Photo AI
Charles investigated the influence of an interference task on recall from short-term memory - AQA - A-Level Psychology - Question 7 - 2020 - Paper 1
Question 7
Charles investigated the influence of an interference task on recall from short-term memory. The same participants had to recall a word list after an interference t... show full transcript
Worked Solution & Example Answer:Charles investigated the influence of an interference task on recall from short-term memory - AQA - A-Level Psychology - Question 7 - 2020 - Paper 1
Step 1
Complete Table 1 and calculate the Wilcoxon Signed Ranks test for Charles’s study
Answer
To complete Table 1, we first need to calculate the difference between the recall after the interference task (Condition A) and the recall with no interference task (Condition B).
Step 1: Calculate the Differences
- For each participant:
- A: 8 - 12 = -4
- B: 9 - 11 = -2
- C: 6 - 12 = -6
- D: 8 - 8 = 0
- E: 10 - 9 = 1
- F: 10 - 11 = -1
- G: 5 - 10 = -5
- H: 4 - 4 = 0
Calculated Differences:
| Participant | Condition A | Condition B | Difference |
|---|---|---|---|
| A | 8 | 12 | -4 |
| B | 9 | 11 | -2 |
| C | 6 | 12 | -6 |
| D | 8 | 8 | 0 |
| E | 10 | 9 | 1 |
| F | 10 | 11 | -1 |
| G | 5 | 10 | -5 |
| H | 4 | 4 | 0 |
Step 2: Rank the Differences
- Exclude zeros first:
- Ranked differences of non-zero values and assign ranks:
- -6, -5 (ranked 1, 2) ... -2 (ranked 4), -1 (ranked 5), 1 (ranked 6)
- The rank for negative differences gets cumulative sums (for T calculation).
Step 3: Calculate T value
- T is the sum of ranks for the negative differences:
- T = 1 + 2 + 4 + 5 = 12
Conclusion of Part (a)
- The completed table and calculated T is T = 4 based on ranking shown.
Step 2
Using the Wilcoxon T value that you calculated for 7(a), determine whether Charles’s data were significant at p≤0.025 for a directional (one-tailed) hypothesis
Answer
Step 4: Determine Significance
To determine significance:
- The critical T value for a one-tailed test should be referenced from a Wilcoxon Signed Ranks table. For 8 participants, we check the critical value.
- Since we calculated T = 4, and let's assume the critical value at p ≤ 0.025 is 2;
Conclusion:
- The calculated T value (4) is greater than the critical value (2), thus Charles’s data are not statistically significant at p≤0.025 for a directional hypothesis.