Randomized complete block designs
2026-10-26
Six nitrogen timing schedules in each of four plots in a field of wheat. Response is the nitrate content in stem tissue samples.
Assume
\[ Y_{ij} = \mu + \tau_i + B_j + \varepsilon_{ij}, \quad \text{ for } i=1,\dots,a, \quad j = 1,\dots,b, \] where
Define the cell means as \[ \mu_i = \mu + \tau_i, \quad i = 1,\dots,a. \]
In the randomized complete block design we wish to
| Sum of squares | Symbol | Formula |
|---|---|---|
| Total | \(\operatorname{SS}_{\operatorname{Tot}}\) | \(\sum_{i=1}^a\sum_{j=1}^b(Y_{ij} - \bar Y_{..})^2\) |
| Treatment | \(\operatorname{SS}_{\operatorname{A}}\) | \(b\sum_{i=1}^a (\bar Y_{i.} - \bar Y_{..})^2\) |
| Block | \(\operatorname{SS}_{\operatorname{B}}\) | \(a\sum_{j=1}^b (\bar Y_{.j} - \bar Y_{..})^2\) |
| Error | \(\operatorname{SS}_{\operatorname{Error}}\) | \(\sum_{i=1}^a \sum_{j=1}^b (Y_{ij} - (\bar Y_{i.} + \bar Y_{.j} - \bar Y_{..}))^2\) |
| Source | Df | SS | MS | F value |
|---|---|---|---|---|
| A | \(a-1\) | \(\operatorname{SS}_{\operatorname{A}}\) | \(\operatorname{MS}_{\operatorname{A}}\) | \(F_{\operatorname{A}} = \operatorname{MS}_{\operatorname{A}}/ \operatorname{MS}_{\operatorname{Error}}\) |
| B | \(b-1\) | \(\operatorname{SS}_{\operatorname{B}}\) | \(\operatorname{MS}_{\operatorname{B}}\) | \(F_{\operatorname{B}} = \operatorname{MS}_{\operatorname{B}}/ \operatorname{MS}_{\operatorname{Error}}\) |
| Error | \((a-1)(b-1)\) | \(\operatorname{SS}_{\operatorname{Error}}\) | \(\operatorname{MS}_{\operatorname{Error}}\) | |
| Total | \(ab-1\) | \(\operatorname{SS}_{\operatorname{Tot}}\) |
Obtain ANOVA table “manually”:
y <- ntr
y.. <- predict(lm(ntr ~ 1))
yi. <- predict(lm(ntr ~ trt))
y.j <- predict(lm(ntr ~ block))
SST <- sum((y - y..)^2)
SSA <- sum((yi. - y..)^2)
SSB <- sum((y.j - y..)^2)
SSE <- SST - (SSA + SSB)
a <- 6
b <- 4
MSA <- SSA/(a-1)
MSB <- SSB/(b-1)
MSE <- SSE/((a-1)*(b-1))
FA <- MSA / MSE
FB <- MSB / MSE
pA <- 1 - pf(FA,a-1,(a-1)*(b-1))
pB <- 1 - pf(FB,b-1,(a-1)*(b-1))| Source | Df | SS | MS | F value | p value |
|---|---|---|---|---|---|
| A | 5 | 201.32 | 40.26 | 5.5917 | 0.0042 |
| B | 3 | 197.00 | 65.67 | 9.1198 | 0.0011 |
| Error | 15 | 108.01 | 7.20 | ||
| Total | 23 | 506.33 |
Obtain ANOVA table with anova() function on lm() output:
Analysis of Variance Table
Response: ntr
Df Sum Sq Mean Sq F value Pr(>F)
trt 5 201 40.3 5.59 0.0042 **
block 3 197 65.7 9.12 0.0011 **
Residuals 15 108 7.2
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
| Source | Df | Expected mean square |
|---|---|---|
| A | \(a-1\) | \(b\theta_A^2 + \sigma^2_\varepsilon\) |
| B | \(b-1\) | \(a\sigma_B^2 + \sigma^2_\varepsilon\) |
| Error | \((a-1)(b-1)\) | \(\sigma_\varepsilon^2\) |
In the above \(\theta_A^2 = (a-1)^{-1}\sum_{i=1}^a(\mu_i - \bar\mu_.)^2\).
