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# Using a Linear Model

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- Let's do another example where we have some data.
- We want to fit a line to that data, and maybe use that to
- estimate some of the data that we didn't
- collect, so to speak.
- So the first thing they ask us, let's look at the chart
- first of all.
- This is birth year.
- They started in 1930.
- They go all the way up to the year 2000 in
- increments of 10 years.
- They tell us the life expectancy in years.
- So a good thing for us.
- I think this is the United States.
- The life expectancy has been steadily increasing.
- The first thing they ask us to do is to make a scatter plot
- of the data.
- I did this on Excel ahead of time.
- This is what I got.
- As you can see in 1930 it was 59.7.
- In 1940 it was 62.9.
- These blue dots are just plotting each
- of these data points.
- 1930 is 57.9.
- It's that data point right there.
- In 1940 it was 62.9.
- It's that data point right there.
- So I just went ahead and plotted all of them.
- Then I let Excel figure out a line of best fit.
- I didn't go into the math of how to do this.
- We'll do that in future videos.
- It's out of the scope of introductory algebra.
- But Excel figured out that the line of best fit, the line
- that is least far away from all of these points, that
- minimizes the distance from all of these points, is y is
- equal to 0.2395x minus 400.99.
- I'm not going to go into what r squared is just yet.
- But it's a measure of how good this line is actually fitting.
- So let's use that.
- Let's use this table.
- Let's use our line of best fit to answer the
- rest of their questions.
- So the second question is, use the line of best fit to
- estimate the life expectancy of a person born in 1955.
- So we have two options.
- We could just eyeball it.
- 1955 is right there.
- So if you go up from 1955 you're going to hit the line
- of best fit.
- Let me do it in different color, because that's the same
- color as the dots.
- In 1955 you're going to hit the line of
- best fit right there.
- If I were to just eyeball it, it looks
- like it's right there.
- This is 1, 2, 3, 4, 5.
- So these are increments of 2.
- So it looks like about 77 years.
- So 1955, it looks roughly equal to 77-- no, no sorry--
- 67 years-- let me be clear, it sounded suspicious-- a 67 year
- life expectancy.
- Now we can use the equation of this line of best fit to come
- up with the exact value.
- When x is equal to 1955, what is y?
- What is the life expectancy in years?
- Let's get our calculator out for this.
- Actually let me use a graphing calculator for
- this one right now.
- So let me clear this.
- Let me clear everything.
- Then let's just punch in 1955 for x.
- So we have 0.2395 times 1955, that's our x-value, minus
- 400.99 is equal to 67.23.
- So our line of best fit gives us the exact
- answer of 67.23 years.
- So our eyeballing it wasn't that far off.
- So that's what our line of best fit would predict, if we
- assume that a linear model is a good model to
- describe this data.
- All we know right now are linear models, so
- that's what we'll use.
- It actually looks pretty good.
- It seems like it fits the data pretty, pretty good.
- Let's do the next part.
- Use linear interpolation to estimate the life expectancy
- of a person born in 1955.
- Now linear interpolation means let's take the data points
- around 1955.
- So you have this data point from 1950.
- You have this data point from 1960.
- Let's draw a line between those two data points and
- assume that the 1955 number would be right in between on
- the line there.
- Now we could figure out the equation of that line.
- We have two points.
- We have the 1950 data and we have the 1960 data.
- We could figure out the slope of the line.
- We could figure out its y-intercept.
- Then we can substitute 1955 in there.
- But this is a line, and 1955 is right in
- between these two points.
- It's 5 from either one.
- So on this point right here, that's on that linear
- interpolation, on that orange, will just be halfway
- between these two.
- So we can even average those.
- So let's do that.
- So if we just average those, 68.2 plus 69.7 is equal, and
- then we want to divide it.
- That answer means our previous answer.
- So that number divided by 2 is equal to 68.95.
- So part 2, through linear interpolation we get 68.95.
- Is that what I got?
- 68.95, so that is this point right here.
- So linear interpolation is giving us a slightly higher
- estimate than our line of best fit across all of the data.
- You can even eyeball it if you look at it right there.
