IBDP Maths SL 4.4 Linear correlation of bivariate data AA HL Paper 2- Exam Style Questions- New Syllabus
Question
| Play time (\(x\)) | 11 | 13 | 14 | 17 | 22 | 24 |
|---|---|---|---|---|---|---|
| Sleep time (\(y\)) | 62 | 65 | 68 | 75 | 84 | 87 |
The relationship between these variables is modeled by the regression line of \(y\) on \(x\), given by the equation \(y = ax + b\).
(a) Calculate the values of the coefficients \(a\) and \(b\).
(b) Estimate the weekly sleep duration for a child whose weekly play time is 20 hours based on this model.
Syllabus Topic Codes (IB Mathematics AA HL):
▶️ Answer/Explanation
(a)
Using the data:
\(\bar{x} = \frac{11+13+14+17+22+24}{6} = 16.8333\ldots\)
\(\bar{y} = \frac{62+65+68+75+84+87}{6} = 73.5\)
\(S_{xx} = \sum (x_i – \bar{x})^2 \approx 141.667\)
\(S_{xy} = \sum (x_i – \bar{x})(y_i – \bar{y}) \approx 280\)
Then:
\(a = \frac{S_{xy}}{S_{xx}} \approx \frac{280}{141.667} \approx 1.9765\)
\(b = \bar{y} – a\bar{x} \approx 73.5 – 1.9765 \times 16.8333 \approx 40.2286\)
\(a \approx \boxed{1.98},\; b \approx \boxed{40.2}\).
(b)
Substitute \(x = 20\) into \(y = 1.9765x + 40.2286\):
\(y = 1.9765 \times 20 + 40.2286 \approx 79.7589\)
Approximately \(\boxed{79.8}\) hours.