STATS 330: Lecture 6

04/21/23 330 Lecture 6 1

STATS 330: Lecture 6

04/21/23 330 Lecture 6 2

Inference for the Regression model

Aim of today’s lecture:To discuss how we assess the significance of

variables in the regression

Key concepts:• Standard errors• Confidence intervals for the coefficients• Tests of significance

Reference: Coursebook Section 3.2

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Variability of the regression coefficients Imagine that we keep the x’s fixed, but

resample the errors and refit the plane. How much would the plane (estimated coefficients) change?

This gives us an idea of the variability (accuracy) of the estimated coefficients as estimates of the coefficients of the true regression plane.

04/21/23 330 Lecture 6 4

The regression model (cont)

The data is scattered above and below the plane:

Size of “sticks” is random, controlled by 2, doesn’t depend on

x1, x2

aaa

Y

X1

O

X2

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Variability of coefficients (2)

Variability depends on• The arrangement of the x’s (the more

correlation, the more change, see Lecture 8)• The error variance (the more scatter about the

true plane, the more the fitted plane changes)

Measure variability by the standard error of the coefficients

04/21/23 330 Lecture 6 6

Call:lm(formula = volume ~ diameter + height, data = cherry.df)

Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -57.9877 8.6382 -6.713 2.75e-07 ***diameter 4.7082 0.2643 17.816 < 2e-16 ***height 0.3393 0.1302 2.607 0.0145 * ---Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1

Residual standard error: 3.882 on 28 degrees of freedomMultiple R-Squared: 0.948, Adjusted R-squared: 0.9442 F-statistic: 255 on 2 and 28 DF, p-value: < 2.2e-16

Standard errors of coefficients

Cherries

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Confidence interval

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Confidence interval (2)

A 95% confidence interval for a regression coefficient is of the form Estimated coefficient +/- standard error twhere t is the 97.5% point of the appropriate t-distribution. The degrees of freedom aren-k-1 where n=number of cases (observations) in the regression, and k is the number of variables (assuming we have a constant term)

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Example: cherry trees

Use function confint

> confint(cherry.lm)

2.5% 97.5%(Intercept) -75.68226247 -40.2930554

diameter 4.16683899 5.2494820

height 0.07264863 0.6058538

Object created by lm

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Hypothesis test Often we ask “do we need a particular

variable, given the others are in the model?”

Note that this is not the same as asking “is a particular variable related to the response?”

Can test the former by examining the ratio of the coefficient to its standard error

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Hypothesis test (2) This is the t-statistic t

The bigger t , the more we need the variable

Equivalently, the smaller the p-value, the more we need the variable

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Call:lm(formula = volume ~ diameter + height, data = cherry.df)Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -57.9877 8.6382 -6.713 2.75e-07 ***diameter 4.7082 0.2643 17.816 < 2e-16 ***height 0.3393 0.1302 2.607 0.0145 * ---Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1 Residual standard error: 3.882 on 28 degrees of freedomMultiple R-Squared: 0.948, Adjusted R-squared: 0.9442 F-statistic: 255 on 2 and 28 DF, p-value: < 2.2e-16

t-values

p-values

Cherries

All variables required since p=values small (<0.05)

P-value

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-4 -2 0 2 4

0.0

0.1

0.2

0.3

0.4

P-value: total area is 0.0145

2.607-2.607

Density curve for t with 28 degrees of freedom

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Other hypotheses Overall significance of the regression: do none

of the variables have a relationship with the response?

Use the F statistic: the bigger F, the more evidence that at least one variable has a relationship • equivalently, the smaller the p-value, the more

evidence that at least one variable has a relationship

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Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -57.9877 8.6382 -6.713 2.75e-07 ***diameter 4.7082 0.2643 17.816 < 2e-16 ***height 0.3393 0.1302 2.607 0.0145 * ---Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1

Residual standard error: 3.882 on 28 degrees of freedomMultiple R-Squared: 0.948, Adjusted R-squared: 0.9442 F-statistic: 255 on 2 and 28 DF, p-value: < 2.2e-16

F-valuep-value

Cherries

04/21/23 330 Lecture 6 16

Testing if a subset is required

Often we want to test if a subset of variables is unnecessary

Terminology:Full model: model with all the variablesSub-model: model with a set of variables deleted.

