类似于SAS的均值ANCOVA差异的估计置信区间

Question

I need to present an ANCOVA model for some medical data. I'm used to work with SAS proc GLM and I wonder if it's possible to obtain confidence intervals for the difference of means as you get with SAS. I'm using LSmatrix from doBy package to obtain the group means.

model <- aov(yield ~ block + N * P + K, npk)
library(doBy)
k1 <- LSmatrix(model, effect="block")
linest(model,K=k1)

  estimate       se df   t.stat      p.value block
1   54.025 1.977116 14 27.32515 1.510775e-13     1
2   57.450 1.977116 14 29.05747 6.479499e-14     2
3   60.775 1.977116 14 30.73922 2.981292e-14     3
4   50.125 1.977116 14 25.35258 4.229573e-13     4
5   50.525 1.977116 14 25.55490 3.792520e-13     5
6   56.350 1.977116 14 28.50111 8.457748e-14     6

I would like to get the difference of means block 1 - bloc 2 and a confidence interval (or a std. error), but I'm lost on how to proceed from here.

The result could be something like this:

    estimate     se upper.CI lower.CI
1-2   -3.425 ??????    ?????    ?????
1-3   -6.750 ??????    ?????    ?????
...

If anyone can tell me how to calculate the first, I can manange the rest :)

Thanks in advance.

Answer 1

I . Solution 1

> pacman::p_load(lsmeans, multcompView)
> 
> eindzl      <- lm(yield ~ block + N * P + K, npk)
> anova(eindzl)
Analysis of Variance Table

Response: yield
          Df Sum Sq Mean Sq F value   Pr(>F)   
block      5 343.29  68.659  4.3911 0.012954 * 
N          1 189.28 189.282 12.1055 0.003684 **
P          1   8.40   8.402  0.5373 0.475637   
K          1  95.20  95.202  6.0886 0.027114 * 
N:P        1  21.28  21.282  1.3611 0.262841   
Residuals 14 218.90  15.636                    
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
> eindzl.rg   <- ref.grid(eindzl)
> eindzl.lsm  <- lsmeans(eindzl.rg, "block")
> cld(eindzl.lsm, alpha = .05)
 block lsmean       SE df lower.CL upper.CL .group
 4     50.125 1.977116 14 45.88451 54.36549  1    
 5     50.525 1.977116 14 46.28451 54.76549  1    
 1     54.025 1.977116 14 49.78451 58.26549  12   
 6     56.350 1.977116 14 52.10951 60.59049  12   
 2     57.450 1.977116 14 53.20951 61.69049  12   
 3     60.775 1.977116 14 56.53451 65.01549   2   

Results are averaged over the levels of: N, P, K 
Confidence level used: 0.95 
P value adjustment: tukey method for comparing a family of 6 estimates 
significance level used: alpha = 0.05 

pairs(eindzl.lsm)
 contrast estimate       SE df t.ratio p.value
 1 - 2      -3.425 2.796064 14  -1.225  0.8180
 1 - 3      -6.750 2.796064 14  -2.414  0.2160
 1 - 4       3.900 2.796064 14   1.395  0.7295
 1 - 5       3.500 2.796064 14   1.252  0.8049
 1 - 6      -2.325 2.796064 14  -0.832  0.9564
 2 - 3      -3.325 2.796064 14  -1.189  0.8348
 2 - 4       7.325 2.796064 14   2.620  0.1560
 2 - 5       6.925 2.796064 14   2.477  0.1960
 2 - 6       1.100 2.796064 14   0.393  0.9985
 3 - 4      10.650 2.796064 14   3.809  0.0190
 3 - 5      10.250 2.796064 14   3.666  0.0248
 3 - 6       4.425 2.796064 14   1.583  0.6217
 4 - 5      -0.400 2.796064 14  -0.143  1.0000
 4 - 6      -6.225 2.796064 14  -2.226  0.2856
 5 - 6      -5.825 2.796064 14  -2.083  0.3485

Results are averaged over the levels of: N, P, K 
P value adjustment: tukey method for comparing a family of 6 estimates 
> 

II. Solution 2

Feel free to use your own equation for calculating the upper and lower bounds or visit CrossValidated if you need help determining the formula. This example is based on the Standard Error and shows the code you need. CrossValidated is for statistics.

pacman::p_load(data.table,doBy)

model <- aov(yield ~ block + N * P + K, npk)
k1    <- LSmatrix(model, effect="block")
linest(model,K=k1)

tmp <- linest(model,K=k1)

tmp <- cbind(as.data.frame(tmp$coef), as.data.frame(tmp$grid))

tmp2 <- as.data.frame(matrix(ncol=4,nrow=9))
setnames(tmp2, c("group","estimate","lower bound of CI", "upper bound of CI"))

for(i in 1:5){
  n                  <- i + 1
  tmp2$estimate[i]   <- tmp$estimate[tmp$block == i] -  tmp$estimate[tmp$block == n]
  tmp2$group[i]      <- paste(i,n,sep="-")
  tmp2[i,3]          <- (tmp$estimate[tmp$block == i] - tmp$se[tmp$block == i]) - (tmp$estimate[tmp$block == n] + tmp$se[tmp$block == n])
  tmp2[i,4]          <- (tmp$estimate[tmp$block == i] + tmp$se[tmp$block == i]) - (tmp$estimate[tmp$block == n] - tmp$se[tmp$block == n])

}


for(i in 1:4){
  n                   <- i + 2
  nn                  <- 5+i    
  tmp2$estimate[nn]   <- tmp$estimate[tmp$block == i] -  tmp$estimate[tmp$block == n]
  tmp2$group[nn]      <- paste(i,n,sep="-")
  tmp2[nn,3]          <- (tmp$estimate[tmp$block == i] - tmp$se[tmp$block == i]) - (tmp$estimate[tmp$block == n] + tmp$se[tmp$block == n])
  tmp2[nn,4]          <- (tmp$estimate[tmp$block == i] + tmp$se[tmp$block == i]) - (tmp$estimate[tmp$block == n] - tmp$se[tmp$block == n])
}

tmp2

  group   estimate lower bound of CI upper bound of CI
1   1-2     -3.425         -7.379232         0.5292322
2   2-3     -3.325         -7.279232         0.6292322
3   3-4     10.650          6.695768        14.6042322
4   4-5     -0.400         -4.354232         3.5542322
5   5-6     -5.825         -9.779232        -1.8707678
6   1-3     -6.750        -10.704232        -2.7957678
7   2-4      7.325          3.370768        11.2792322
8   3-5     10.250          6.295768        14.2042322
9   4-6     -6.225        -10.179232        -2.2707678

Note that if you weren't choosing this particular package and function that you'd be able to get results like that without as much custom manipulation. Usually you type way less in R than in SAS.