# StatQuest: Multiple Regression in R

```## Here's the data
mouse.data <- data.frame(
size = c(1.4, 2.6, 1.0, 3.7, 5.5, 3.2, 3.0, 4.9, 6.3),
weight = c(0.9, 1.8, 2.4, 3.5, 3.9, 4.4, 5.1, 5.6, 6.3),
tail = c(0.7, 1.3, 0.7, 2.0, 3.6, 3.0, 2.9, 3.9, 4.0))

mouse.data

#######################################################
##
## Let's start by reviewing simple regression by
## modeling mouse size with mouse weight.
##
#######################################################

## STEP 1: Draw a graph of the data to make sure the relationship make sense
plot(mouse.data\$weight, mouse.data\$size, pch=16, cex=2)

## STEP 2: Do the regression
simple.regression <- lm(size ~ weight, data=mouse.data)

## STEP 3: Look at the R^2, F-value and p-value
summary(simple.regression)

abline(simple.regression, lwd=5, col="red")

## now let's verify that our formula for R^2 is correct..
ss.mean <- sum((mouse.data\$size - mean(mouse.data\$size))^2)
ss.simple <- sum(simple.regression\$residuals^2)

(ss.mean - ss.simple) / ss.mean # this is the R^2 value

## now let's verify the our formula for F is correct...
f.simple <- ((ss.mean - ss.simple) / (2 - 1)) /
(ss.simple / (nrow(mouse.data) - 2))

f.simple # this is the F-value

## Now let's draw a figure that shows how to calculate the p-value from the
## F-value
##
## First, draw the correct f-distribution curve with df1=1 and df2=7
x <- seq(from=0, to=15, by=0.1)
y <- df(x, df1=1, df2=7)
plot(x, y, type="l")

## now draw a verticle line where our F-value, f.simple, is.
abline(v=f.simple, col="red")

## color the graph on the left side of the line blue
x.zero.to.line <- seq(from=0, to=f.simple, by=0.1)
y.zero.to.line <- df(x.zero.to.line, df1=1, df2=7)
polygon(x=c(x.zero.to.line, 0), y=c(y.zero.to.line, 0), col="blue")

## color the graph on the right side of the line red
x.line.to.20 <- seq(from=f.simple, to=20, by=0.1)
y.line.to.20 <- df(x.line.to.20, df1=1, df2=7)
polygon(x=c(x.line.to.20, f.simple), y=c(y.line.to.20, 0), col="red")

pf(f.simple, df1=1, df2=7) ## the area under the curve that is blue

1-pf(f.simple, df1=1, df2=7) ## the area under the curve that is red

## lastly, let's compare this p-value to the one from the
## original regression
summary(simple.regression)

#######################################################
##
## Now let's do multiple regression by adding an extra term, tail length
##
#######################################################

## STEP 1: Draw a graph of the data to make sure the relationship make sense
## This graph is more complex because it shows the relationships between all
## of the columns in "mouse.data".
plot(mouse.data)

## STEP 2: Do the regression
multiple.regression <- lm(size ~ weight + tail, data=mouse.data)

## STEP 3: Look at the R^2, F-value and p-value
summary(multiple.regression)

## again, we can verify that our R^2 value is what we think it is
ss.multiple <- sum(multiple.regression\$residuals^2)

(ss.mean - ss.multiple) / ss.mean

## we can also verify that the F-value is what we think it is
f.multiple <- ((ss.mean - ss.multiple) / (3 - 1)) /
(ss.multiple / (nrow(mouse.data) - 3))

f.multiple

## Again let's draw a figure that shows how to calculate the p-value from the
## F-value
##
## First, draw the correct f-distribution curve with df1=2 and df2=6
x <- seq(from=0, to=20, by=0.1)
y <- df(x, df1=2, df2=6)
plot(x, y, type="l")

## now draw a verticle line where our f.value is for this test
abline(v=f.multiple, col="red")

## color the graph on the left side of the line blue
x.zero.to.line <- seq(from=0, to=f.multiple, by=0.1)
y.zero.to.line <- df(x.zero.to.line, df1=2, df2=6)
polygon(x=c(x.zero.to.line, 0), y=c(y.zero.to.line, 0), col="blue")

## color the graph on the right side of the line red
x.line.to.20 <- seq(from=f.multiple, to=20, by=0.1)
y.line.to.20 <- df(x.line.to.20, df1=2, df2=6)
polygon(x=c(x.line.to.20, f.multiple), y=c(y.line.to.20, 0), col="red")

pf(f.multiple, df1=2, df2=6) ## the area under the curve that is blue

1-pf(f.multiple, df1=2, df2=6) ## the area under the curve that is red

## lastly, let's compare this p-value to the one from the
## original regression
summary(multiple.regression)

#######################################################
##
## Now, let's see if "tail" makes a significant controbution by
## comparing the "simple" fit (which does not include the tail data)
## to the "multiple" fit (which has the extra term for the tail data)
##
#######################################################

f.simple.v.multiple <- ((ss.simple - ss.multiple) / (3-2)) /
(ss.multiple / (nrow(mouse.data) - 3))

1-pf(f.simple.v.multiple, df1=1, df2=6)

## Notice that this value is the same as the p-value next to the term for
## for "tail" in the summary of multiple regression:
summary(multiple.regression)

## Thus, the summary already calculated this F-value and p-value for us.
## this line tells us that including the "tail" term makes a statistically
## significant difference. The magnitude can be determined by looking
## at the change in R^2 between the simple and multiple regressions.
```

