Nature and determinants of the continued influence effect

The main aim of this report is given as below:

Aim:

1. To assess whether replacing the misinformation at retraction (B at Time 2) with different types of information reduces the CIE (B is less likely to be present at Time 3).
2. To assess whether the type of retrieval task (free recall or recognition) has an impact on the strength of the CIE (likelihood of B present at Time 3).

Hypothesis

• Hypothesis : If a complete causal structure results in a stronger event model, then retraction alternative (1) should result in less CIE at Time 3 than retraction-only
• Hypothesis : If retractions are strengthened by intentional content and/or mutual exclusion of the misinformation, then retraction-alternative (2) should result in a lower CIE than retraction-alternative.
• Hypothesis : If free recall results in more accurate retrieval in a Cognitive Interview, it would be expected that free recall will also be associated with a smaller CIE than will be the case for recognition.

METHOD

Given data:

Table 1:

Recognition Group	Task	Â	CIE
4	2	1	2
4	2	1	2
4	2	1	1
4	2	1	1
4	2	1	4
4	2	1	5
4	2	1	2
4	2	1	5
4	2	1	2
4	2	1	2
4	2	1	5
4	2	1	1
4	2	1	6
4	2	1	2
4	2	1	7
5	2	2	1
5	2	2	1
5	2	2	1
5	2	2	2
5	2	2	1
5	2	2	2
5	2	2	1
5	2	2	2
5	2	2	1
5	2	2	1
5	2	2	1
5	2	2	2
5	2	2	1
5	2	2	2
5	2	2	2
6	2	3	0
6	2	3	1
6	2	3	2
6	2	3	1
6	2	3	1
6	2	3	1
6	2	3	0
6	2	3	2
6	2	3	1
6	2	3	0
6	2	3	2
6	2	3	0
6	2	3	1
6	2	3	0
6	2	3	2

Calculation required for independent t-test:

One-way ANNOVA test for Recognition condition

Table 2:

C4 Recognition only		C5 Recognition alternative (1)
CIE	Score Squared	CIE	Score Squared
2	4	1	1
2	4	1	1
1	1	1	1
1	1	2	4
4	16	1	1
5	25	2	4
2	4	1	1
5	25	2	4
2	4	1	1
2	4	1	1
5	25	1	1
1	1	2	4
6	36	1	1
2	4	2	4
7	49	2	4
Sum X1 = 47	Sum X12 = 203	Sum X2 = 21	Sum X22 = 33
Mean (X1) = 3.11	Â	Mean (X2) = 1.4	Â

Now, after identifying the mean (𝑋̅) for each condition of recognition, next step is to calculate Sums of Squares (SS) for each condition by using following formula :-

𝑆𝑆₁ = ∑X₁² − (∑ 𝑋₁)² / 𝑁

      = 203 − (47)² / 15

     = 203 – 147.26

    = 55.74

𝑆𝑆₂ = ∑X₂² − (∑ 𝑋₂)² / 𝑁

      = 33 − (21)² / 15

     = 33 - 29.4

    = 3.6

Now, the combined variance at n-1 degrees of freedom, i.e. (15-1 = 14), can be calculated by using following formula:

𝑠_{𝑝}² = 𝑆𝑆₁ + 𝑆𝑆₂

           𝑑𝑓₁ + 𝑑𝑓₂    

          = 55.74 + 3.6

           14 + 14

         = 59.34 / 28 = 2.11

 

This combined variance is further used for calculating the sampling error, that provides an estimate of non-systematic or error variance in following way:

 S_{𝑋̅  - 𝑋̅  Â}  = √ 𝑠_{𝑝}²   + 𝑠_{𝑝}²

                     n_{1          Â}  n₂

_{                   =} √ 2.11   + 2.11

                      15          15

                = √0.28   = 0.53

 At last, independent measures, i.e., the t-test, which can be defined as a ratio, can be calculated by dividing the mean difference from estimate of error variance in following way:  

                                      𝑡 = (𝑋̅₁ - 𝑋̅₂)

                                          S_{𝑋̅  - 𝑋̅}

_{                                                          Â} Â = 3.11 - 1.4

                                                        0.53

                                                  = 1.71 / 0.53 = 3.22

At t_28 (t _critical) = 2.048, the mean difference between C₄ and C₅ is 1.71, which is greater than p value (p > 0.05), but the calculated value of t is greater than the critical value, so, under this case, the null hypothesis is rejected. 

Turning now towards second recognition condition, t-test can be measured in following way:

Table 3:

C5 Recognition only		C6 Recognition alternative (1)
CIE	Score Squared	CIE	Score Squared
1	1	0	0
1	1	1	1
1	1	2	4
2	4	1	1
1	1	1	1
2	4	1	1
1	1	0	0
2	4	2	4
1	1	1	1
1	1	0	0
1	1	2	4
2	4	0	0
1	1	1	1
2	4	0	0
2	4	2	4
Sum X1 = 21	Sum X12 = 33	Sum X2 = 14	Sum X22 = 22
Mean (X1) = 1.4	Â	Mean (X2) = 0.93	Â

Sums of Squares (SS) for each condition can be calculated:

𝑆𝑆₁ = ∑X₁² − (∑ 𝑋₁)² / 𝑁

      = 33 − (21)² / 15

     = 33 - 29.4

    = 3.6

𝑆𝑆₂ = ∑X₂² − (∑ 𝑋₂)² / 𝑁

      = 22 − (14)² / 15

     = 22 – 13.06

    = 8.94

Now, the pooled or combined variance with n-1 degrees of freedom, i.e. (15-1 = 14), is calculated as

