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a mathematics teacher wanted to see the correlation between test scores and homework. the homework grade (x) and test grade (y) are given in the accompanying table. write the linear regression equation that represents this set of data, rounding all coefficients to the nearest tenth. using this equation, find the projected test grade, to the nearest integer, for a student with a homework grade of 35.
| homework grade (x) | test grade (y) |
|---|---|
| 88 | 87 |
| 88 | 85 |
| 83 | 78 |
| 81 | 74 |
| 81 | 76 |
| 52 | 41 |
| 62 | 55 |
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Step1: List data points
Data pairs: (90,78), (88,87), (88,85), (83,78), (81,74), (81,76), (52,41), (62,55) (n=8)
Step2: Calculate sums
$\sum x = 90+88+88+83+81+81+52+62 = 625$
$\sum y = 78+87+85+78+74+76+41+55 = 574$
$\sum xy = (90×78)+(88×87)+(88×85)+(83×78)+(81×74)+(81×76)+(52×41)+(62×55) = 46368$
$\sum x^2 = 90^2+88^2+88^2+83^2+81^2+81^2+52^2+62^2 = 49337$
Step3: Compute slope (m)
$m = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2} = \frac{8×46368 - 625×574}{8×49337 - 625^2} ≈ 1.0$
Step4: Compute intercept (b)
$b = \frac{\sum y - m\sum x}{n} = \frac{574 - 1.0×625}{8} ≈ -6.4$
Step5: Regression equation
$y = 1.0x - 6.4$
Step6: Predict y for x=35
$y = 1.0×35 - 6.4 = 28.6 ≈ 29$
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