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Problem:

The following table below shows the marks scored by seven students on two different mathematics tests.

$$\begin{array}{c|c|c|} \text{Test 1 (x)} & 15 & 23 & 25 & 30 & 34 & 34 & 40 \\ \hline \text{Test 2 (y)} & 20 & 26 & 27 & 32 & 35 & 37 & 35 \\ \hline \end{array}$$

Let $L_{1}$ be the regression line of x on y. The equation of the line $L_{1}$ can be written in the form $x = ay + b$.

(a) Find the value of $a$ and the value of $b$.

Let $L_{2}$ be the regression line of $y$ on $x$. The lines $L_{1}$ and $L_{2}$ pass through the same point with coordinates $(p,q)$.

(b) Find the value of $p$ and the value of $q$.

My solution:

I inputted both sets of data into the calculator (TI-84) and got a line of best fit for both.

$x$ on $y$: $y=1.2908291457286x-10.379396984925$

$y$ on $x$: $y=0.69993188010899x+10.187670299727$

These two lines intersect at $(34.81,34.55)$

The actual solution: The lines should meet at $(x̄, ȳ) = (28.7,30.3)$

Can anyone point me in the correct direction? I just don't see how these two lines could intersect at the mean x,y values when those values are obviously flipped in the two lines.

Edit: This problem is from the IB Math AA HL curriculum, exam code SPEC/5/MATAA/SP2/ENG/TZ0/XX/M . This is a practice test provided by IB.

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  • $\begingroup$ Welcome to CV! Is this a question from a course or textbook? If so, please add the [self-study] tag & read its wiki. $\endgroup$ – Stochastic Aug 12 at 19:20
  • $\begingroup$ This is called the point of averages; the link goes to a search that will show you many answers. $\endgroup$ – whuber Aug 12 at 19:23
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Any linear regression goes through $\bar x,\bar y$, hence, these two should intersect too.

Consider regressions $y=a+bx+e$ and $x=c+dy+u$, take the expectations of both sides of equations: $$E[y]=a+bE[x]+E[e]$$ $$E[y]=a+bE[x]$$ similarly $$E[x]=c+dE[y]$$

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  • $\begingroup$ For the 'x on y' case, wouldn't the linear regression pass through $\bar y, \bar x$? $\endgroup$ – maydc Aug 13 at 15:15
  • $\begingroup$ @maydc, yes, that's the point. except then you draw y on x-axis and vice versa, hence, the intersection $\endgroup$ – Aksakal Aug 13 at 15:39
  • $\begingroup$ A little unintentional pun there, haha. Thanks for the help $\endgroup$ – maydc Aug 13 at 21:51
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Ok, so I think I see where I messed up. In this line:

$x$ on $y$: $y=1.2908291457286x-10.379396984925$

The position of the x and y are reversed. The equation should be written as $x=1.2908291457286y-10.379396984925$

In order to put these two regression lines on the same graph, both equations need to be solved for the same variable (even if in the first equation, Y is the independent).

Now when I graph, I see that these two lines of regression do indeed intersect at the $(\bar x,\bar y)$ .

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