190728_学力差の検討_標準偏差にしてみる
今回の課題
- 学級規模に関係なく，どこで学力差が生じるのか
- (a)は学級規模で説明されるのか
- 学力の変動の大きさは，prior achievementで説明されるのか
- (a)に対する(b)と(c)の交互作用は見られるか
とりあえず国語をやってる
NRTデータを読み込む
setwd("/Users/koyo/Dropbox/000078_CSKAKEN/04_NRT/SY2018")
nrt.all <- read.csv("nrt.csv", fileEncoding = "Shift_JIS")
setwd("/Users/koyo/Dropbox/000078_CSKAKEN/190700_Anal")国語のデータに限定する
nrt.koku <- nrt.all %>%
dplyr::select(c("renban", "sho.sid",
"koku.ss.1213", #1-2年生追加
"koku.ss.1314",
"koku.ss.1415",
"koku.ss.1516",
"koku.ss.1617",
"koku.ss.1718" #6-7年生追加
))
colnames(nrt.koku) <- c("renban", "sid",
"g12.koku",
"g23.koku",
"g34.koku",
"g45.koku",
"g56.koku",
"g67.koku"
)
nrt.koku <- nrt.koku%>%
mutate(id = row_number())
# write.csv(nrt.koku, "nrt_koku.csv")全体について，学年間の分散比を求めてみる
library(psych)
g12.ds <- describe(nrt.koku$g12.koku)
g23.ds <- describe(nrt.koku$g23.koku)
g34.ds <- describe(nrt.koku$g34.koku)
g45.ds <- describe(nrt.koku$g45.koku)
g56.ds <- describe(nrt.koku$g56.koku)
g67.ds <- describe(nrt.koku$g67.koku)
#平均差と分散比
d23 <- matrix(c(g23.ds[,3] - g12.ds[,3], g23.ds[,4]^2 / g12.ds[,4]^2), nrow=1, ncol=2)
d34<- matrix(c(g34.ds[,3] - g23.ds[,3], g34.ds[,4]^2 / g23.ds[,4]^2), nrow=1, ncol=2)
d45<- matrix(c(g45.ds[,3] - g34.ds[,3], g45.ds[,4]^2 / g34.ds[,4]^2), nrow=1, ncol=2)
d56 <- matrix(c(g56.ds[,3] - g45.ds[,3], g56.ds[,4]^2 / g45.ds[,4]^2), nrow=1, ncol=2)
d67 <- matrix(c(g67.ds[,3] - g56.ds[,3], g67.ds[,4]^2 / g56.ds[,4]^2), nrow=1, ncol=2)
diff.1 <- rbind(d23, d34, d45, d56, d67)
colnames(diff.1) <- c("M.Diff", "V.Ratio")
rownames(diff.1) <- c("d23", "d34", "d45", "d56", "d67")
diff.1## M.Diff V.Ratio
## d23 -0.394416929 0.9156668
## d34 0.122917605 1.3135990
## d45 -0.330789788 0.9198229
## d56 -0.509411453 0.7934490
## d67 0.003895468 1.4200221標準偏差を求めてみる
sd2 <- g12.ds[,4]
sd3 <- g23.ds[,4]
sd4 <- g34.ds[,4]
sd5 <- g45.ds[,4]
sd6 <- g56.ds[,4]
sd7 <- g67.ds[,4]
sd.1 <- matrix(c(sd2, sd3, sd4, sd5, sd6, sd7), nrow=6, ncol=1)
colnames(sd.1) <- c("sd")
rownames(sd.1) <- c("1-2","2-3","3-4","4-5","5-6","6-7")
sd.1## sd
## 1-2 8.804191
## 2-3 8.424773
## 3-4 9.655830
## 4-5 9.260656
## 5-6 8.248999
## 6-7 9.829893標準偏差の学年間の比を求めてみる
sd.ratio23 <- sd.1[2]/ sd.1[1]
sd.ratio34 <- sd.1[3]/ sd.1[2]
sd.ratio45 <- sd.1[4]/ sd.1[3]
sd.ratio56 <- sd.1[5]/ sd.1[4]
sd.ratio67 <- sd.1[6]/ sd.1[5]
sd.ratio1 <- matrix(c(sd.ratio23, sd.ratio34, sd.ratio45, sd.ratio56, sd.ratio67), nrow=5, ncol=1)
colnames(sd.ratio1) <- c("sd.rario")
rownames(sd.ratio1) <- c("2-3", "3-4","4-5","5-6", "6-7")
sd.ratio1## sd.rario
