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Louise is interested to see whether there is a difference between the number of pictures recalled by children with dyslexia and by those who do not have dyslexia - AQA - A-Level Psychology - Question 5 - 2017 - Paper 1
Question 5
Louise is interested to see whether there is a difference between the number of pictures recalled by children with dyslexia and by those who do not have dyslexia. Th... show full transcript
Worked Solution & Example Answer:Louise is interested to see whether there is a difference between the number of pictures recalled by children with dyslexia and by those who do not have dyslexia - AQA - A-Level Psychology - Question 5 - 2017 - Paper 1
Step 1
Calculate the range of scores Louise gathered in both conditions of her study and complete Table 2 above.
Answer
To calculate the range, subtract the lowest score from the highest score for each condition.
For children with dyslexia:
- Highest score: 17
- Lowest score: 5
- Range = 17 - 5 = 12
For children without dyslexia:
- Highest score: 17
- Lowest score: 4
- Range = 17 - 4 = 13
Therefore, the ranges are:
- Range for children with dyslexia: 12
- Range for children without dyslexia: 13
Step 2
Calculate the standard deviation for the number of pictures recalled by children with dyslexia. Show your working and give your answer to two decimal places.
Answer
To calculate the standard deviation, follow these steps:
-
Find the Mean (ar{x}):
ar{x} = rac{ ext{sum of scores}}{ ext{number of scores}} = rac{16 + 8 + 5 + 12 + 14 + 17 + 11 + 11 + 13 + 13}{10} = rac{129}{10} = 12.9
-
Calculate each score's deviation from the Mean and square it:
- (16 - 12.9)² = (3.1)² = 9.61
- (8 - 12.9)² = (-4.9)² = 24.01
- (5 - 12.9)² = (-7.9)² = 62.41
- (12 - 12.9)² = (-0.9)² = 0.81
- (14 - 12.9)² = (1.1)² = 1.21
- (17 - 12.9)² = (4.1)² = 16.81
- (11 - 12.9)² = (-1.9)² = 3.61
- (11 - 12.9)² = (-1.9)² = 3.61
- (13 - 12.9)² = (0.1)² = 0.01
- (13 - 12.9)² = (0.1)² = 0.01
-
Find the Variance:
ext{Variance} = rac{ ext{sum of squared deviations}}{N} = rac{9.61 + 24.01 + 62.41 + 0.81 + 1.21 + 16.81 + 3.61 + 3.61 + 0.01 + 0.01}{10} = rac{121.29}{10} = 12.129
-
Find the Standard Deviation ( ext{SD}):
ext{SD} = ext{sqrt(Variance)} = ext{sqrt(12.129)} ≈ 3.48
Therefore, the standard deviation for the number of pictures recalled by children with dyslexia is approximately 3.48.