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    Nature and determinants of the continued influence effect

    University: N/A

    • Unit No: N/A
    • Level: High school
    • Pages: 16 / Words 4111
    • Paper Type: Case Study
    • Course Code: PSYC206
    • Downloads: 236

    INTRODUCTION

    People sometimes encounter with information from inte-way rnet and other media that they subsequently learn is incorrect. With the continued influence effect, they learn "facts" about a particular event which later turn out to be unfounded or false, even after the same is corrected but discredited information continues influence their reasoning and understanding. The current report work shows how Continued Influence Effect (CIE) that consider as a memory phenomenon under which the misinformation, instead of retraction, impact on reasoning and future memories for that event. For this purpose, a case analysis is chosen which is based upon plane crash statements, where 90 participants are selected from 400 respondents, to conduct a test for recall and recognition. Under this test, 15 are distributed in each of 6 groups to arrive the result. To conclude the result, ANNOVA test method will be used to test the hypothesis for the recognition and recall condition.
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    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.

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    METHOD

    Given data:

    Table 1:

    Recognition Group

    Task

     

    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

     

    Mean (X2) = 1.4

     

    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:

    𝑠𝑝2 = 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:

     S𝑋̅  - 𝑋̅    = √ 𝑠𝑝2   + 𝑠𝑝2

                         n1            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)

                                              S𝑋̅  - 𝑋̅

                                                                = 3.11 - 1.4

                                                            0.53

                                                      = 1.71 / 0.53 = 3.22

    At t28 (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

     

    Mean (X2) = 0.93

     

    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

    𝑠𝑝2 = 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

    S𝑋̅  - 𝑋̅    = √ 𝑠𝑝2   + 𝑠𝑝2

                         n1            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

                        = 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  t28   (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

     

    n = 15

     

    n = 15

     

    Sum (X1) =  47

     

    Sum (X2) = 21

     

    Sum (X3) = 14

     

    Mean = 3.13

     

    Mean = 1.4

     

    Mean = 0.93

     

     

    Sum (X12) = 203

     

    Sum (X22) = 33

     

    Sum (X22) = 22

    SS = 55.74

     

    SS = 3.6

     

    SS = 8.94

     

    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:

                SST = ∑X2 – G2 / N

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

                            = 258 – 149.42

                           = 108.58

    At this level, dftotal = 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

     

    44

     

     

     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 √ MSerror

                                                                                  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

     

    C2

    1.33 (0.98)

    t(14) = 5.29, p<0.001

     

    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

     

    C6

    0.93 (0.80)

    t(14) = 4.53, p<0.001

    Discussion

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    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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