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## Statistics and probability

### Course: Statistics and probability > Unit 5

Lesson 6: More on regression- Squared error of regression line
- Proof (part 1) minimizing squared error to regression line
- Proof (part 2) minimizing squared error to regression line
- Proof (part 3) minimizing squared error to regression line
- Proof (part 4) minimizing squared error to regression line
- Regression line example
- Second regression example
- Calculating R-squared
- Covariance and the regression line

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# Regression line example

Regression Line Example. Created by Sal Khan.

## Want to join the conversation?

- I don't really understand the meaning of the word "regression" being used as a noun in this context. I thought it made sense in the phrase "regression to the mean", as in "returning to the mean". And I can find clear definitions of "regression line" or "regression analysis" but none of the word "regression" on its own.(15 votes)
- It's a quirk of history. The "regression" part of the name came from its early application by Sir Francis Galton who used the technique doing work in genetics during the 19th century. He was looking at how an offspring's characteristics tended to be between those of the parents (i.e. they regressed to the mean of the parents). The "regression" part just ended up stuck as part of the name from then on. Other than that, linear regression has nothing to do with regression to the mean.(27 votes)

- What is the difference between this method of figuring out the formula for the regression line and the one we had learned previously? that is: slope = r*(Sy/Sx) and since we know the line goes through the mean of the Xs and the mean of the Y's we can figure out the y-intercept by substituting on the formula y= mx +b.(17 votes)
- When would you learn this? Algebra 1 or 2? Or something else?(3 votes)
- You would probably learn this in HS statistics. If not, then definitely in college stats, especially since the proof in the previous videos uses some calculus. Algebra I/II wouldn't go into this area.(3 votes)

- How can we derive m = r*(std y / stdx) from the formula m in this lesson?(3 votes)
- My textbook uses 'least squares regression line', and has this really complicated equation (or at least it looks like it's complicated) that involves summations and such. Is this the same thing? It has the x's and y's in nearly the same spots, but it's got those different symbols.(1 vote)
- Yes.
`N`

∑ : x(1) + x(2) + x(3) + ... + x(N)

n=1(3 votes)

- At3:30and4:04, why do you divide the "Y's" and the "X's" by 3? Where did the "three" come from?(2 votes)
- he was just calculating mean by adding up all the numbers then DIVIDING BY THE TOTAL NUMBER OF DATA POINTS, which is 3(1 vote)

- Where can I find the video for the derivation of the slope and y-intercept of the best fitting regression line using least squares? It doesn't seem to follow the introduction video or precede this example.(2 votes)
- Which videos does it refer to at the beginning, i.e. to derive these formulas? Could you please share the link?(2 votes)
- How can we calculate "m" and "b" when we have multiple independent variables? Can I use the above formula for multiple independent variable also?(1 vote)
- You might be able to use the same process that Sal used in the sequence of videos titled "Proof (part x) Minimizing squared error to regression line" to get slope estimates when there are multiple independent variables. More frequently, matrix algebra is used to get the slopes. If you are familiar with linear algebra, the idea it so say that:
`Y = Xβ + e`

Where:

Y is a*vector*containing all the values from the dependent variables

X is a*matrix*where each column is all of the values for a given independent variable.

e is a*vector*of residuals.

Then we say that a predicted point is`Yhat = Xβ`

, and using matrix algebra we get to`β = (X'X)^(-1) (X'Y)`

(3 votes)

- Could you please explain orthogonal regression with an example?(1 vote)
- An engineer at a medical device company wants to determine whether the company's new blood pressure monitor is equivalent to a similar monitor that is made by a different company. The engineer measures the systolic blood pressure of a random sample of 60 people using both monitors.

To determine whether the two monitors are equivalent, the engineer uses orthogonal regression. Previous to the data collection for the orthogonal regression, the engineer did separate studies on each monitor to estimate the variances. The variance for the new monitor was 1.08. The variance for the other company's monitor was 1.2. The engineer decides to assign the new monitor to be the response variable and the other company's monitor to be the predictor variable. With these assignments, the error variance ratio is 1.08 / 1.2 = 0.9.(3 votes)