𝑠_{𝑝}² = 𝑆𝑆₁ + 𝑆𝑆₂

           𝑑𝑓₁ + 𝑑𝑓₂

               = 3.6 + 8.94

                  14 + 14

                = 12.54 / 28 = 0.45

Now, using this combined variance to calculate sampling error as

S_{𝑋̅  - 𝑋̅  Â}  = √ 𝑠_{𝑝}²   + 𝑠_{𝑝}²

                     n_{1          Â}  n₂

_{                          =} √ 0.45   + 0.45

                      15          15

                     = √0.06   = 0.25

 Then, value of t will be

              𝑡 = (𝑋̅₁ - ð'‹Ì…2) ð¡

               ð'‹Ì_{…  - 𝑋̅}

_{                  Â} Â = 1.4 – 0.93

                     0.25

                = 0.47 / 0.25 = 1.88

Now, mean difference between C₅ and C₆ is 0.47; therefore, p > 0.05, while at  t_28   (t _critical) = 2.048, the calculated value of t is less than critical value, so, under this condition, null hypothesis is accepted.

Table 4:

Calculation for one-way ANOVA for independent samples

C4 Recognition only		C5 Recognition alternative (1)		C6 Recognition alternative (2)
CIE	Score Squared	CIE	Score Squared	CIE	Score Squared
2	4	1	1	0	0
2	4	1	1	1	1
1	1	1	1	2	4
1	1	2	4	1	1
4	16	1	1	1	1
5	25	2	4	1	1
2	4	1	1	0	0
5	25	2	4	2	4
2	4	1	1	1	1
2	4	1	1	0	0
5	25	1	1	2	4
1	1	2	4	0	0
6	36	1	1	1	1
2	4	2	4	0	0
7	49	2	4	2	4
n = 15	Â	n = 15	Â	n = 15	Â
Sum (X1) =  47	Â	Sum (X2) = 21	Â	Sum (X3) = 14	Â
Mean = 3.13	Â	Mean = 1.4	Â	Mean = 0.93	Â
Â	Sum (X12) = 203	Â	Sum (X22) = 33	Â	Sum (X22) = 22
SS = 55.74	Â	SS = 3.6	Â	SS = 8.94	Â

Total number . of participants at recognition condition is N (15 + 15 +15) = 45, while G i.e. total number of whether references, is 82. Now,

Sum of squares between groups,

SS _between =  ∑ T²  - G²

                        n     

then,

SSb = (472 + 212 + 142) - 822

         15       15         15        45

            = (189.73 - 149.42)

           = 40.31

Here, degrees of freedom (df) = k – 1 = 3 – 1 = 2

Sum of squares within groups,

SS _within =  ∑ SS

             = 55.74 + 3.6 + 8.94

             = 68.28

 Here, degrees of freedom (df) = N – k = 45 – 3 = 42

 Now, total sum of squares:

            SS_T = ∑X² – G² / N

                         = (203 + 33 + 22) – 82² / 45

                        = 258 – 149.42

                       = 108.58

At this level, df_total = N – 1 = 45 – 1 = 44

Now, F value can be calculated by

ANOVA Summary Table

Source	SS	df	MS	F
Between	40.31	2	SS/df = 40.31/2 = 20.15	MS between = 20.15          MS within         1.62                          = 12.40
Within	68.28	42	SS/df = 68.28/42 = 1.62
Total	Â	44	Â	Â

 Here, calculated value of F(2,42) is 12.40, p < .05. Considering the Fishers’ table (F table), at p = .01, critical value at this level is 5.06. So, actual F value as shown in above table is greater than the critical value, therefore, null hypothesis is rejected at this level.

Again, as unlike t-test, there is a significant effect between more than two groups – C₄ and C₅, C₅ and C₆, or C₄ and C₆. Therefore, to determine where the significant effects lie, Tukey’s HSD in following way –

                               HSD = q √ MS_error

_{                                                                            Â}  n

                                                      = 3.44 √1.62 /15

                                                     = 3.44 x 0.33

                                                     = 1.1352 = 1.14 (approx.)

Now, C₄ (3.13) – C₅ (1.40) = 1.73

         C₅ (1.40) – C₆ (0.93) = 0.47

As value of HSD is smaller than first difference, it can fit between C₄ and C₅, hence a significant difference will lie between them. Therefore, one-way ANOVA shows the difference between these two retractions, including alternative conditions at recognition.

Result

Mean CIE values (Standard Deviations in parentheses) and one-sample ANNOVA test results (comparing means to 0) for

Each level of Retrieval Task by retrieval type, N = 90 (15 per cell), is given as below:

Retrieval Task	Retraction Type
	Retraction Only	Retraction alternative (1)	Retraction Alternative (2)
Free Recall	C1 1.60 (0.91) t(14) = 6.81, p <.001 Â	C2 1.33 (0.98) t(14) = 5.29, p<0.001 Â	C3 0.53 (0.52) t(14) = 4.00, p = 0.001
Recognition	C4  3.13 (2.00) t(14) = 6.08, p<.001	C5 1.40 (0.51) t(14) = 10.69, p<.001 Â	C6 0.93 (0.80) t(14) = 4.53, p<0.001

Discussion

By addressing the hypothesis for the recall and recognition condition, a t-test (one-way  ANOVA) is used in order to identify the differences among the six conditional statements that were made for the plane crash story. Through applying the statistical method, it has been identified that a significant difference lies between C₁ and C₂ (0.27) and C₄ and C₅ (1.73). Therefore, both null hypotheses, H₁ and H2,_ at the recognition condition are rejected, while at the recall condition both are accepted.

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