1
Chapter 8 Conclusion
Load the data:
teachdata =
read.csv("http://home.cc.umanitoba.ca/~godwinrt/3040/data/str4.csv"
)
attach(teachdata)
Variables:
head(teachdata)
sublunch score str avginc hiel
1 2.0408 690.80 17.88991 22.690001 0
2 47.9167 661.20 21.52466 9.824000 0
3 76.3226 643.60 18.69723 8.978000 1
4 77.0492 647.70 17.35714 8.978000 0
5 78.4270 640.85 18.67133 9.080333 1
6 86.9565 605.55 21.40625 10.415000 1
hiel = 1 if 10% or more of the class is learning English; = 0 otherwise
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In a previous lecture, it was argued that avginc might have a non-linear
relationship with score:
plot(avginc, score, xlim = c(5,60), ylim = c(600,710))
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avginc
sco
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Two ways we can deal with this non-linear effect is by using a (i) polynomial
regression model, and (ii) by taking logs.
(i) Polynomials. Create the new variables: avginc2 = avginc^2
avginc3 = avginc^3
eqcubic = lm(score ~ avginc + avginc2 + avginc3)
summary(eqcubic)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 6.001e+02 5.830e+00 102.937 < 2e-16 ***
avginc 5.019e+00 8.595e-01 5.839 1.06e-08 ***
avginc2 -9.581e-02 3.736e-02 -2.564 0.0107 *
avginc3 6.855e-04 4.720e-04 1.452 0.1471
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 12.71 on 416 degrees of freedom
Multiple R-squared: 0.5584, Adjusted R-squared: 0.5552
F-statistic: 175.4 on 3 and 416 DF, p-value: < 2.2e-16
4
Let’s plot the cubic regression function:
par(new = TRUE)
curve(600.1 + 5.019*x - 0.09581*x^2 + 0.0006855*x^3, xlim =
c(5,60), ylim = c(600,710), ylab = "", xlab = "", col = 2)
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(ii) Logarithms:
eqlog = lm(score ~ log(avginc))
summary(eqlog)
Estimate Std. Error t value Pr(>|t|)
(Intercept) 557.832 4.200 132.81 <2e-16 ***
log(avginc) 36.420 1.571 23.18 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 12.62 on 418 degrees of freedom
Multiple R-squared: 0.5625, Adjusted R-squared: 0.5615
F-statistic: 537.4 on 1 and 418 DF, p-value: < 2.2e-16
This is the “lin-log” model. Interpretation: a 1% increase in avginc is
associated with a 0.364 increase in score.
Add this regression to the plot:
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par(new = TRUE)
curve(557.832 + 36.42*log(x), xlim = c(5,60), ylim = c(600,710),
ylab = "", xlab = "", col = 3)
legend("bottomright", c("Cubic", "Lin-Log"), pch ="__", col=c(2,3))
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_
_Cubic
Lin-Log
7
Do you like the cubic or lin-log model better? What are the
advantages/disadvantages?
We will proceed by using log(avginc) – this is a simpler model (fewer estimated
𝛽𝑠) and fits better (higher �̅�2). But first, to revise omitted variable bias, let’s see
what happens if we leave log(avginc) out of the regression.
eq1 = lm(score ~ str + hiel + sublunch)
summary(eq1)
Estimate Std. Error t value Pr(>|t|)
(Intercept) 701.3732 4.6496 150.846 < 2e-16 ***
str -1.0312 0.2379 -4.334 1.84e-05 ***
hiel -3.8601 1.0470 -3.687 0.000257 ***
sublunch -0.5636 0.0192 -29.355 < 2e-16 ***
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Now add log(avginc):
eq2 = lm(score ~ str + pctel + sublunch + log(avginc))
summary(eq2)
Estimate Std. Error t value Pr(>|t|)
(Intercept) 659.35563 7.62805 86.438 < 2e-16 ***
str -0.77244 0.22934 -3.368 0.000828 ***
hiel -5.79113 1.03517 -5.594 4.03e-08 ***
sublunch -0.41701 0.02835 -14.708 < 2e-16 ***
log(avginc) 11.81992 1.74920 6.757 4.77e-11 ***
How have the results changed? What is going on here?
O.V.B. occurs if:
(i) avginc is associated with score (It is! Note the small p-value)
(ii) avginc is associated with str
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Regressor (1) (2)
str -1.03**
(0.24)
-0.77**
(0.23)
hiel -3.86**
(1.05)
-5.79**
(1.04)
sublunch -0.56**
(0.02)
-0.42**
(0.03)
log(avginc)
11.81**
(1.75)
Intercept 701.37**
(4.65)
659.36**
(7.63)
�̅�2 0.7726 0.7946
** significant at 1% level, * significant at 5% level
10
Does str have a different effect for classes with a high % of English learners?