## 2-3 0.9569048
## 3-4 1.1461235
## 4-5 0.9590740
## 5-6 0.8907575
## 6-7 1.1916468学校ごとに標準偏差を求める
g12.mean <- nrt.koku[c("sid", "g12.koku")] %>% na.omit() %>% group_by(sid) %>% summarise(avg.12 = mean(g12.koku))
g12.sd <- nrt.koku[c("sid", "g12.koku")] %>% na.omit() %>% group_by(sid) %>% summarise(sd.12 = sd(g12.koku))
g12.msd <-data.frame(dplyr::inner_join(g12.mean, g12.sd, by = "sid"))
g23.mean <- nrt.koku[c("sid", "g23.koku")] %>% na.omit() %>% group_by(sid) %>% summarise(avg.23 = mean(g23.koku))
g23.sd <- nrt.koku[c("sid", "g23.koku")] %>% na.omit() %>% group_by(sid) %>% summarise(sd.23 = sd(g23.koku))
g23.msd <-data.frame(dplyr::inner_join(g23.mean, g23.sd, by = "sid"))
g34.mean <- nrt.koku[c("sid", "g34.koku")] %>% na.omit() %>% group_by(sid) %>% summarise(avg.34 = mean(g34.koku))
g34.sd <- nrt.koku[c("sid", "g34.koku")] %>% na.omit() %>% group_by(sid) %>% summarise(sd.34 = sd(g34.koku))
g34.msd <-data.frame(dplyr::inner_join(g34.mean, g34.sd, by = "sid"))
g45.mean <- nrt.koku[c("sid", "g45.koku")] %>% na.omit() %>% group_by(sid) %>% summarise(avg.45 = mean(g45.koku))
g45.sd <- nrt.koku[c("sid", "g45.koku")] %>% na.omit() %>% group_by(sid) %>% summarise(sd.45 = sd(g45.koku))
g45.msd <-data.frame(dplyr::inner_join(g45.mean, g45.sd, by = "sid"))
g56.mean <- nrt.koku[c("sid", "g56.koku")] %>% na.omit() %>% group_by(sid) %>% summarise(avg.56 = mean(g56.koku))
g56.sd <- nrt.koku[c("sid", "g56.koku")] %>% na.omit() %>% group_by(sid) %>% summarise(sd.56 = sd(g56.koku))
g56.msd <-data.frame(dplyr::inner_join(g56.mean, g56.sd, by = "sid"))
g67.mean <- nrt.koku[c("sid", "g67.koku")] %>% na.omit() %>% group_by(sid) %>% summarise(avg.67 = mean(g67.koku))
g67.sd <- nrt.koku[c("sid", "g67.koku")] %>% na.omit() %>% group_by(sid) %>% summarise(sd.67 = sd(g67.koku))
g67.msd <-data.frame(dplyr::inner_join(g67.mean, g67.sd, by = "sid"))
g123.msd <- data.frame(dplyr::full_join(g12.msd, g23.msd, by = "sid"))
g1234.msd <- data.frame(dplyr::full_join(g123.msd, g34.msd, by = "sid"))
g12345.msd <- data.frame(dplyr::full_join(g1234.msd, g45.msd, by = "sid"))
g123456.msd <- data.frame(dplyr::full_join(g12345.msd, g56.msd, by = "sid"))
g17.msd <- data.frame(dplyr::full_join(g123456.msd, g67.msd, by = "sid"))
g17.msd <- g17.msd %>% dplyr::mutate(sdr.23 = sd.23 / sd.12)
g17.msd <- g17.msd %>% dplyr::mutate(sdr.34 = sd.34 / sd.23)
g17.msd <- g17.msd %>% dplyr::mutate(sdr.45 = sd.45 / sd.34)
g17.msd <- g17.msd %>% dplyr::mutate(sdr.56 = sd.56 / sd.45)
g17.msd <- g17.msd %>% dplyr::mutate(sdr.67 = sd.67 / sd.56)
