Question

I have created a script like the one below to do something I called as “weighted” regression:

library(plyr)

set.seed(100)

temp.df <- data.frame(uid=1:200,
                      bp=sample(x=c(100:200),size=200,replace=TRUE),
                      age=sample(x=c(30:65),size=200,replace=TRUE),
                      weight=sample(c(1:10),size=200,replace=TRUE),
                      stringsAsFactors=FALSE)

temp.df.expand <- ddply(temp.df,
                        c("uid"),
                        function(df) {
                          data.frame(bp=rep(df[,"bp"],df[,"weight"]),
                                     age=rep(df[,"age"],df[,"weight"]),
                                     stringsAsFactors=FALSE)})

temp.df.lm <- lm(bp~age,data=temp.df,weights=weight)
temp.df.expand.lm <- lm(bp~age,data=temp.df.expand)

You can see that in temp.df, each row has its weight, what I mean is that there is a total of 1178 sample but for rows with same bp and age, they are merge into 1 row and represented in the weight column.

I used the weight parameters in the lm function, then I cross check the result with another dataframe that the temp.df dataframe is “expanded”. But I found the lm outputs different for the 2 dataframe.

Did I misinterpret the weight parameters in lm function, and can anyone let me know how to I run regression properly (i.e. without expanding the dataframe manually) for a dataset presented like temp.df? Thanks.

Answer 1

The problem here is that the degrees of freedom are not being properly added up to get the right Df and mean-sum-squares statistics. This will correct the problem:

temp.df.lm.aov <- anova(temp.df.lm)
temp.df.lm.aov$Df[length(temp.df.lm.aov$Df)] <- 
        sum(temp.df.lm$weights)-   
        sum(temp.df.lm.aov$Df[-length(temp.df.lm.aov$Df)]  ) -1
temp.df.lm.aov$`Mean Sq` <- temp.df.lm.aov$`Sum Sq`/temp.df.lm.aov$Df
temp.df.lm.aov$`F value`[1] <- temp.df.lm.aov$`Mean Sq`[1]/
                                        temp.df.lm.aov$`Mean Sq`[2]
temp.df.lm.aov$`Pr(>F)`[1] <- pf(temp.df.lm.aov$`F value`[1], 1, 
                                      temp.df.lm.aov$Df, lower.tail=FALSE)[2]
temp.df.lm.aov
Analysis of Variance Table

Response: bp
            Df Sum Sq Mean Sq F value   Pr(>F)   
age          1   8741  8740.5  10.628 0.001146 **
Residuals 1176 967146   822.4        

Compare with:

> anova(temp.df.expand.lm)
Analysis of Variance Table

Response: bp
            Df Sum Sq Mean Sq F value   Pr(>F)   
age          1   8741  8740.5  10.628 0.001146 **
Residuals 1176 967146   822.4                    
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

I am a bit surprised this has not come up more often on R-help. Either that or my search strategy development powers are weakening with old age.