Question

Scenario:            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: {x_A} = (4,8,6,6,5,9)

Condition B: {x_B} = (9,10,7,9,11,7)

Condition C: {x_C} = (2,1,2,2,4,3)

Condition D: {x_D} = (0,2,1,1,0,3)

Questions:

1. Express in your own words the null and alternate hypotheses of this ANOVA analysis. [1 mark]

2. What is the critical value of F at α = 0.05? Based on your findings from Question 4, can we reject the null hypothesis? Explain why, and report on these findings. [2 marks]

3. Create and insert a bar chart displaying the means of each of the four groups. Make sure that both your chart and axes are properly labelled. [2 marks]

4. 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. [1 mark]

5. 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. [1 mark]

Answer 1

1.

Null hypothesis- There would be no difference in false memories among all the conditions.

Alternate hypothesis- There would be a difference in false memories among all the conditions.

2.

X1	X2	X3	X4	X12	X22	X32	X42
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
∑ X1= 38	∑ X2=53	∑ X3= 14	∑ X4=7	∑ X12= 258	∑ X22= 481	∑ X32= 38	∑ X42= 15

Here-

Grand sum of X or ∑ X= 112

Grand sum of squares= 792

n₁= 6, n₂=6, n₃=6, n₄=6

       N= 24

df_between= r-1

df_between= 4-1

df_between= 3

df (within group)= N-k

df (within group)= 24-4

df (within group)= 20

Source of variance	Sum of square	df	Mean square = SS/df	f-ratio = Mean square of between group/ Mean square of within group	Level of significance
B/W groups	227	3	75.7	36.1	0.05 & 0.01
Within group	42.3	20	2.1

Critical value of F is significant at 0.05 and 0.01 level as well. It means null hypothesis is rejected and alternative hypothesis is accepted.