#head(g16.msd)
hist(g17.msd$sdr.23, breaks=seq(0,4,0.2), main="sdr.23", xlab="sd_ato / sd_mae", ylim=c(0,120), xlim=c(0,4))
hist(g17.msd$sdr.34, breaks=seq(0,4,0.2), main="sdr.34", xlab="sd_ato / sd_mae", ylim=c(0,120), xlim=c(0,4))
hist(g17.msd$sdr.45, breaks=seq(0,4,0.2), main="sdr.45", xlab="sd_ato / sd_mae", ylim=c(0,120), xlim=c(0,4))
hist(g17.msd$sdr.56, breaks=seq(0,4,0.2), main="sdr.56", xlab="sd_ato / sd_mae", ylim=c(0,120), xlim=c(0,4))
hist(g17.msd$sdr.67, breaks=seq(0,4,0.2), main="sdr.67", xlab="sd_ato / sd_mae", ylim=c(0,120), xlim=c(0,4))
plot(g12.msd$avg.12, g12.msd$sd.12, xlim = c(30, 70), ylim = c(0,20), main = "Correl of SS and SD (g12)")
plot(g23.msd$avg.23, g23.msd$sd.23, xlim = c(30, 70), ylim = c(0,20), main = "Correl of SS and SD (g23)")
plot(g34.msd$avg.34, g34.msd$sd.34, xlim = c(30, 70), ylim = c(0,20), main = "Correl of SS and SD (g34)")
plot(g45.msd$avg.45, g45.msd$sd.45, xlim = c(30, 70), ylim = c(0,20), main = "Correl of SS and SD (g45)")
plot(g56.msd$avg.56, g56.msd$sd.56, xlim = c(30, 70), ylim = c(0,20), main = "Correl of SS and SD (g56)")
plot(g67.msd$avg.67, g67.msd$sd.67, xlim = c(30, 70), ylim = c(0,20), main = "Correl of SS and SD (g67)")
## 学級規模データの読み込み
setwd("/Users/koyo/Dropbox/000078_CSKAKEN/01_CSNC")
csnc.all <- read_excel("sho_csnc.xlsx")
# 学校データ整形
setwd("/Users/koyo/Dropbox/000078_CSKAKEN/190700_Anal")
#### 統廃合のない学校のみを対象 複式設置校を除外
csnc.taisho_ <- dplyr::filter(csnc.all, taisho.g1 == 1 &
togo == 0 & nonrt == 0 & fuku == 0)
# 学校データを数値型にする
csnc.taisho <- select(csnc.taisho_,(c("taisho", "sid.new",
"nc.g1", "csmean.g1",
"nc.g2", "csmean.g2",
"nc.g3", "csmean.g3",
"nc.g4", "csmean.g4",
"nc.g5", "csmean.g5",
"nc.g6", "csmean.g6"
)))
csnc.taisho$taisho <- as.numeric(csnc.taisho$taisho)
csnc.taisho$sid.new <- as.numeric(csnc.taisho$sid.new)
csnc.taisho$nc.g1 <- as.numeric(csnc.taisho$nc.g1)
csnc.taisho$csmean.g1 <- as.numeric(csnc.taisho$csmean.g1)
csnc.taisho$nc.g2 <- as.numeric(csnc.taisho$nc.g2)
csnc.taisho$csmean.g2 <- as.numeric(csnc.taisho$csmean.g2)
csnc.taisho$nc.g3 <- as.numeric(csnc.taisho$nc.g3)
csnc.taisho$csmean.g3 <- as.numeric(csnc.taisho$csmean.g3)
csnc.taisho$nc.g4 <- as.numeric(csnc.taisho$nc.g4)
csnc.taisho$csmean.g4 <- as.numeric(csnc.taisho$csmean.g4)
csnc.taisho$nc.g5 <- as.numeric(csnc.taisho$nc.g5)
csnc.taisho$csmean.g5 <- as.numeric(csnc.taisho$csmean.g5)
csnc.taisho$nc.g6 <- as.numeric(csnc.taisho$nc.g6)
csnc.taisho$csmean.g6 <- as.numeric(csnc.taisho$csmean.g6)
csnc.nona <- na.omit(csnc.taisho)
