B	C	D		A	B	C	D
	4	9	2	0		16	81	4	0
	8	10	1	2		64	100	1	4
	6	7	2	1		36	49	4	1
	6	9	2	1		36	81	4	1
	5	11	4	0		25	121	16	0
	9	7	3	3		81	49	9	9
Total	38	53	14	7	GT=112	258	481	38	15	∑x2=792
Mean	6.333	8.833	2.333	1.167	GM=4.667

We have to test

H₀: all µ_i are same i.e. µ₁= µ₂= µ₃= µ₄

H₁: all µ_i are not same i.e. at least one mean is different than others

We have n= 24

            t=4

            r=6

Correction factor=

Correction factor=

Correction factor=522.667

Total sum of squares= sum of squares of every observation - correction factor

Total sum of squares= 792 – 522.667

Total sum of squares= 269.333

Treatment sum of squares=

Treatment sum of squares=

Treatment sum of squares=

Treatment sum of squares=227

Error sum of squares= Total SS – Treatment SS

Error sum of squares= 269.333 – 227

Error sum of squares= 42.333

ANOVA Table

Source of Variation	df	SS	MS	F	p value
Groups	4-1=3	227	75.667	35.75	0.0000
Error	20	42.333	2.1167
Total	24-1=23	269.333

We have the p-value which is very less (almost 0) which indicates that we have very strong evidence against null hypothesis to reject it, so we reject the null hypothesis and we can conclude that there is significant difference in different ways to implant false memories in people.

Answer(8):

As we have rejected null hypothesis (i.e. we have significant test), so we can proceed for pairwise comparison.

We can use the LSD test to compare the effect of different ways.

The standard error for difference in means is given by

Let us take the level of significance as 0.05 i.e. α=0.05

The least significant difference (LSD) is

LSD =t_{(α/2,error df)}*se

LSD =t_(0.975,20)*se

LSD =2.086*0.84

LSD = 1.75

We can compare the difference in means with the above LSD to know if the pair wise difference in means of the three experimental conditions and the control group is significant. We can do it as below:

pair of treatments	pair wise absolute mean difference	LSD	Decision about mean difference	Significance
A-D	5.167	1.75	Greater than LSD	Significant
B-D	7.667	1.75	Greater than LSD	Significant
C-D	1.167	1.75	Less than LSD	Non significant

Answer(9):

In similar way of above answer, we can answer this question also.

We can compare the difference in means with the above LSD to know if the pair wise difference between the means of the three experimental conditions is significant. We can do it as below:

pair of treatments	pair wise absolute mean difference	LSD	Decision about mean difference	Significance
A-B	2.5	1.75	Greater than LSD	Significant
A-C	4	1.75	Greater than LSD	Significant
B-C	6.5	1.75	Greater than LSD	Significant