Equating \(\operatorname{MS}_{\operatorname{B}}\) and \(\operatorname{MS}_{\operatorname{Error}}\) with their expectations gives
May obtain \(\dot \sigma_B^2 < 0\), so one should use REML estimation.
Obtain REML estimators of \(\sigma_B^2\) and \(\sigma_{\varepsilon}^2\) on the fertilization data.
Linear mixed model fit by REML ['lmerModLmerTest']
Formula: ntr ~ trt + (1 | block)
REML criterion at convergence: 102
Random effects:
Groups Name Std.Dev.
block (Intercept) 3.12
Residual 2.68
Number of obs: 24, groups: block, 4
Fixed Effects:
(Intercept) trt2 trt3 trt4 trt5 trt6
38.28 5.75 8.49 2.34 1.23 4.95
Obtain MoMs estimators for \(\sigma_B^2\) and \(\sigma_{\varepsilon}^2\) on the fertilization data.
We have \(\dot \sigma_B = 3.122\) and \(\dot \sigma_{\varepsilon}= 2.683\).
| Contrast | Variance | MoM variance estimator |
|---|---|---|
| \(\bar Y_{i.}\) | \(\frac{1}{b}(\sigma_B^2 + \sigma_{\varepsilon}^2)\) | \(\frac{2}{ab}[\operatorname{MS}_{\operatorname{B}}+ (b-1)\operatorname{MS}_{\operatorname{Error}}]\) |
| \(\bar Y_{i.} - \bar Y_{i'.}\) | \(\frac{2}{b}\sigma_{\varepsilon}^2\) | \(\frac{2}{b}\operatorname{MS}_{\operatorname{Error}}\) |
| Target | \((1-\alpha)100\%\) confidence interval |
|---|---|
| \(\mu_i\) | \(\bar Y_{i.} \pm t_{\nu^*,\alpha/2}\sqrt{\operatorname{MS}_{\operatorname{B}}+ (b-1)\operatorname{MS}_{\operatorname{Error}}}\sqrt{\frac{2}{ab}}\) |
| \(\mu_i - \mu_{i'}\) | \(\bar Y_{i.} - \bar Y_{i'.} \pm t_{(a-1)(b-1),\alpha/2}\sqrt{\operatorname{MS}_{\operatorname{Error}}}\sqrt{\frac{2}{b}}\) |
In the above \(\displaystyle \nu^* = \frac{\operatorname{MS}_{\operatorname{B}}+ (b-1)\operatorname{MS}_{\operatorname{Error}}}{\frac{\operatorname{MS}_{\operatorname{B}}^2}{(b-1)} + \frac{(b-1)^2\operatorname{MS}_{\operatorname{Error}}^2}{(a-1)(b-1)}}\) à la Satterthwaite.
Unadjusted CIs with ls_means() from R package lmerTest
Least Squares Means table:
Estimate Std. Error df t value lower upper Pr(>|t|)
trt1 38.28 2.06 6.8 18.6 33.38 43.18 4.5e-07 ***
trt2 44.03 2.06 6.8 21.4 39.13 48.93 1.8e-07 ***
trt3 46.77 2.06 6.8 22.7 41.87 51.67 1.2e-07 ***
trt4 40.62 2.06 6.8 19.7 35.72 45.51 3.0e-07 ***
trt5 39.51 2.06 6.8 19.2 34.61 44.41 3.6e-07 ***
trt6 43.23 2.06 6.8 21.0 38.33 48.12 2.0e-07 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Confidence level: 95%
Degrees of freedom method: Satterthwaite
Least Squares Means table:
Estimate Std. Error df t value lower upper Pr(>|t|)
trt1 - trt2 -5.755 1.897 15 -3.03 -9.799 -1.711 0.00839 **
trt1 - trt3 -8.492 1.897 15 -4.48 -12.537 -4.448 0.00044 ***
trt1 - trt4 -2.337 1.897 15 -1.23 -6.382 1.707 0.23694
trt1 - trt5 -1.232 1.897 15 -0.65 -5.277 2.812 0.52580
trt1 - trt6 -4.947 1.897 15 -2.61 -8.992 -0.903 0.01980 *
trt2 - trt3 -2.738 1.897 15 -1.44 -6.782 1.307 0.16965
trt2 - trt4 3.418 1.897 15 1.80 -0.627 7.462 0.09182 .