- Let's do the next one.
- Use a line of best fit to estimate the life expectancy
- of a person born in 1976.
- Once again, we can use the equation of the line of best
- fit, plug in 1976 here and see what it gets us.
- We can eyeball it.
- 1975 is there.
- 1976 is there.
- It'll be right around there.
- Let's actually use the actual equation.
- So in 1976, this is going to be the line of best fit, the
- line of best fit is going to give us, if we actually figure
- out what this line value gives us, so we get 0.2395 times
- 1976 minus 400.99.
- So 72.26 years, so it gives us 72.26 years.
- So it equals 72.26.
- Let me make this clear here.
- These columns, this one right here, is line of best fit.
- This one right here is interpolation.
- Now they want to use linear interpolation to estimate the
- life expectancy of a person born in 1976.
- So what they want us to do is draw a line between these two
- points, and then use that line, not the whole line of
- best fit, to figure out what 1976 might have been.
- Now before, 1955 was right in between those two values.
- 1976 is not right in between those values.
- So let's actually figure out the equation of this line
- using the points, these two points, 1970 comma 70.8, and
- 1980 comma 73.7.
- Let's figure out the slope of that line, of
- this line right here.
- So the slope is the change in y over change in x.
- It's going to be 73.7 minus 70.8.
- That's how much we change in the y direction, divided by
- how much we change in the x direction, 1980 minus 1970.
- So this is going to be equal to 3.7 minus 0.8 is 2.9.
- So it's 2.9 over 10.
- So it's equal to 0.29.
- We changed 2.9 years in life expectancy every 10 years over
- that time period right there, or 0.29 per year.
- Now what's the equation of that line?
- Well we know that the point 1970 and 70.8 is on that line.
- So if we say the equation is y is equal to mx plus b, we know
- that y is equal to 70.8 when-- we know m is 0.29-- when x is
- equal to, or when the year is equal to 1970,
- times 1970 plus b.
- Now we can calculate our y-intercept.
- So what is our y-intercept?
- We'll get our calculator out.
- We have to subtract this from both sides.
- So b is equal to 70.8 minus 0.29 times 1970.
- Let's get our calculator out.
- There we go.
- All right, 70.8 minus 0.29 times 1970 is
- equal to minus 500.5.
- So this is equal to negative 500.5.
- So the equation of our interpolation line-- remember
- we're just finding the equation of this line between
- 1970 and 1980-- is, that equation is y is equal to
- 0.29x minus 500.5.
- Now if we want to figure out what that line, if we want to
- interpolate 1976, we just have to put 1976 in there for x.
- So what is life expectancy in 1976?
- Get our calculator out again.
- 0.29 times 1976 minus 500.5 is equal to 72.54.
- So we get 72.54 is what happens?
- It's the number we get when we do linear interpolation
- between those two data points.
- All we did is we figured out the equation of the line
- between those two points.
- Then we plugged in the number 1976 to get 72.54.
- That's a slightly, slightly higher number than what the
- line of best fit gave us.
- But it's pretty, pretty close in either scenario.
- Now let's do this last part.
- Use the line of best fit to estimate the life expectancy
- of a person born 2012.
- This is what's cool about the line of best fit.
- We don't have 2012 here.
- 2012 goes off the table.
- But we could just keep continuing on with this line
- right here-- now I'm using the wrong color-- keep continuing
- with this line right here and see what does this line equal
- when the year is equal to 2012?
- So all we have to do is put 2012 in for x and we'll get
- what the line of best fits tells us.
- We'll get the calculator out again.
- So we have 0.2395 times 2012 minus
- 400.99 is equal to 80.88.
- So this last part, in 2012 if you were to use a line of best
- fit, the average life expectancy in the U.S. will be
- 80.88 years.
- Anyway, hopefully you found that to be interesting.
- Sometimes dealing with data can be a little bit hairy.
- These aren't nice, clean numbers.
- But it gives you a little sense of why slope, and y-
- intercept, and interpolation, and fitting a line, is
- actually very important to analyzing data and coming to
- conclusions from that data.

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