Linear Regression for Prediction

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En provenance du cours de Johns Hopkins University

Regression Models

2066 notes

Johns Hopkins University

Regression Models

2066 notes

Cours 7 sur 10 dans Specialization Data Science

Linear models, as their name implies, relates an outcome to a set of predictors of interest using linear assumptions. Regression models, a subset of linear models, are the most important statistical analysis tool in a data scientist’s toolkit. This course covers regression analysis, least squares and inference using regression models. Special cases of the regression model, ANOVA and ANCOVA will be covered as well. Analysis of residuals and variability will be investigated. The course will cover modern thinking on model selection and novel uses of regression models including scatterplot smoothing.

À partir de la leçon

Week 2: Linear Regression & Multivariable Regression

This week, we will work through the remainder of linear regression and then turn to the first part of multivariable regression.

Statistical Linear Regression Models2:38

Interpreting Coefficients3:47

Linear Regression for Prediction10:34

Rencontrer les enseignants

Brian Caffo, PhD
Professor, Biostatistics
Bloomberg School of Public Health
Roger D. Peng, PhD
Associate Professor, Biostatistics
Bloomberg School of Public Health
Jeff Leek, PhD
Associate Professor, Biostatistics
Bloomberg School of Public Health

Also, I'd like to remind everyone that we can get a prediction for

Y by taking beta not hat plus beta nought hat and

multiplying it by the X that we want to predict at.

Now if we plug in the observed X's, then we get the fitted values.

The ones who were trying to make as close as possible to the observed data, but

that doesn't mean that we can only predict at the fitted values.

We can predict at any value of X, we'd like to plug into the equation.

However, we're going to have more reasonable predictions if the value of

x that we plugin is in the cloud of data that we're, we used to build the model.

Later on, we'll also talk about how to account for

that kind of uncertainty with prediction intervals.

But for the time being, let's just talk about how we get a prediction.

And in the next couple slides, we'll go through an example of interpreting

the intercept, interpreting the slope and

generating predictions from a specific regression model and setting.

All right. Let's go through some code now

to interpret our regression coefficients and

show you running of the regression coefficient sort of in real time.

The dataset is the diamond dataset from the UsingR package.

The data is diamond prices in Singapore dollars and diamond weight in carats,

which is a standard measure of diamond mass.

To get the data, you need to start out by using

library UsingR to get the UsingR package and

data diamond and then we want, I want to do ggplot2,

because I'm going to do a ggplot first.

So here, let me go through my ggplot commands.

So, I would like to assign to the variable g my ggplot, the dataset is diamond.

My aesthetic has the horizontal axis variable as carat and

the y-axis variable as price.

I'd like to label, I'd like to get in the practice of labeling my axes,

so my plot, I add a layer where the xlab isM ass and carats.

And where the y label, ylab is pr, price in Singapore dollars.

So let me run those, [SOUND] let me run those lines and then I'd like to

add the points, I'd like to add the points of the black background and

then a, a light alpha blending color on top and

then it quite easy to add the regression line.

So, I'm going to add a layer that is geom_smooth and method equals,

method equals lm will add the regression line.

And if you omit any arguments,

it's just going to assume the regression line with y's the outcome and

x is the predictor and that I want my regression line color to be black.

So let me run that line and then call my plot.

[SOUND] There's my plot, you can see on my x-axis,

I have mass, on my y-axis I have price.

And now what I'm plotting is the fitted line, the line that minimizes the sum

of the squared vertical distances between these points and the lines.

Now let's actually go through and get our fitted line.

Just to remind us, the function lm is r is linear model procedure,

so it includes regression as a special case.

Y on the left-hand side of our tilde is the outcome price,

then tilde, think of that as sort of the equal sign in the model.

Our x variable carat.

By default, lm includes an intercept.

So, if you don't want an intercept, you have to explicitly force it in the model.

Then the comma,

we want the dataset that we're looking at to be the diamond dataset.

So, in other words, we have to give it the data frame.

Otherwise, it looks in the regular r environment for variables in the model and

we're going to assign that to, to the variable named fit.

So let's see what that output looks like after we run it.