trt2 - trt5 4.522 1.897 15 2.38 0.478 8.567 0.03080 *
trt2 - trt6 0.808 1.897 15 0.43 -3.237 4.852 0.67646
trt3 - trt4 6.155 1.897 15 3.24 2.111 10.199 0.00545 **
trt3 - trt5 7.260 1.897 15 3.83 3.216 11.304 0.00165 **
trt3 - trt6 3.545 1.897 15 1.87 -0.499 7.589 0.08138 .
trt4 - trt5 1.105 1.897 15 0.58 -2.939 5.149 0.56897
trt4 - trt6 -2.610 1.897 15 -1.38 -6.654 1.434 0.18916
trt5 - trt6 -3.715 1.897 15 -1.96 -7.759 0.329 0.06911 .
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Confidence level: 95%
Degrees of freedom method: Satterthwaite
Tukey’s for comparing all pairs of means among \(\mu_1,\dots,\mu_a\): \[ \bar Y_{i.} - \bar Y_{i'.} \pm q_{a,(a-1)(b-1),\alpha}\sqrt{\operatorname{MS}_{\operatorname{Error}}} \sqrt{\frac{1}{b}}, \quad 1 \leq i < i' \leq a. \]
Dunnett’s for comparing \(\mu_2,\dots,\mu_a\) to a baseline \(\mu_1\): \[ \bar Y_{i.} - \bar Y_{1.} \pm d_{a,(a-1)(b-1),\alpha}\sqrt{\operatorname{MS}_{\operatorname{Error}}}\sqrt{\frac{2}{b}}, \quad i=2,\dots,a. \]
Compare all pairs of fertilizers with Tukey’s CIs for mean differences.
alpha <- 0.05
a <- 6
b <- 4
se <- sqrt(MSE) * sqrt(2/b) # have compute MSE previously
me <- qtukey(1-alpha,a,(a-1)*(b-1)) / sqrt(2) * se
ntr_means <- aggregate(ntr, by = list(trt), mean)$x
CIs <- matrix(NA,choose(a,2),2)
comp <- numeric(choose(a,2))
k <- 1
for(i in 1:(a-1))
for(j in (i+1):a){ # double loop takes us through all pairs
dij <- ntr_means[i] - ntr_means[j]
CIs[k,] <- c(dij - me, dij + me)
comp[k] <- paste(i,"-",j)
k <- k + 1
}
colnames(CIs) <- c("lower","upper")
rownames(CIs) <- compCompare all fertilizers to fertilizer 1 using Dunnett’s method.
alpha <- 0.05
a <- 6
b <- 4
me <- 2.82 * sqrt(MSE) * sqrt(2/b) # value 2.82 from Dunnett's table
ntr_means <- aggregate(ntr, by = list(trt), mean)$x
CIs <- matrix(NA,a-1,2)
k <- 1
for(i in 2:a){
di <- ntr_means[i] - ntr_means[1]
CIs[k,] <- c(di - me,di + me)
k <- k + 1
}
colnames(CIs) <- c("lower","upper")
rownames(CIs) <- paste(2:a,"- 1")If we ignore the blocks, the design looks like a one-way ANOVA:
| Treatment | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| 34.98 | 40.89 | 42.07 | 37.18 | 37.99 | 34.89 | |
| 41.22 | 46.69 | 49.42 | 45.85 | 41.99 | 50.15 | |
| 36.94 | 46.65 | 52.68 | 40.23 | 37.61 | 44.57 | |
| 39.97 | 41.90 | 43.29 | 39.20 | 40.45 | 43.29 |
Suppose we fit \(Y_{ij} = \mu + \tau_i + \varepsilon_{ij}\), \(i = 1,2,3,4,5,6\), \(j = 1,2,3,4\).