colnames(csnc.nona) <- c("taisho", "sid",
"nc.g1", "cs.g1",
"nc.g2", "cs.g2",
"nc.g3", "cs.g3",
"nc.g4", "cs.g4",
"nc.g5", "cs.g5",
"nc.g6", "cs.g6"
)
csnc <- csnc.nona[,2:14]
## 学級規模を中心化する
### 各学年での平均
cs.m.g1 <- mean(csnc$cs.g1)
cs.m.g2 <- mean(csnc$cs.g2)
cs.m.g3 <- mean(csnc$cs.g3)
cs.m.g4 <- mean(csnc$cs.g4)
cs.m.g5 <- mean(csnc$cs.g5)
cs.m.g6 <- mean(csnc$cs.g6)
### 各学年の平均の平均
csm <- matrix(c(cs.m.g1, cs.m.g2, cs.m.g3, cs.m.g4, cs.m.g5, cs.m.g6), nrow=6, ncol=1)
csnc$cs.c.g1 <- csnc$cs.g1 - mean(csm)
csnc$cs.c.g2 <- csnc$cs.g2 - mean(csm)
csnc$cs.c.g3 <- csnc$cs.g3 - mean(csm)
csnc$cs.c.g4 <- csnc$cs.g4 - mean(csm)
csnc$cs.c.g5 <- csnc$cs.g5 - mean(csm)
csnc$cs.c.g6 <- csnc$cs.g6 - mean(csm)
## 学級規模変動差分データ列作成
csnc$cs.d12 <- csnc$cs.g2 - csnc$cs.g1
csnc$cs.d23 <- csnc$cs.g3 - csnc$cs.g2
csnc$cs.d34 <- csnc$cs.g4 - csnc$cs.g3
csnc$cs.d45 <- csnc$cs.g5 - csnc$cs.g4
csnc$cs.d56 <- csnc$cs.g6 - csnc$cs.g5 学級規模データと学校ごとの標準偏差データを結合する
# 2-3年生
cs.23 <- csnc[c("sid", "nc.g2", "cs.g2", "cs.c.g2", "cs.d12")]
sdr.23 <- g17.msd[c("sid", "avg.12", "avg.23", "sdr.23")]
cs.sdr.23<-na.omit(data.frame(dplyr::inner_join(cs.23, sdr.23, by = "sid")))
#3-4年生
cs.34 <- csnc[c("sid", "nc.g3", "cs.g3", "cs.c.g3", "cs.d23")]
sdr.34 <- g17.msd[c("sid", "avg.23", "avg.34", "sdr.34")]
cs.sdr.34<-na.omit(data.frame(dplyr::inner_join(cs.34, sdr.34, by = "sid")))
# 4-5年生
cs.45 <- csnc[c("sid", "nc.g4", "cs.g4", "cs.c.g4", "cs.d34")]
sdr.45 <- g17.msd[c("sid", "avg.34", "avg.45", "sdr.45")]
cs.sdr.45<-na.omit(data.frame(dplyr::inner_join(cs.45, sdr.45, by = "sid")))
# 5-6年生
cs.56 <- csnc[c("sid", "nc.g5", "cs.g5", "cs.c.g5", "cs.d45")]
sdr.56 <- g17.msd[c("sid", "avg.45", "avg.56", "sdr.56")]
cs.sdr.56<-na.omit(data.frame(dplyr::inner_join(cs.56, sdr.56, by = "sid")))
# 6-7年生
cs.67 <- csnc[c("sid", "nc.g6", "cs.g6", "cs.c.g6", "cs.d56")]
sdr.67 <- g17.msd[c("sid", "avg.56", "avg.67", "sdr.67")]
cs.sdr.67<-na.omit(data.frame(dplyr::inner_join(cs.67, sdr.67, by = "sid")))学級規模と標準偏差比の散布図を描いてみる
#plot(cs.cvr.12$cs.g2, cs.cvr.12$cvr.12, xlim = c(0, 50), ylim = c(0,3), main = "cs.cvr.23")
plot(cs.sdr.23$cs.g2, cs.sdr.23$sdr.23, xlim = c(0, 50), ylim = c(0,3), main = "cs.sdr.23")
plot(cs.sdr.34$cs.g3, cs.sdr.34$sdr.34, xlim = c(0, 50), ylim = c(0,3), main = "cs.sdr.34")
plot(cs.sdr.45$cs.g4, cs.sdr.45$sdr.45, xlim = c(0, 50), ylim = c(0,3), main = "cs.sdr.45")
plot(cs.sdr.56$cs.g5, cs.sdr.56$sdr.56, xlim = c(0, 50), ylim = c(0,3), main = "cs.sdr.56")
