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1. 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
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 a... show full transcript
Worked Solution & Example Answer:1. 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:
( S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n} )
Where:
- ( n = 10 ) (the number of students)
- ( \sum x^2 = 57.22 ) and ( \sum x = 21.4 )
So,
( S_{xx} = 57.22 - \frac{(21.4)^2}{10} = 57.22 - 45.756 = 11.464 )
Next, to calculate ( S_{yy} ), we use:
( S_{yy} = \sum y^2 - \frac{(\sum y)^2}{n} )
Where:
- ( \sum y^2 = 313.7 ) and ( \sum y = 96 )
( S_{yy} = 313.7 - \frac{(96)^2}{10} = 313.7 - 921.6 = -607.9 )
Hence, ( S_{xx} \approx 11.464 ) and ( S_{yy} \approx -607.9 ).
Step 2
Find the equation of the least squares regression line of $y$ on $x$ in the form $y = a + bx$.
Answer
The slope ( b ) of the regression line can be calculated using:
( b = \frac{S_{xy}}{S_{xx}} )
We first need to calculate ( S_{xy} ):
( S_{xy} = \sum xy - \frac{(\sum x)(\sum y)}{n} )
Using the given data:
- ( S_{xy} = 313.7 )
( = 313.7 - \frac{(21.4)(96)}{10} = 313.7 - 205.44 = 108.26 )
Hence, we can find ( b ):
( b = \frac{108.26}{11.464} \approx 9.45 )
Next, we calculate the y-intercept ( a ):
( a = \bar{y} - b\bar{x} )
Where:
- ( \bar{y} = \frac{96}{10} = 9.6 )
- ( \bar{x} = \frac{21.4}{10} = 2.14 )
Calculating:
( a = 9.6 - 9.45(2.14) \approx 9.6 - 20.22 = -10.62 )
The equation of the regression line is:
( y = -10.62 + 9.45x ).
Step 3
Give an interpretation of the gradient of your regression line.
Answer
The gradient ( b \approx 9.45 ) indicates that for each additional hour spent in the revision course, the expected improvement in marks is approximately 9.45 marks. This shows a positive relationship between hours spent and marks obtained, suggesting effective revision.
Step 4
Rosemary spends 3.3 hours using the revision course. Predict her improvement in marks.
Answer
To predict Rosemary's improvement in marks, substitute ( x = 3.3 ) into the regression equation:
( y = -10.62 + 9.45(3.3) )
Calculating:
( y \approx -10.62 + 31.185 = 20.565. )
Therefore, her predicted improvement in marks is approximately 20.57 marks.
Step 5
Comment on Lee's claim.
Answer
Lee's claim of achieving over 60 marks after spending 8 hours on the revision course may not be supported by the regression model, as we calculate:
( y = -10.62 + 9.45(8) \approx 62.78. )
However, since the model suggests improved performance above 60 marks, it is important to note that such predictions could be invalid if outside the data range, particularly when considering that the original data showed large variances in results.