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Example 8. At <strong>the</strong> right <strong>is</strong> tabulated <strong>the</strong> mid-term test mark (out <strong>of</strong><br />

10) and <strong>the</strong> final marks for four <strong>of</strong> Jason’s friends who took <strong>the</strong> course<br />

last year. Jason’s mid-year test mark <strong>is</strong> 7. He decides to use a <strong>linear</strong><br />

model<br />

y = mx + b<br />

to predict h<strong>is</strong> final mark. What does he d<strong>is</strong>cover?<br />

Solution.<br />

We use a “least squares” model. The equations are:<br />

90 = 8m + b<br />

85 = 7m + b<br />

95 = 9m + b<br />

75 = 7m + b<br />

In <strong>matrix</strong> form:<br />

8<br />

<br />

<br />

7<br />

9<br />

<br />

7<br />

Xb = y<br />

1<br />

90<br />

1<br />

<br />

<br />

m<br />

<br />

85<br />

.<br />

1<br />

<br />

b<br />

95<br />

<br />

1<br />

75<br />

The least squares solution b^ <strong>is</strong> <strong>the</strong> solution <strong>of</strong> <strong>the</strong> equation:<br />

243<br />

<br />

31<br />

m^<br />

243<br />

<br />

b^<br />

31<br />

The best fit recursive equation <strong>is</strong><br />

(X T X)b^ = X T y.<br />

31m^<br />

2695<br />

<br />

4<br />

<br />

<br />

b^<br />

345 <br />

31<br />

4<br />

<br />

<br />

1<br />

2695<br />

<br />

345 <br />

y = (85x + 290)/11<br />

1 85 <br />

.<br />

11 290<br />

Jason’s estimate <strong>of</strong> h<strong>is</strong> final mark <strong>is</strong> y = (85×7 + 290)/11 = 80.4.<br />

student Mid Final<br />

Brenda 8 90<br />

Alnoor 7 87<br />

Kit 9 94<br />

Tim 7 75<br />

23C approximation and best-fit. 8

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