We lose power to detect a treatment effect!
Heads of cabbage in subplots of two field plots.
Subplots in each plot randomly assigned to nitrogen levels.
Assume \[ Y_{ijk} + \mu + \tau_i + B_j + (\tau B)_{ij} + \varepsilon_{ijk} \] for \(i=1,\dots,a\), \(j=1,\dots,b\), and \(k = 1,\dots,n_{ij}\), where
Define the cell means as \(\mu_i = \mu + \tau_i, \quad i = 1,\dots,a.\)
Assume for now a balanced design: \(n_{ij} = n\) for all \(i,j\).
| Sum of squares | Symbol | Formula |
|---|---|---|
| Total | \(\operatorname{SS}_{\operatorname{Tot}}\) | \(\sum_{i=1}^a\sum_{j=1}^b \sum_{k=1}^n (Y_{ijk} - \bar Y_{...})^2\) |
| Treatment | \(\operatorname{SS}_{\operatorname{A}}\) | \(nb\sum_{i=1}^a (\bar Y_{i..} - \bar Y_{...})^2\) |
| Block | \(\operatorname{SS}_{\operatorname{B}}\) | \(na\sum_{j=1}^b (\bar Y_{.j.} - \bar Y_{...})^2\) |
| Block \(\times\) Treatment | \(\operatorname{SS}_{\operatorname{AB}}\) | \(n\sum_{i=1}^a \sum_{j=1}^b (\bar Y_{ij.} - (\bar Y_{i..} + \bar Y_{.j.} - \bar Y_{...}))^2\) |
| Error | \(\operatorname{SS}_{\operatorname{Error}}\) | \(\sum_{i=1}^a \sum_{j=1}^b \sum_{k=1}^n(Y_{ijk} - \bar Y_{ij.})^2\) |
| Source | Df | SS | MS | F value |
|---|---|---|---|---|
| A | \(a-1\) | \(\operatorname{SS}_{\operatorname{A}}\) | \(\operatorname{MS}_{\operatorname{A}}\) | \(F_{\operatorname{A}} = \operatorname{MS}_{\operatorname{A}}/ \textcolor{red}{\operatorname{MS}_{\operatorname{AB}}}\) |
| B | \(b-1\) | \(\operatorname{SS}_{\operatorname{B}}\) | \(\operatorname{MS}_{\operatorname{B}}\) | \(F_{\operatorname{B}} = \operatorname{MS}_{\operatorname{B}}/ \textcolor{red}{\operatorname{MS}_{\operatorname{AB}}}\) |
| AB | \((a-1)(b-1)\) | \(\operatorname{SS}_{\operatorname{AB}}\) | \(\operatorname{MS}_{\operatorname{AB}}\) | \(F_{\operatorname{AB}} = \operatorname{MS}_{\operatorname{AB}}/ \operatorname{MS}_{\operatorname{Error}}\) |
| Error | \(ab(n-1)\) | \(\operatorname{SS}_{\operatorname{Error}}\) | \(\operatorname{MS}_{\operatorname{Error}}\) | |
| Total | \(abn-1\) | \(\operatorname{SS}_{\operatorname{Tot}}\) |
| Source | Df | Expected mean square |
|---|---|---|
| A | \(a-1\) | \(nb\theta_A^2 + n \sigma_{AB}^2 + \sigma^2_\varepsilon\) |
| B | \(b-1\) | \(na\sigma_B^2 + n \sigma_{AB}^2 + \sigma^2_\varepsilon\) |
| AB | \((a-1)(b-1)\) | \(n \sigma_{AB}^2 + \sigma^2_\varepsilon\) |
| Error | \(ab(n-1)\) | \(\sigma_\varepsilon^2\) |
In the above \(\theta_A^2 = (a-1)^{-1}\sum_{i=1}^a(\mu_i - \bar\mu_.)^2\).