plot(cs.sdr.67$cs.g6, cs.sdr.67$sdr.67, xlim = c(0, 50), ylim = c(0,3), main = "cs.sdr.67")
学級規模の変動と標準偏差比の散布図を描いてみる
plot(cs.sdr.23$cs.d12, cs.sdr.23$sdr.23, xlim = c(-15, 15), ylim = c(0,3), main = "cs_d.sdr.23")
plot(cs.sdr.34$cs.d23, cs.sdr.34$sdr.34, xlim = c(-15, 15), ylim = c(0,3), main = "cs_d.sdr.34")
plot(cs.sdr.45$cs.d34, cs.sdr.45$sdr.45, xlim = c(-15, 15), ylim = c(0,3), main = "cs_d.sdr.45")
plot(cs.sdr.56$cs.d45, cs.sdr.56$sdr.56, xlim = c(-15, 15), ylim = c(0,3), main = "cs_d.sdr.56")
plot(cs.sdr.67$cs.d56, cs.sdr.67$sdr.67, xlim = c(-15, 15), ylim = c(0,3), main = "cs_d.sdr.56")
以下のことを検討する
- 標準偏差比は学級規模で説明されるのか
- 標準偏差比は，prior achievementで説明されるのか
- 標準偏差比の大きさに対する(a)と(b)の交互作用は見られるか
# 学級規模の中心化
cs.sdr.23$cs.c.g2 <- cs.sdr.23$cs.g2 - mean(cs.sdr.23$cs.g2)
cs.sdr.34$cs.c.g3 <- cs.sdr.34$cs.g3 - mean(cs.sdr.34$cs.g3)
cs.sdr.45$cs.c.g4 <- cs.sdr.45$cs.g4 - mean(cs.sdr.45$cs.g4)
cs.sdr.56$cs.c.g5 <- cs.sdr.56$cs.g5 - mean(cs.sdr.56$cs.g5)
cs.sdr.67$cs.c.g6 <- cs.sdr.67$cs.g6 - mean(cs.sdr.67$cs.g6)
# Prior achievementの中心化
cs.sdr.23$avg.c.12 <- cs.sdr.23$avg.12 - mean(cs.sdr.23$avg.12)
cs.sdr.34$avg.c.23 <- cs.sdr.34$avg.23 - mean(cs.sdr.34$avg.23)
cs.sdr.45$avg.c.34 <- cs.sdr.45$avg.34 - mean(cs.sdr.45$avg.34)
cs.sdr.56$avg.c.45 <- cs.sdr.56$avg.45 - mean(cs.sdr.56$avg.45)
cs.sdr.67$avg.c.56 <- cs.sdr.67$avg.56 - mean(cs.sdr.67$avg.56)
head(cs.sdr.23)## sid nc.g2 cs.g2 cs.c.g2 cs.d12 avg.12 avg.23 sdr.23
## 34 18034 2 32.5 10.0368056 0 49.71429 51.42857 1.0038162
## 35 18035 1 23.0 0.5368056 0 50.31818 52.36364 0.8472441
## 36 18036 4 27.5 5.0368056 0 53.63636 52.81818 0.9290338
## 37 18037 1 18.0 -4.4631944 0 54.35714 53.85714 0.9225696
## 38 18050 3 26.0 3.5368056 0 51.43662 48.88732 0.9391956
## 39 18051 2 25.5 3.0368056 0 49.60417 52.66667 0.7812879
## avg.c.12
## 34 -4.2219832
## 35 -3.6180871
## 36 -0.2999053
## 37 0.4208740
## 38 -2.4996492
## 39 -4.3321022Prior achievementと標準偏差比の散布図を描いてみる
plot(cs.sdr.23$avg.c.12, cs.sdr.23$sdr.23, xlim = c(-15, 15), ylim = c(0,3), main = "Prior_1, sdr.23")
plot(cs.sdr.34$avg.c.23, cs.sdr.34$sdr.34, xlim = c(-15, 15), ylim = c(0,3), main = "Prior_2, sdr.34")
plot(cs.sdr.45$avg.c.34, cs.sdr.45$sdr.45, xlim = c(-15, 15), ylim = c(0,3), main = "Prior_3, sdr.45")
plot(cs.sdr.56$avg.c.45, cs.sdr.56$sdr.56, xlim = c(-15, 15), ylim = c(0,3), main = "Prior_4, sdr.56")
plot(cs.sdr.67$avg.c.56, cs.sdr.67$sdr.67, xlim = c(-15, 15), ylim = c(0,3), main = "Prior_5, sdr.56")
回帰分析をしてみる
# 2年生終了時
library(brms)## Loading required package: Rcpp## Loading required package: ggplot2##