Test is based on comparing the RSS of the submodel with the RSS of the full model. Full model RSS is always smaller (why?)

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Testing if a subset is adequate (2)

If the full model RSS is not much smaller than the submodel RSS, the submodel is adequate: we don’t need the extra variables.

To do the test, we • Fit both models, get RSS for both.• Calculate test statistic (see next slide)• If the test statistic is large, (equivalently the p-

value is small) the submodel is not adequate

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Test statistic Test statistic is

d is the number of variables dropped s2 is the estimate of 2 from the full model (the

residual mean square) R has a function anova to do the calculations

2

/)(

s

dRSSRSSF FULLSUB

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P-values When the smaller model is correct, the test

statistic has an F-distribution with d and n-k-1 degrees of freedom

We assess if the value of F calculated from the sample is a plausible value from this distribution by means of a p-value

If the p-value is too small, we reject the hypothesis that the submodel is ok

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P-values (cont)

0 2 4 6 8

0.0

0.2

0.4

0.6

0.8

1.0

Density of F-distribution with 2 and 16 df

F

Den

sity

Value of F

P-value

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Example Free fatty acid data: use physical measures

to model a biochemical parameter in overweight children

Variables are • FFA: free fatty acid level in blood (response)• Age (months)• Weight (pounds)• Skinfold thickness (inches)

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Dataffa age weight skinfold0.759 105 67 0.960.274 107 70 0.520.685 100 54 0.620.526 103 60 0.760.859 97 61 1.000.652 101 62 0.740.349 99 71 0.761.120 101 48 0.621.059 107 59 0.561.035 100 51 0.44… 20 observations in all

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Analysis (1)

This suggests that age is not required if weight, skinfold retained, skinfold is not required if weight, age retainedCan we get away with just weight?

> model.full<- lm(ffa~age+weight+skinfold,data=fatty.df)> summary(model.full)Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 3.95777 1.40138 2.824 0.01222 * age -0.01912 0.01275 -1.499 0.15323 weight -0.02007 0.00613 -3.274 0.00478 **skinfold -0.07788 0.31377 -0.248 0.80714

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Analysis (2)> model.sub<-lm(ffa~weight,data=fatty.df)> anova(model.sub,model.full)Analysis of Variance Table

Model 1: ffa ~ weightModel 2: ffa ~ age + weight + skinfold Res.Df RSS Df Sum of Sq F Pr(>F)1 18 0.91007 2 16 0.79113 2 0.11895 1.2028 0.3261

Small F, large p-value suggest weight alone is adequate. But test should be interpreted with caution, as we “pretested”

Testing a combination of coefficients

Cherry trees: Our model is V = c DH

or

log(V) = + log(D) + log(H)

Dimension analysis suggests +

How can we test this? Test statistic is

P value is area under t-curve beyond +/- t 04/21/23 330 Lecture 6 25

)ˆˆ(

3ˆˆ

21

21

se

t

Testing a combination (cont)

We can use the “R330” function test.lc to compute the value of t:

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> cherry.lm = lm(log(volume)~log(diameter)+log(height),data=cherry.df)> cc = c(0,1,1)> c = 3> test.lc(cherry.lm,cc,c)$est[1] 3.099773$std.err[1] 0.1765222$t.stat[1] 0.5652165$df[1] 28$p.val[1] 0.5764278

The “R330 package”

A set of functions written for the course, in the form of an R package

Install the package using the R packages menu (see coursebook for details). Then type

library(R330)

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Testing a combination (cont)

In general, we might want to testc + c+ cc

(in our example c = 0, c=1, c=1, c = 3)

Estimate is

Test statistic is

04/21/23 330 Lecture 6 28

221100ˆˆˆ ccc

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ccces

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