## 4 thoughts on “StatQuest: Multiple Regression in R”

1. GVista says:

#Python Code
“””
Created on Mon May 20 18:42:39 2019

@author: Guru

## Here’s the data

import pandas as pd
from plotly.graph_objs import *
from sklearn.linear_model import LinearRegression
import statsmodels.api as sm
import seaborn as sns
from scipy import stats
import matplotlib.pyplot as plt
## Here’s the data from the example:
mouse = pd.DataFrame({“weight”:[0.9, 1.8, 2.4, 3.5, 3.9, 4.4, 5.1, 5.6, 6.3],
“sizes”:[1.4, 2.6, 1.0, 3.7, 5.5, 3.2, 3.0, 4.9, 6.3],
“tail” :[0.7, 1.3, 0.7, 2.0, 3.6, 3.0, 2.9, 3.9, 4.0]})
print(mouse)

#######################################################
##
## Let’s start by reviewing simple regression by
## modeling mouse size with mouse weight.
##
#######################################################

## STEP 1: Draw a graph of the data to make sure the relationship make sense
#Plotting Scatter Matrix using pandas

trace0 = Scatter(
x=mouse.weight,
y=mouse.sizes,
mode=’markers’)

## STEP 2: Do the regression
est = sm.OLS(mouse.iloc[:,1].values, X2)
est2 = est.fit()

## STEP 3: Look at the R^2, F-value and p-value
print(est2.rsquared,est2.pvalues)

# Plot using plotly
# add the regression line to our x/y scatter plot
trace2 = Scatter(
x = mouse.weight,
y = est2.predict(X2)
)

data = [trace0,trace2]

layout = Layout(
showlegend=True,
height=600,
width=600,
)

fig = dict( data=data, layout=layout )
#plot(fig)

sns.lmplot(x=”weight”, y=”sizes”, data=mouse);

# Plot using Seaborn
#Plotting Scatter Matrix using seaborn

“””sns.set()
sns.pairplot(mouse)
#Plotting Scatter Matrix using pandas
pd.plotting.scatter_matrix(mouse);
plt.show()”””

# cumulative distribution function
from scipy.stats import f, norm
def plot_f_distrubiton(fvalue,dfn,dfd):
# Set figure
plt.figure(figsize=(8, 6))

# Set degrees of freedom

rejection_reg = f.ppf(q=.95, dfn=dfn, dfd=dfd)
mean, var, skew, kurt = f.stats(dfn, dfd, moments=’mvsk’)

x = np.linspace(f.ppf(0.01, dfn, dfd),
f.ppf(0.99, dfn, dfd), 100)