## Attaching package: 'ggplot2'## The following objects are masked from 'package:psych':
##
## %+%, alpha## Loading 'brms' package (version 2.7.0). Useful instructions
## can be found by typing help('brms'). A more detailed introduction
## to the package is available through vignette('brms_overview').
## Run theme_set(theme_default()) to use the default bayesplot theme.##
## Attaching package: 'brms'## The following object is masked from 'package:psych':
##
## csres.23 <- brm(sdr.23 ~ cs.c.g2 + avg.c.12 + cs.c.g2:avg.c.12,
data =cs.sdr.23,
prior = c(set_prior("normal(0,10)", class = "b")),
chains = 4,
iter = 10000,
warmup = 3000
)## Compiling the C++ model## Start sampling##
## SAMPLING FOR MODEL 'd35359081d7733aebc9e00ac9119bde7' NOW (CHAIN 1).
## Chain 1:
## Chain 1: Gradient evaluation took 3.6e-05 seconds
## Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.36 seconds.
## Chain 1: Adjust your expectations accordingly!
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## Chain 1:
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## Chain 1:
##
## SAMPLING FOR MODEL 'd35359081d7733aebc9e00ac9119bde7' NOW (CHAIN 2).
## Chain 2:
## Chain 2: Gradient evaluation took 1.3e-05 seconds
## Chain 2: 1000 transitions using 10 leapfrog steps per transition would take 0.13 seconds.
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## SAMPLING FOR MODEL 'd35359081d7733aebc9e00ac9119bde7' NOW (CHAIN 3).
## Chain 3:
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## SAMPLING FOR MODEL 'd35359081d7733aebc9e00ac9119bde7' NOW (CHAIN 4).
## Chain 4:
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## Chain 4:print(res.23, digits = 3)## Family: gaussian
## Links: mu = identity; sigma = identity
## Formula: sdr.23 ~ cs.c.g2 + avg.c.12 + cs.c.g2:avg.c.12
## Data: cs.sdr.23 (Number of observations: 120)
## Samples: 4 chains, each with iter = 10000; warmup = 3000; thin = 1;
## total post-warmup samples = 28000
##
## Population-Level Effects:
## Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat
## Intercept 0.976 0.019 0.939 1.013 31735 1.000
## cs.c.g2 0.006 0.003 0.001 0.012 32639 1.000
## avg.c.12 0.031 0.007 0.018 0.044 32720 1.000
## cs.c.g2:avg.c.12 -0.002 0.001 -0.004 0.000 37856 1.000
##
## Family Specific Parameters:
## Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat
## sigma 0.204 0.014 0.180 0.233 29314 1.000
##
## Samples were drawn using sampling(NUTS). For each parameter, Eff.Sample
## is a crude measure of effective sample size, and Rhat is the potential
## scale reduction factor on split chains (at convergence, Rhat = 1).# 3年生終了時
res.34 <- brm(sdr.34 ~ cs.c.g3 + avg.c.23 + cs.c.g3:avg.c.23,
data =cs.sdr.34,
prior = c(set_prior("normal(0,10)", class = "b")),
chains = 4,
iter = 10000,
warmup = 3000
)## Compiling the C++ model## recompiling to avoid crashing R session## Start sampling##
## SAMPLING FOR MODEL 'd35359081d7733aebc9e00ac9119bde7' NOW (CHAIN 1).
## Chain 1:
## Chain 1: Gradient evaluation took 3.3e-05 seconds
## Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.33 seconds.
## Chain 1: Adjust your expectations accordingly!
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## SAMPLING FOR MODEL 'd35359081d7733aebc9e00ac9119bde7' NOW (CHAIN 2).
## Chain 2:
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## SAMPLING FOR MODEL 'd35359081d7733aebc9e00ac9119bde7' NOW (CHAIN 3).
## Chain 3:
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## SAMPLING FOR MODEL 'd35359081d7733aebc9e00ac9119bde7' NOW (CHAIN 4).