## Video transcript

In the last several videos,
we did some fairly hairy mathematics. And you might have even
skipped them. But we got to a pretty
neat result. We got to a formula for the
slope and y-intercept of the best fitting regression line
when you measure the error by the squared distance
to that line. And our formula is, and I'll
just rewrite it here just so we have something
neat to look at. So the slope of that line is
going to be the mean of x's times the mean of the y's minus
the mean of the xy's. And don't worry, this seems
really confusing, we're going to do an example of this
actually in a few seconds. Divided by the mean of x squared
minus the mean of the x squareds. And if this looks a little
different than what you see in your statistics class or your
textbook, you might see this swapped around. If you multiply both the
numerator and denominator by negative 1, you could see this
written as the mean of the xy's minus the mean of x times
the mean of the y's. All of that over the mean of the
x squareds minus the mean of the x's squared. These are obviously
the same thing. You're just multiplying the
numerator and denominator by negative 1, which is same thing
as multiplying the whole thing by 1. And of course, whatever you get
for m, you can then just substitute back in this
to get your b. Your b is going to be equal
to the mean of the y's minus your m. Let me write that in yellow
so it's very clear. You solved for the m value. Minus m times the
mean of the x's. And this is all you need. So let's actually put
that into practice. So let's say I have three
points, and I'm going to make sure that these points
aren't colinear. Because, otherwise, it wouldn't
be interesting. So let me draw three
points over here. Let's say that to one point
is the point 1 comma 2. So this 1, 2. And then we also have
the point 2 comma 1. And then, let's say we also
have the point, let's do something a little bit
crazy, 4 comma 3. So this is 4, 3. So those are our three points. And what we want to do is find
it the best fitting regression line, which we suspect
is going to look something like that. We'll see what it actually looks
like using our formulas, which we have proven. So a good place to start is just
to calculate these things ahead of time, and then
to substitute them back in the equation. So what's the mean of our x's? The mean of our x's is going
to be 1 plus 2 plus 4 divided by 3. And what's this going to be? 1 plus 2 is 3, plus 4
is 7 divided by 3. It is equal to 7/3. Now, what is the mean
of our y's? The mean of our y's is equal
to 2 plus 1 plus 3. All of that over 3. So this is 2 plus 1 is 3. Plus 3 is 6. Divided by 3 is equal to 2. This is 6 divided by
3 is equal to 2. Now, what is the mean
of our xy's? So our first xy over
here is 1 times 2. Plus 2 times 1 plus 4 times 3. And we have three
of these xy's. So divided by 3. So what's this going
to be equal to? We have 2 plus 2, which is 4. 4 plus 12, which is 16. So it's going to be 16/3. And then the last one we have
to calculate is the mean of the x squareds. So what's the mean of
the x squareds? The first x squared is just
going to be 1 squared. Plus this 2 squared, plus
this 4 squared. And we have three data
points again. So this is 1 plus
4, which is 5. Plus 16. Is equal to 21/3, which
is equal to 7. So that worked out to a
pretty neat number. So let's actually find
our m's and our b's. So our slope, our optimal slope
for our regression line, the mean of the x's is
going to be 7/3. Times the mean of the y's. The mean of the y's is 2. Minus the mean of the xy's. Well, that's 16/3. And then, all of that over
the mean of the x's. The mean of the x's
is 7/3 squared. Minus the mean of
the x squareds. So it's going to be minus
this 7 right over here. And we just have to do a little
bit of mathematics. I'm tempted to get out my
calculator, but i'll resist the temptation. It's nice to keep things
as fractions. Let's see if I can
calculate this. This is 14/3 minus 16/3. All of that over,
this is 49/9. And then minus 7. If I wanted to express that as
something over 9, that's the same thing as 63/9. So in our numerator, we
get negative 2/3. And then in our denominator,
what's 49 minus 63? That's negative 14/9. And this is the same thing
as negative 2/3 times negative 9/ 14. Divide numerator and
denominator by 3. Well, the negatives are going
to cancel out first of all. You divide by 3. That becomes a 1. That becomes a 3. Divide by 2. Becomes a 1. That becomes a 7. So our slope is 3/7. Not too bad. Now, we can go back and figure
out our y-intercept. So let's figure out our
y-intercept using this right over here. So our y-intercept, b, is going
to be equal to the mean of the y's, the mean of the
y's is 2, minus our slope. We just figured out our
slope to be 3/7. Times the mean of the
x's, which is 7/3. These just are the reciprocal
of each other, so they cancel out. That just becomes 1. So our y-intercept is literally
just 2 minus 1. So it equals 1. So we have the equation
for our line. Our regression line is going
to be y is equal to-- We figured out m. m is 3/7. y is equal to 3/7 x plus,
our y-intercept is 1. And we are done. So let's actually try
to graph this. So our y-intercept
is going to be 1. It's going to be right
over there. And the slope of our
line is 3/7. So for every 7 we
run, we rise 3. Or another way to think of
it, for every 3.5 we run, we rise 1.5. So we're going to go 1.5
right over here. So this line, if you were to
graph it, and obviously I'm hand drawing it, so it's not
going to be that exact, is going to look like that
right over there. And it actually won't go
directly through that line. So I don't want to give
you that impression. So it might look something
like this. And this line, we have shown,
that this formula minimizes the squared distances
from each of these points to that line. Anyway, that was, at least
in my mind, pretty neat.