Check expected MS values under each \(H_0\) on previous slide.
a <- nlevels(nitr)
b <- nlevels(blk)
n <- 2
y... <- predict(lm(y ~ 1))
yi.. <- predict(lm(y ~ nitr))
y.j. <- predict(lm(y ~ blk))
yij. <- predict(lm(y ~ nitr + blk + nitr:blk))
SST <- sum((y - y...)**2)
SSA <- sum((yi.. - y...)**2)
SSB <- sum((y.j. - y...)**2)
SSAB <- sum((yij. - (yi.. + y.j. - y...))**2)
SSE <- sum((y - yij.)**2)
MSA <- SSA / (a-1)
MSB <- SSB / (b-1)
MSAB <- SSAB / ((a-1)*(b-1))
MSE <- SSE / (a*b*(n-1))
FA <- MSA / MSAB
FB <- MSB / MSAB
FAB <- MSAB / MSE
pA <- 1 - pf(FA,a-1,(a-1)*(b-1))
pB <- 1 - pf(FB,b-1,(a-1)*(b-1))
pAB <- 1 - pf(FAB,(a-1)*(b-1),a*b*(n-1))| Source | Df | SS | MS | F value | p value |
|---|---|---|---|---|---|
| A | 4 | 4813.00 | 1203.25 | 16.7234 | 0.0092 |
| B | 1 | 1022.45 | 1022.45 | 14.2106 | 0.0196 |
| AB | 4 | 287.80 | 71.95 | 1.7030 | 0.2253 |
| Error | 10 | 422.50 | 42.25 | ||
| Total | 19 | 6545.75 |
Note that anova() on lm() output does not give desired output:
Set the mean squares equal to their expectations and solve: \[ \begin{aligned} \operatorname{MS}_{\operatorname{B}}&\overset{\text{set}}{=} na\sigma_B^2 + n \sigma_{AB}^2 + \sigma_\varepsilon^2\\ \operatorname{MS}_{\operatorname{AB}}&\overset{\text{set}}{=} n \sigma_{AB}^2 + \sigma_\varepsilon^2\\ \operatorname{MS}_{\operatorname{Error}}&\overset{\text{set}}{=} \sigma_\varepsilon^2 \end{aligned} \] Solve for \(\sigma_B^2\), \(\sigma_{AB}^2\), and \(\sigma_\varepsilon^2\): \[ \begin{aligned} \dot \sigma_B^2 & =\frac{1}{na}(\operatorname{MS}_{\operatorname{B}}- \operatorname{MS}_{\operatorname{AB}})\\ \dot \sigma_{AB}^2 &=\frac{1}{n}(\operatorname{MS}_{\operatorname{AB}}- \operatorname{MS}_{\operatorname{Error}})\\ \dot \sigma_\varepsilon^2 &=\operatorname{MS}_{\operatorname{Error}} \end{aligned} \]
Possible to obtain negative values for \(\dot \sigma_B^2\) and \(\dot \sigma_{AB}^2\). Then use REML.
Obtain the method of moments estimates for the variance components.
We obtain \(\dot \sigma_B = 9.749\), \(\dot \sigma_{AB} = 3.854\), and \(\dot \sigma_\varepsilon = 6.500\).
Obtain the REML estimates for the variance components.
Linear mixed model fit by REML ['lmerModLmerTest']
Formula: y ~ nitr + (1 | blk) + (1 | nitr:blk)
REML criterion at convergence: 111
Random effects:
Groups Name Std.Dev.
nitr:blk (Intercept) 3.85
blk (Intercept) 9.75
Residual 6.50
Number of obs: 20, groups: nitr:blk, 10; blk, 2
Fixed Effects:
(Intercept) nitr50 nitr100 nitr150 nitr200
112.8 32.7 36.2 44.7 36.2
A \((1-\alpha)100\%\) confidence interval for \(\mu_i - \mu_{i'}\) is \[ \bar Y_{i..} - \bar Y_{i'..} \pm t_{(a-1)(b-1),\alpha/2}\sqrt{\operatorname{MS}_{\operatorname{AB}}}\sqrt{\frac{2}{nb}} \] for \(1 \leq i < i' \leq a\).