# Plot values
plt.plot(x, f.pdf(x, dfn, dfd), alpha=0.6,
label=’ X ~ F({}, {})’.format(dfn, dfd))
plt.axvline(x=fvalue)
plt.vlines(rejection_reg, 0.0, 1.0,
linestyles=”dashdot”, label=”Crit. Value: {:.2f}”.format(rejection_reg))
plt.legend()
plt.ylim(0.0, 1.0)
plt.xlim(0.0, 20.0,5)
plt.title(‘F-Distribution dfn:{}, dfd:{}’.format(dfn, dfd))

plot_f_distrubiton(est2.fvalue,1,7);

from sklearn.metrics import r2_score
print(“r2_score”,r2_score(mouse.sizes,est2.predict(X2)))
print(“fvalue”,est2.fvalue)
print(“f_pvalue”,est2.f_pvalue)

print(stats.f.cdf(est2.fvalue,1,7))
print(1-stats.f.cdf(est2.fvalue,1,7))
ss_mean = sum((mouse.sizes – np.mean(mouse.sizes))**2)
ss_simple = sum((mouse.sizes – est2.fittedvalues)**2)
f_simple = ((ss_mean – ss_simple) / (2 – 1))/ (ss_simple/ (len(mouse) – 2))
print(est2.summary())

#######################################################
##
## Now let’s do multiple regression by adding an extra term, tail length
##
#######################################################

## STEP 1: Draw a graph of the data to make sure the relationship make sense
## This graph is more complex because it shows the relationships between all
## of the columns in “mouse.data”.
pd.plotting.scatter_matrix(mouse);
plt.show()

sns.set()
sns.pairplot(mouse)

#plot(mouse.data)

## STEP 2: Do the regression

X2 = mouse[[‘weight’, ‘tail’]]
y = mouse[‘sizes’]
est = sm.OLS(y, X2).fit()

print(“r2_score,pvalues”,est.rsquared,est.pvalues)

print(“r2_score”,r2_score(mouse.sizes,est.predict(X2)))
print(“fvalue”,est.fvalue)
print(“f_pvalue”,est.f_pvalue)

plot_f_distrubiton(est.fvalue,2,6);

print(stats.f.cdf(est.fvalue,2,6))
print(“f_pvalue”,1-stats.f.cdf(est.fvalue,2,6))

## lastly, let’s compare this p-value to the one from the
## original regression

#ss_mean = sum((mouse.sizes – np.mean(mouse.sizes))**2)
ss_multiple = sum((mouse.sizes – est.fittedvalues)**2)

f_multiple = ((ss_mean – ss_multiple) / (len(est.params) – 1)) / (ss_multiple / (len(mouse) – len(est.params)))
print(est.summary())

## we can also verify that the F-value is what we think it is

#######################################################
##
## Now, let’s see if “tail” makes a significant controbution by
## comparing the “simple” fit (which does not include the tail data)
## to the “multiple” fit (which has the extra term for the tail data)
##
#######################################################

f_simple_v_multiple = ((ss_simple – ss_multiple) / (len(est.params)-2)) / (ss_multiple / (len(mouse) – len(est.params)))

print(f_simple_v_multiple)
print(1-stats.f.cdf(f_simple_v_multiple,1,6))

## Notice that this value is the same as the p-value next to the term for
## for “tail” in the summary of multiple regression:
print(est.summary())

## Thus, the summary already calculated this F-value and p-value for us.
## this line tells us that including the “tail” term makes a statistically
## significant difference. The magnitude can be determined by looking
## at the change in R^2 between the simple and multiple regressions.

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2. […] via StatQuest: Multiple Regression in R […]

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3. Felix says:

why df has x value from 0 o 20

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• Josh says:

Those are just possible values for the ‘f’ statistic. In theory, x should go from 0 to positive infinity, but the line gets really close to 0 for values greater than 20.

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