We can use an interaction term to allow, and test, for this possibility.
Create the interaction term:
hielstr = hiel*str
Add this to the model:
eq3 = lm(score ~ str + hiel + hielstr + sublunch + log(avginc))
summary(eq3)
Estimate Std. Error t value Pr(>|t|)
(Intercept) 653.66612 8.89113 73.519 < 2e-16 ***
str -0.53103 0.30039 -1.768 0.0778 .
hiel 5.49821 9.13897 0.602 0.5478
hielstr -0.57767 0.46463 -1.243 0.2145
sublunch -0.41138 0.02869 -14.337 < 2e-16 ***
log(avginc) 12.12447 1.76513 6.869 2.38e-11 ***
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Which coefficient should we be testing to see if str has a different effect for
classes with many English learners? What do we conclude?
𝐻0: str has the same effect on score for classes with a high and low % of English
learners
𝐻𝐴: str has a different effect on score for classes with a high and low % of
English learners
Or…
𝐻0: 𝛽ℎ𝑖𝑒𝑙×𝑠𝑡𝑟 = 0
𝐻𝐴: 𝛽ℎ𝑖𝑒𝑙×𝑠𝑡𝑟 ≠ 0
t = -1.243, p-val = 0.2145
We fail to reject 𝐻0.
How would you test the hypothesis that str has no effect on score, using model
(3)?
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Regressor (1) (2) (3)
str -1.03**
(0.24)
-0.77**
(0.23)
-0.53
(0.30)
hiel -3.86**
(1.05)
-5.79**
(1.04)
5.50
(9.1)
hiel×str
-0.58
(0.47)
sublunch -0.56**
(0.02)
-0.42**
(0.03)
-0.411**
(0.029)
log(avginc)
11.81**
(1.75)
12.12**
(1.8)
Intercept 701.37**
(4.65)
659.36**
(7.63)
653.7**
(8.9)
�̅�2 0.7726 0.7946 0.7949
** significant at 1% level, * significant at 5% level
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The relationship between str and score may be non-linear. We will use a
polynomial regression model to allow, and test, for this. I leave out the
interaction term hiel×str (for now) since it was previously found to be
insignificant.
str2 = str^2
str3 = str^3
eq4 = lm(score ~ str + str2 + str3 + hiel + sublunch + log(avginc))
summary(eq4)
Estimate Std. Error t value Pr(>|t|)
(Intercept) 252.05089 165.82433 1.520 0.12928
str 64.33886 25.46223 2.527 0.01188 *
str2 -3.42388 1.29374 -2.646 0.00844 **
str3 0.05929 0.02174 2.728 0.00665 **
hiel -5.47399 1.03187 -5.305 1.84e-07 ***
sublunch -0.42006 0.02814 -14.928 < 2e-16 ***
log(avginc) 11.74818 1.73446 6.773 4.34e-11 ***
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Regressor (1) (2) (3) (4)
str -1.03**
(0.24)
-0.77**
(0.23)
-0.53
(0.30)
64.33**
(25.5)
str2
-3.42**
(1.29)
str3
0.06**
(0.02)
hiel -3.86**
(1.05)
-5.79**
(1.04)
5.50
(9.10)
-5.47**
(1.03)
hiel×str
-0.58
(0.47)
sublunch -0.56**
(0.02)
-0.42**
(0.03)
-0.41**
(0.03)
-0.42**
(0.03)
log(avginc)
11.81**
(1.75)
12.12**
(1.77)
11.75**
(1.73)
Intercept 701.37**
(4.65)
659.36**
(7.63)
653.66**
(8.89)
252.05
(165.82)
�̅�2 0.7726 0.7946 0.7949 0.7982
𝑅2 0.7742 0.7966 0.7974 0.8011
** significant at 1% level, * significant at 5% level
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Test if the relationship between str and score is linear or non-linear:
𝐻0: 𝛽𝑠𝑡𝑟2 = 0 and 𝛽𝑠𝑡𝑟3 = 0
𝐻𝐴: 𝛽𝑠𝑡𝑟2 ≠ 0 and/or 𝛽𝑠𝑡𝑟3 ≠ 0
We must use an F-test since we have multiple restrictions!
Formula for F-statistic:
𝐹 =(𝑅𝑈
2 − 𝑅𝑅2) 𝑞⁄
(1 − 𝑅𝑈2 ) (𝑛 − 𝑘𝑈 − 1)⁄
What is the unrestricted model?
What is the restricted model?
What is q?
What is n?
What is 𝑘𝑈?