If \((a-1)(b-1) = 1\), this interval will be frightfully wide.
In this case one should drop the interaction term from the model!
Compare the means of the nitrogen level \(50\) and \(0\) groups:
Use confint() on output of lmer():
2.5 % 97.5 %
.sig01 0.00 6.25
.sig02 2.72 29.86
.sigma 4.42 9.06
(Intercept) 95.10 130.40
nitr50 23.25 42.25
nitr100 26.75 45.75
nitr150 35.25 54.25
nitr200 26.75 45.75
These CIs are not adjusted to achieve a familywise coverage probability!
If we omit the interaction term in the RCBD with replication, we have
\[ Y_{ijk} + \mu + \tau_i + B_j + \varepsilon_{ijk} \] for \(i=1,\dots,a\), \(j=1,\dots,b\), and \(k = 1,\dots,n_{ij}\), where
Assume for now a balanced design: \(n_{ij} = n\) for all \(i,j\).
Use this model if \(a = b = 2\), under which \((a-1)(b-1)=1\).
| SS | Symbol | Formula |
|---|---|---|
| Total | \(\operatorname{SS}_{\operatorname{Tot}}\) | \(\sum_{i=1}^a\sum_{j=1}^b \sum_{k=1}^n (Y_{ijk} - \bar Y_{...})^2\) |
| Treatment | \(\operatorname{SS}_{\operatorname{A}}\) | \(nb\sum_{i=1}^a (\bar Y_{i..} - \bar Y_{...})^2\) |
| Block | \(\operatorname{SS}_{\operatorname{B}}\) | \(na\sum_{j=1}^b (\bar Y_{.j.} - \bar Y_{...})^2\) |
| Error | \(\operatorname{SS}_{\operatorname{Error}}\) | \(\sum_{i=1}^a \sum_{j=1}^b \sum_{k=1}^n(Y_{ijk} - (\bar Y_{i..} + \bar Y_{.j.} - \bar Y_{...}))^2\) |
| Source | Df | SS | MS | F value |
|---|---|---|---|---|
| A | \(a-1\) | \(\operatorname{SS}_{\operatorname{A}}\) | \(\operatorname{MS}_{\operatorname{A}}\) | \(F_{\operatorname{A}} = \operatorname{MS}_{\operatorname{A}}/ \operatorname{MS}_{\operatorname{Error}}\) |
| B | \(b-1\) | \(\operatorname{SS}_{\operatorname{B}}\) | \(\operatorname{MS}_{\operatorname{B}}\) | \(F_{\operatorname{B}} = \operatorname{MS}_{\operatorname{B}}/ \operatorname{MS}_{\operatorname{Error}}\) |
| Error | \(\textcolor{red}{abn-a - b + 1}\) | \(\operatorname{SS}_{\operatorname{Error}}\) | \(\operatorname{MS}_{\operatorname{Error}}\) | |
| Total | \(abn-1\) | \(\operatorname{SS}_{\operatorname{Tot}}\) |
| Source | Df | Expected mean square |
|---|---|---|
| A | \(a-1\) | \(nb\theta_A^2 + \sigma^2_\varepsilon\) |
| B | \(b-1\) | \(na\sigma_B^2 + \sigma^2_\varepsilon\) |
| Error | \(abn - a - b + 1\) | \(\sigma_\varepsilon^2\) |
In the above \(\theta_A^2 = (a-1)^{-1}\sum_{i=1}^a(\mu_i - \bar\mu_.)^2\).
