I started following the debate on differential minimum wage for youth (1519 year old) and adults in New Zealand. Eric Crampton has written a nice series of blog posts, unhealthy making the data from Statistics New Zealand available. I will use the nzunemployment.csv data file (with quarterly data from March 1986 to June 2011) and show an example of multiple linear regression with autocorrelated residuals in R.
A first approach could be to ignore autocorrelation and fit a linear model that attempts to predict youth unemployment with two explanatory variables: adult unemployment (continuous) and minimum wage rules (categorical: equal or different). This can be done using:
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setwd('~/Dropbox/quantumforest') un = read.csv('nzunemployment.csv', header = TRUE) # Create factor for minimum wage, which was different for youth # and adults before quarter 90 (June 2008) un$minwage = factor(ifelse(un$q < 90, 'Different', 'Equal')) mod1 = lm(youth ~ adult*minwage, data = un) summary(mod1) Coefficients: Estimate Std. Error t value Pr(>t) (Intercept) 8.15314 0.43328 18.817 < 2e16 *** adult 1.53334 0.07506 20.428 < 2e16 *** minwageEqual 5.69192 2.19356 2.595 0.0109 * adult:minwageEqual 2.85518 0.46197 6.180 1.47e08 *** Residual standard error: 1.447 on 98 degrees of freedom Multiple Rsquared: 0.8816, Adjusted Rsquared: 0.878 Fstatistic: 243.3 on 3 and 98 DF, pvalue: < 2.2e16 
Remember that adult*minwage
is expanded to adult + minwage + adult:minwage
. We can make the coefficients easier to understand if we center adult unemployment on the mean of the first 80 quarters. Notice that we get the same slope, AdjR2, etc. but now the intercept corresponds to the youth unemployment for the average adult unemployment before changing minimum wage rules. All additional analyses will use the centered version.
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un$cadult = with(un, adult  mean(adult)) mod2 = lm(youth ~ cadult*minwage, data = un) summary(mod2) Coefficients: Estimate Std. Error t value Pr(>t) (Intercept) 16.28209 0.15352 106.06 < 2e16 *** cadult 1.53334 0.07506 20.43 < 2e16 *** minwageEqual 9.44472 0.52629 17.95 < 2e16 *** cadult:minwageEqual 2.85518 0.46197 6.18 1.47e08 ***  Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 1.447 on 98 degrees of freedom Multiple Rsquared: 0.8816, Adjusted Rsquared: 0.878 Fstatistic: 243.3 on 3 and 98 DF, pvalue: < 2.2e16 plot(mod2) # Plots residuals for the model fit acf(mod2$res) # Plots autocorrelation of the residuals 
In the centered version, the intercept corresponds to youth unemployment when adult unemployment rate is 5.4 (average for the first 89 quarters). The coefficient of minwageEqual corresponds to the increase of youth unemployment (9.44%) when the law moved to have equal minimum wage for youth and adults. Notice that the slopes did not change at all.
I will use the function gls()
from the nlme package (which comes by default with all R installations) to take into account the serial correlation. First we can fit a model equivalent to mod2, just to check that we get the same results.
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library(nlme) mod3 = gls(youth ~ cadult*minwage, data = un) summary(mod3) Generalized least squares fit by REML Model: youth ~ cadult * minwage Data: un AIC BIC logLik 375.7722 388.6971 182.8861 Coefficients: Value Std.Error tvalue pvalue (Intercept) 16.282089 0.1535237 106.05585 0 cadult 1.533341 0.0750595 20.42834 0 minwageEqual 9.444719 0.5262926 17.94576 0 cadult:minwageEqual 2.855184 0.4619725 6.18042 0 Correlation: (Intr) cadult mnwgEq cadult 0.048 minwageEqual 0.292 0.014 cadult:minwageEqual 0.008 0.162 0.568 Standardized residuals: Min Q1 Med Q3 Max 2.96256631 0.53975848 0.02071559 0.63499262 2.35900240 Residual standard error: 1.446696 Degrees of freedom: 102 total; 98 residual 
Yes, they are identical. Notice that the model fitting is done using Restricted Maximum Likelihood (REML). Now we can add an autoregressive process of order 1 for the residuals and compare the two models:
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mod4 = gls(youth ~ cadult*minwage, correlation = corAR1(form=~1), data = un) summary(mod4) Generalized least squares fit by REML Model: youth ~ cadult * minwage Data: un AIC BIC logLik 353.0064 368.5162 170.5032 Correlation Structure: AR(1) Formula: ~1 Parameter estimate(s): Phi 0.5012431 Coefficients: Value Std.Error tvalue pvalue (Intercept) 16.328637 0.2733468 59.73598 0 cadult 1.522192 0.1274453 11.94389 0 minwageEqual 9.082626 0.8613543 10.54459 0 cadult:minwageEqual 2.545011 0.5771780 4.40940 0 Correlation: (Intr) cadult mnwgEq cadult 0.020 minwageEqual 0.318 0.007 cadult:minwageEqual 0.067 0.155 0.583 Standardized residuals: Min Q1 Med Q3 Max 2.89233359 0.55460580 0.02419759 0.55449166 2.29571080 Residual standard error: 1.5052 Degrees of freedom: 102 total; 98 residual anova(mod3, mod4) Model df AIC BIC logLik Test L.Ratio pvalue mod3 1 5 375.7722 388.6971 182.8861 mod4 2 6 353.0064 368.5162 170.5032 1 vs 2 24.7658 <.0001 
There is a substantial improvement for the log likelihood (from 182 to 170). We can take into account the additional parameter (autocorrelation) that we are fitting by comparing AIC, which improved from 375.77 (2*(182.8861) + 2*5) to 368.52 (2*(170.5032) + 2*6). Remember that AIC is 2*logLikelihood + 2*number of parameters.
The file unemployment.txt contains the R code used in this post (I didn’t use the .R extension as WordPress complains).