A simple ANOVA

是 否 +2 论坛币

k人 参与回答

经管之家送您一份

应届毕业生专属福利!

求职就业群

赵安豆老师微信：zhaoandou666

经管之家联合CDA

送您一个全额奖学金名额~ !

立即领取

感谢您参与论坛问题回答

经管之家送您两个论坛币！

+2 论坛币

(This article was first published on Wiekvoet, and kindly contributed to R-bloggers)

I was browsing Davies Design and Analysis of Industrial Experiments (second edition, 1967). Published by for ICI in times when industry did that kind of thing. It is quite an applied book. On page 107 there is an example where the variance of a process is estimated.

Data

Data is from nine batches from which three samples were selected (A, B and C) and each a duplicate measurement. I am not sure about copyright of these data, so I will not reprint the data here. The problem is to determine the measurement ans sampling error in a chemical process.
ggplot(r4,aes(x=Sample,y=x))+
    geom_point()+
    facet_wrap(~  batch )

Analysis

At the time of writing the book, the only approach was to do a classical ANOVA and calculate the estimates from there.
aov(x~ batch + batch:Sample,data=r4) %>%
  anova
Analysis of Variance Table

Response: x
             Df Sum Sq Mean Sq  F value  Pr(>F)   
batch         8 792.88  99.110 132.6710 < 2e-16 ***
batch:Sample 18  25.30   1.406   1.8818 0.06675 .  
Residuals    27  20.17   0.747                     
—
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
In this case the residual variation is 0.75. The batch:Sample variation estimates is, due to the design, twice the sapling variation plus residual variation. Hence it is estimated as 0.33. How lucky we are to have tools (lme4) which can do this estimate directly. In this case, as it was a well designed experiment, these estimates are the same as from the ANOVA.
l1 <- lmer(x ~1+ (1 | batch) + (1|batch:Sample) ,data=r4 )

summary(l1)
Linear mixed model fit by REML [‘lmerMod’]
Formula: x ~ 1 + (1 | batch) + (1 | batch:Sample)
   Data: r4

REML criterion at convergence: 189.4

Scaled residuals:
     Min       1Q   Median       3Q      Max
-1.64833 -0.50283 -0.06649  0.55039  1.57801

Random effects:
Groups       Name        Variance Std.Dev.
batch:Sample (Intercept)  0.3294  0.5739  
batch        (Intercept) 16.2841  4.0354  
Residual                  0.7470  0.8643  
Number of obs: 54, groups: batch:Sample, 27; batch, 9

Fixed effects:
            Estimate Std. Error t value

(Intercept)   47.148      1.355    34.8
A next step is confidence intervals around the estimates. Davies uses limits from a Chi-squared distribution for the residual variation, leading to a 90% interval 0.505  to 1.25. In contrast lme4 has two estimators, profile (computing a likelihood profile and finding the appropriate cutoffs based on the likelihood ratio test;) and bootstrap (perform parametric bootstrapping with confidence intervals computed from the bootstrap distribution according to boot.type). Each of these takes one or few second on my laptop, not feasible for the pre computer age. The estimates are different, to my surprise more narrow:
Computing profile confidence intervals …
                   5 %       95 %
.sig01       0.0000000  0.9623748
.sig02       2.6742109  5.9597328
.sigma       0.7017849  1.1007261
(Intercept) 44.8789739 49.4173227

Computing bootstrap confidence intervals …
                                  5 %       95 %
sd_(Intercept)|batch:Sample  0.000000  0.8880414
sd_(Intercept)|batch         2.203608  5.7998348
sigma                        0.664149  1.0430984

(Intercept)                 45.140652 49.4931109
Davies continues to estimate the ratio to residual for sampling variation, which was the best available for that time. This I won’t repeat.

扫码加我 拉你入群

请注明：姓名-公司-职位

以便审核进群资格，未注明则拒绝

分享0 收藏0 回帖

关键词：simple ANOVA anov MPL sim simple

thanks for sharing