## Chain 4:
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## Chain 4:print(res.34, digits = 3)## Family: gaussian
## Links: mu = identity; sigma = identity
## Formula: sdr.34 ~ cs.c.g3 + avg.c.23 + cs.c.g3:avg.c.23
## Data: cs.sdr.34 (Number of observations: 120)
## Samples: 4 chains, each with iter = 10000; warmup = 3000; thin = 1;
## total post-warmup samples = 28000
##
## Population-Level Effects:
## Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat
## Intercept 1.191 0.016 1.159 1.222 29863 1.000
## cs.c.g3 -0.000 0.002 -0.005 0.004 35506 1.000
## avg.c.23 0.043 0.006 0.031 0.056 29353 1.000
## cs.c.g3:avg.c.23 -0.001 0.001 -0.003 0.001 35942 1.000
##
## Family Specific Parameters:
## Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat
## sigma 0.172 0.012 0.151 0.197 26503 1.000
##
## Samples were drawn using sampling(NUTS). For each parameter, Eff.Sample
## is a crude measure of effective sample size, and Rhat is the potential
## scale reduction factor on split chains (at convergence, Rhat = 1).# 4年生終了時
res.45 <- brm(sdr.45 ~ cs.c.g4 + avg.c.34 + cs.c.g4:avg.c.34,
data =cs.sdr.45,
prior = c(set_prior("normal(0,10)", class = "b")),
chains = 4,
iter = 10000,
warmup = 3000
)## Compiling the C++ model## recompiling to avoid crashing R session## Start sampling##
## SAMPLING FOR MODEL 'd35359081d7733aebc9e00ac9119bde7' NOW (CHAIN 1).
## Chain 1:
## Chain 1: Gradient evaluation took 3.4e-05 seconds
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## SAMPLING FOR MODEL 'd35359081d7733aebc9e00ac9119bde7' NOW (CHAIN 2).
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## Chain 2: Gradient evaluation took 3.1e-05 seconds
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## SAMPLING FOR MODEL 'd35359081d7733aebc9e00ac9119bde7' NOW (CHAIN 3).
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## SAMPLING FOR MODEL 'd35359081d7733aebc9e00ac9119bde7' NOW (CHAIN 4).
## Chain 4:
## Chain 4: Gradient evaluation took 1.4e-05 seconds
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## Chain 4:print(res.45, digits = 3)## Family: gaussian
## Links: mu = identity; sigma = identity
## Formula: sdr.45 ~ cs.c.g4 + avg.c.34 + cs.c.g4:avg.c.34
## Data: cs.sdr.45 (Number of observations: 120)
## Samples: 4 chains, each with iter = 10000; warmup = 3000; thin = 1;
## total post-warmup samples = 28000
##
## Population-Level Effects:
## Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat
## Intercept 0.958 0.016 0.926 0.990 22015 1.000
## cs.c.g4 -0.001 0.002 -0.005 0.004 29249 1.000
## avg.c.34 0.009 0.005 -0.001 0.019 23749 1.000
## cs.c.g4:avg.c.34 0.000 0.001 -0.001 0.002 35218 1.000
##
## Family Specific Parameters:
## Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat
## sigma 0.174 0.012 0.153 0.199 22084 1.000
##
## Samples were drawn using sampling(NUTS). For each parameter, Eff.Sample
## is a crude measure of effective sample size, and Rhat is the potential
## scale reduction factor on split chains (at convergence, Rhat = 1).# 5年生終了時
res.56 <- brm(sdr.56 ~ cs.c.g5 + avg.c.45 + cs.c.g5:avg.c.45,
data =cs.sdr.56,
prior = c(set_prior("normal(0,10)", class = "b")),
chains = 4,
iter = 10000,
warmup = 3000
)## Compiling the C++ model## recompiling to avoid crashing R session## Start sampling##
## SAMPLING FOR MODEL 'd35359081d7733aebc9e00ac9119bde7' NOW (CHAIN 1).
## Chain 1:
## Chain 1: Gradient evaluation took 3.6e-05 seconds
## Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.36 seconds.
## Chain 1: Adjust your expectations accordingly!
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## SAMPLING FOR MODEL 'd35359081d7733aebc9e00ac9119bde7' NOW (CHAIN 2).
## Chain 2:
## Chain 2: Gradient evaluation took 2.4e-05 seconds
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## SAMPLING FOR MODEL 'd35359081d7733aebc9e00ac9119bde7' NOW (CHAIN 3).
## Chain 3:
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## SAMPLING FOR MODEL 'd35359081d7733aebc9e00ac9119bde7' NOW (CHAIN 4).
