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A teacher is monitoring the progress of students using a computer based revision course - Edexcel - A-Level Maths Statistics - Question 1 - 2009 - Paper 1
Question 1
A teacher is monitoring the progress of students using a computer based revision course. The improvement in performance, y marks, is recorded for each student along ... show full transcript
Worked Solution & Example Answer:A teacher is monitoring the progress of students using a computer based revision course - Edexcel - A-Level Maths Statistics - Question 1 - 2009 - Paper 1
Step 1
Calculate \( S_{xx} \) and \( S_{yy} \)
Answer
To calculate ( S_{xx} ) and ( S_{yy} ), we can use the formulas:
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Calculate ( S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n} )
- Here, ( \sum x^2 = 57.22 )
- ( \sum x = 21.4 )
- With ( n = 10 ):
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Calculate ( S_{yy} = \sum y^2 - \frac{(\sum y)^2}{n} )
- Here, ( \sum y = 96 )
- And using the individual ( y ) values:
- ( \sum y^2 = (5^2 + 20^2 + 21^2 + 13^2 + 17^2 + (-3)^2 + (-5)^2 + 15^2 + (-10)^2 + 20^2) = 313.7 )
- Using this we find:
Step 2
Find the equation of the least squares regression line of y on x in the form \( y = a + bx \)
Answer
To find the equation of the least squares regression line:
-
Calculate the slope ( b ):
- Formula: ( b = \frac{S_{xy}}{S_{xx}} ) where ( S_{xy} = \sum xy - \frac{(\sum x)(\sum y)}{n} )
- Yielding the calculation as:
S_{xy} = 313.7 - \frac{(21.4)(96)}{10} = 313.7 - 204.48 = 109.22 \
- So, ( b = \frac{109.22}{11.464} = 9.525 $$
-
Calculate the intercept ( a ):
- Formula: ( a = \frac{\sum y - b \sum x}{n} )
-
Therefore, the regression equation is:
Step 3
Give an interpretation of the gradient of your regression line
Answer
The gradient of the regression line, ( b = 9.525 ), indicates that for each additional hour spent in the revision course, the students' marks are expected to improve by approximately 9.5 marks. This shows a positive correlation between time spent on the revision and performance.
Step 4
Step 5
Lee spends 8 hours using the revision course claiming that this should give him an improvement in performance of over 60 marks. Comment on Lee's claim.
Answer
To assess Lee's claim, we substitute ( x = 8 ) into our regression equation:
Calculating:
This indicates that Lee is predicted to improve by approximately 65.8 marks; therefore, his claim holds true as it exceeds 60 marks. However, the model's validity may be questioned since it is based on a limited dataset.