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We need 𝑅2, not �̅�2!
𝐹 =(0.8011 − 0.7966) 2⁄
(1 − 0.8011) (420 − 6 − 1)⁄= 4.67
The 5% critical values for an F-test with large n are:
q 5% crit. value
1 3.84
2 3.00
3 2.60
4 2.37
5 2.21
We reject the null at the 5% level.
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Use R to calculate the F-stat and p-val:
anova(eq4, eq2)
Model 1: score ~ str + str2 + str3 + hiel + sublunch + log(avginc)
Model 2: score ~ str + hiel + sublunch + log(avginc)
Res.Df RSS Df Sum of Sq F Pr(>F)
1 413 30257
2 415 30939 -2 -682.05 4.6549 0.01002 *
So far, model (4) seems the most appropriate.
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Let’s reconsider if the effect of str on score is different for classes with a high
percentage of English learners. Before, we found no difference (using model 3),
but this might change now that we have model 4.
Again, the strategy is:
• have the dummy variable hiel interact with all terms involving str
• this allows for the “marginal effect” to differ between the two groups
• testing to see if the estimated coeffecients on the interaction terms are
jointly equal to zero is equivalent to testing that there is no difference
between the two groups
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Create the new interaction terms:
hielstr2 = hiel*str2
hielstr3 = hiel*str3
Add the interaction terms to model (4):
eq5 = lm(score ~ str + str2 + str3 + hiel + hielstr + hielstr2 +
hielstr3 + sublunch + log(avginc))
summary(eq5)
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Regressor (1) (2) (3) (4) (5)
str -1.03**
(0.24)
-0.77**
(0.23)
-0.53
(0.30)
64.33**
(25.46)
83.70**
(29.69)
str2
-3.42**
(1.29)
-4.38**
(1.51)
str3
0.06**
(0.02)
0.08**
(0.03)
hiel -3.86**
(1.05)
-5.79**
(1.04)
5.50
(9.10)
-5.47**
(1.03)
816.08
(434.61)
hiel×str
-0.58
(0.47)
-123.28
(66.35)
hiel×str2
6.12
(3.35)
hiel×str3
-0.10
(0.06)
sublunch -0.56**
(0.02)
-0.42**
(0.03)
-0.41**
(0.03)
-0.42**
(0.03)
-0.42**
(0.03)
log(avginc)
11.81**
(1.75)
12.12**
(1.77)
11.75**
(1.73)
11.80**
(1.75)
Intercept 701.37**
(4.65)
659.36**
(7.63)
653.66**
(8.89)
252.05
(165.82)
122.35
(192.18)
�̅�2 0.7726 0.7946 0.7949 0.7982 0.7988
𝑅2 0.7742 0.7966 0.7974 0.8011 0.8031
** significant at 1% level, * significant at 5% level
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Test to see if str has a different effect for classes with many English learners.
We must test all terms involving hiel and str together.
𝐻0: 𝛽ℎ𝑖𝑒𝑙×𝑠𝑡𝑟 = 0 and 𝛽ℎ𝑖𝑒𝑙×𝑠𝑡𝑟2 = 0 and 𝛽ℎ𝑖𝑒𝑙×𝑠𝑡𝑟3 = 0
𝐻𝐴: not 𝐻0
Or…
𝐻0: model (4)
𝐻𝐴: model (5)
𝐹 =(0.8031 − 0.8011) 3⁄
(1 − 0.8031) (420 − 9 − 1)⁄= 1.38
Compared to the 2.60 critical value, we fail to reject the null. Model (4) is
preferred.
How would you test if str matters, using model (4)?
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Summary
There doesn’t appear to be a substantial difference in the effect of str on score
for classes with many English learners.
A hypothesis test involving model (4) indicates the relationship between str and
score is non-linear.
Using F-tests, the null hypothesis that str has no effect on score is rejected in all
models.
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Interpreting the model
Note: using the values from the table will induce too much rounding error.
If str = 20, then reducing str to 19 would improve score by a predicted:
𝑠𝑐𝑜𝑟�̂�|𝑠𝑡𝑟=20 − 𝑠𝑐𝑜𝑟�̂�|𝑠𝑡𝑟=19
= [64.33886 × (20) − 3.42388 × (20)2 + 0.05929 × (20)3]
− [64.33886 × (19) − 3.42388 × (20)2 + 0.05929 × (19)3]
= 1.54
If str = 15, then reducing str to 14 would improve score by a predicted 2.46.
Going from str = 20 to str = 10 would improve score by a predicted 31.26, etc.
How much does it cost to double the number of teachers? Note that for every
1% increase in the budget, score is predicted to increase by 0.1175.