## Chain 4:
## Chain 4: Gradient evaluation took 1.4e-05 seconds
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## Chain 4:print(res.56, digits = 3)## Family: gaussian
## Links: mu = identity; sigma = identity
## Formula: sdr.56 ~ cs.c.g5 + avg.c.45 + cs.c.g5:avg.c.45
## Data: cs.sdr.56 (Number of observations: 170)
## Samples: 4 chains, each with iter = 10000; warmup = 3000; thin = 1;
## total post-warmup samples = 28000
##
## Population-Level Effects:
## Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat
## Intercept 0.898 0.009 0.881 0.915 21474 1.000
## cs.c.g5 -0.001 0.001 -0.004 0.001 33082 1.000
## avg.c.45 0.003 0.003 -0.003 0.010 23171 1.000
## cs.c.g5:avg.c.45 -0.000 0.000 -0.001 0.001 30062 1.000
##
## Family Specific Parameters:
## Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat
## sigma 0.113 0.006 0.101 0.126 20098 1.000
##
## Samples were drawn using sampling(NUTS). For each parameter, Eff.Sample
## is a crude measure of effective sample size, and Rhat is the potential
## scale reduction factor on split chains (at convergence, Rhat = 1).# 6年生終了時
res.67 <- brm(sdr.67 ~ cs.c.g6 + avg.c.56 + cs.c.g6:avg.c.56,
data =cs.sdr.67,
prior = c(set_prior("normal(0,10)", class = "b")),
chains = 4,
iter = 10000,
warmup = 3000
)## Compiling the C++ model## recompiling to avoid crashing R session## Start sampling##
## SAMPLING FOR MODEL 'd35359081d7733aebc9e00ac9119bde7' NOW (CHAIN 1).
## Chain 1:
## Chain 1: Gradient evaluation took 2.9e-05 seconds
## Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.29 seconds.
## Chain 1: Adjust your expectations accordingly!
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## Chain 1:
## Chain 1: Elapsed Time: 0.412625 seconds (Warm-up)
## Chain 1: 0.728523 seconds (Sampling)
## Chain 1: 1.14115 seconds (Total)
## Chain 1:
##
## SAMPLING FOR MODEL 'd35359081d7733aebc9e00ac9119bde7' NOW (CHAIN 2).
## Chain 2:
## Chain 2: Gradient evaluation took 1.5e-05 seconds
## Chain 2: 1000 transitions using 10 leapfrog steps per transition would take 0.15 seconds.
## Chain 2: Adjust your expectations accordingly!
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## Chain 2:
## Chain 2: Elapsed Time: 0.450087 seconds (Warm-up)
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## Chain 2: 1.11261 seconds (Total)
## Chain 2:
##
## SAMPLING FOR MODEL 'd35359081d7733aebc9e00ac9119bde7' NOW (CHAIN 3).
## Chain 3:
## Chain 3: Gradient evaluation took 1.5e-05 seconds
## Chain 3: 1000 transitions using 10 leapfrog steps per transition would take 0.15 seconds.
## Chain 3: Adjust your expectations accordingly!
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## Chain 3: 1.06241 seconds (Total)
## Chain 3:
##
## SAMPLING FOR MODEL 'd35359081d7733aebc9e00ac9119bde7' NOW (CHAIN 4).
## Chain 4:
## Chain 4: Gradient evaluation took 1.3e-05 seconds
## Chain 4: 1000 transitions using 10 leapfrog steps per transition would take 0.13 seconds.
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## Chain 4: Elapsed Time: 0.451808 seconds (Warm-up)
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## Chain 4: 0.91416 seconds (Total)
## Chain 4:print(res.67, digits = 3)## Family: gaussian
## Links: mu = identity; sigma = identity
## Formula: sdr.67 ~ cs.c.g6 + avg.c.56 + cs.c.g6:avg.c.56
## Data: cs.sdr.67 (Number of observations: 170)
## Samples: 4 chains, each with iter = 10000; warmup = 3000; thin = 1;
## total post-warmup samples = 28000
##
## Population-Level Effects:
## Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat
## Intercept 1.218 0.014 1.191 1.245 24008 1.000
## cs.c.g6 -0.001 0.002 -0.005 0.002 32114 1.000
## avg.c.56 0.006 0.006 -0.005 0.018 24127 1.000
## cs.c.g6:avg.c.56 -0.000 0.001 -0.001 0.001 34979 1.000
##
## Family Specific Parameters:
## Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat
## sigma 0.171 0.009 0.154 0.191 22950 1.000
##
## Samples were drawn using sampling(NUTS). For each parameter, Eff.Sample
## is a crude measure of effective sample size, and Rhat is the potential
## scale reduction factor on split chains (at convergence, Rhat = 1).