So there I've run the model and

now I'm going to type fit just to see what it prints out.

It basically just prints out the coefficients beta nought and beta one.

I would note that you can get a much more detailed printout

by doing summary of the outputted variable from the l,

lm fit and you get this more elaborate printout.

And we're going to go through and detail all of the numbers on this printout,

you'll be able to interpret everything on this

printout at some point after the class.

But for the time being, lets just talk about the coefficients.

If you just want to grab the coefficients as a vector, lets do coef fit and

then we get the intercept and labels it as Intercept and the regression variable for

the carat, the slope for the carat regression variable.

So let's look at this 3,721 variable and try to interpret it.

It's saying that we have an expected 3,721 Singapore dollar

increase in price for every carat increase in mass of the diamond.

The intercept, negative 259 is the expected price of a 0 carat diamond.

So not very interested, interesting,

because we're not interested in zero carat diamonds.

Now let's mean center our x variable, so

that the intercept is on a more interpretable scale.

The first thing I'd like to do is assign it to a different variable,

fit2 instead of fit, because I don't want to overwrite the original fit.

Lm is again, my linear model procedure.

My outcome stays the same and

now I want to main center my predictor variable, carat.

So carat minus mean carat.

If you want to do arithmetic operations inside equation statements in lm,

you actually have to surround them by this I function and

then we still want our dataset to be the diamond dataset.

So let's run that code.

[NOISE] So, I've run fit2 [NOISE] and

there are my new coefficients.

Notice of course, the slope stays the same,

3,721, but my intercept has changed to 500.

So $500, Singapore dollars is the expected price of the average sized diamond.

In this case, the average diamond is about 0.2 carats.

A one carat increase is actually kind of big.

What about changing the units to one-tenth of a carat?

We can do this just by dividing the coefficient by ten.

So we know that we would expect to see a $372 increase in price for

every one-tenth of a carat increase in the mass of a diamond.

But let's actually show in r how this works, as well.

Here I am, now assigning to the ver, to the variable fit3.

The linear model fit, where now instead of putting in carat,

I'm putting in carat times 10.

So the units of this new variable is one-tenth of a carat.

The data is of course, still the diamond dataset.

So let me run that and then let me find the coefficient.

And you get, of course,

that it is now 372 rather than 3,721.

So, imagine if someone came to you with three new diamonds that they had.

0.16 carats, 0.27 carats and 0.35 carats,

so here they are right here and let me assign those.

And they wanted to know what you would estimate the price would be.

Well, you could do it manually by grabbing the two coefficients in multiplying

the intercept or adding the intercept plus the slope times these new values.

Let's do that.

And so you would predict 336, 745 and $1,006 for these three diamonds,

respectively based on your fitted linear regression model.

Which by the way, from the scatter plot seems to fit pretty well.

Often, you don't want to do even that much coding, you want to more general method,

especially when you get lots of regression variables.

So there's this general method called predict that will take the output from

several different kinds of model fits.

Linear models are one example, but predict is a generic function, you know, or, and

it applies to several different prediction models.

So we predict from the output of our lm fit and

then you need to give it some new data to predict that.

So new data is a data.frame that has the new values of x for the carat variable.

Then when we do that, what you'll see is of course, it gives you the same answer.

The, now in a way that scales up when we have lots of regressors

in much more complicated settings.

So you generally, want to predict using the predict function.

If you omit this new data statement, if you just do predict fit,

I'll show it to you.

[SOUND] It predicts at the observed x values,

so it gives you the y hat values.

If you want it at new x values, you have to give it this new data argument.

I just wanted to briefly illustrate what the prediction was accomplishing.

Here's our observe data points in blue.

The fitted values when we do the predict command, the fitted values in red all of

the observed x values and their associated fitted points on the line.

These are if we were to draw vertical lines from the observed data points on to

the fitted line, they would occur on these red points.

When we predicted a new value of x,

what we're doing is we're finding a point along this horizontal axis.

That here, I'm giving the three values that we want, 0.16, 0.27 and 0.34.

We're drawing a line up to the fitted regression line and

then over to dollars and those are our predicted dollar amounts.

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