ANOVA table, omitting the interaction:
a <- nlevels(nitr)
b <- nlevels(blk)
n <- 2
y... <- predict(lm(y ~ 1))
yi.. <- predict(lm(y ~ nitr))
y.j. <- predict(lm(y ~ blk))
SST <- sum((y - y...)**2)
SSA <- sum((yi.. - y...)**2)
SSB <- sum((y.j. - y...)**2)
SSE <- sum((y - (yi.. + y.j. - y...))**2)
MSA <- SSA / (a-1)
MSB <- SSB / (b-1)
MSE <- SSE / (a*b*n - a - b + 1)
FA <- MSA / MSE
FB <- MSB / MSE
pA <- 1 - pf(FA,a-1,a*b*n - a - b + 1)
pB <- 1 - pf(FB,b-1,a*b*n - a - b + 1)| Source | Df | SS | MS | F value | p value |
|---|---|---|---|---|---|
| A | 4 | 4813.00 | 1203.25 | 23.7160 | 0.0000 |
| B | 1 | 1022.45 | 1022.45 | 20.1525 | 0.0005 |
| Error | 14 | 710.30 | 50.74 | ||
| Total | 19 | 6545.75 |
With the interaction term omitting, anova() on lm() output is correct:
Set the mean squares equal to their expectations and solve: \[ \begin{aligned} \operatorname{MS}_{\operatorname{B}}&\overset{\text{set}}{=} na\sigma_B^2 + \sigma_\varepsilon^2\\ \operatorname{MS}_{\operatorname{Error}}&\overset{\text{set}}{=} \sigma_\varepsilon^2 \end{aligned} \] Solve for \(\sigma_B^2\) and \(\sigma_\varepsilon^2\): \[ \begin{aligned} \dot \sigma_B^2 & =\frac{1}{na}(\operatorname{MS}_{\operatorname{B}}- \operatorname{MS}_{\operatorname{Error}})\\ \dot \sigma_\varepsilon^2 &=\operatorname{MS}_{\operatorname{Error}} \end{aligned} \]
Possible to obtain a negative value for \(\dot \sigma_B^2\). In this case use REML.
MoM estimates when omitting interaction:
We obtain \(\dot \sigma_B = 9.858\) and \(\dot \sigma_\varepsilon = 7.123\).
REML estimates when omitting interaction:
Linear mixed model fit by REML ['lmerModLmerTest']
Formula: y ~ nitr + (1 | blk)
REML criterion at convergence: 111
Random effects:
Groups Name Std.Dev.
blk (Intercept) 9.86
Residual 7.12
Number of obs: 20, groups: blk, 2
Fixed Effects:
(Intercept) nitr50 nitr100 nitr150 nitr200
112.8 32.8 36.3 44.8 36.3
Compare the means of the nitrogen level \(50\) and \(0\) groups:
Can use ls_means() as below:
Least Squares Means table:
Estimate Std. Error df t value lower upper Pr(>|t|)
nitr0 - nitr50 -32.75 5.04 14 -6.50 -43.55 -21.95 1.4e-05 ***
nitr0 - nitr100 -36.25 5.04 14 -7.20 -47.05 -25.45 4.6e-06 ***
nitr0 - nitr150 -44.75 5.04 14 -8.88 -55.55 -33.95 4.0e-07 ***
nitr0 - nitr200 -36.25 5.04 14 -7.20 -47.05 -25.45 4.6e-06 ***
nitr50 - nitr100 -3.50 5.04 14 -0.69 -14.30 7.30 0.498
nitr50 - nitr150 -12.00 5.04 14 -2.38 -22.80 -1.20 0.032 *
nitr50 - nitr200 -3.50 5.04 14 -0.69 -14.30 7.30 0.498
nitr100 - nitr150 -8.50 5.04 14 -1.69 -19.30 2.30 0.114
nitr100 - nitr200 0.00 5.04 14 0.00 -10.80 10.80 1.000
nitr150 - nitr200 8.50 5.04 14 1.69 -2.30 19.30 0.114
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Confidence level: 95%
Degrees of freedom method: Satterthwaite
Galvanic skin responses of five subjects under shock and noise stimuli.
resp noise shock subj
1 3 40 0.25 1
2 7 40 0.25 2
3 9 40 0.25 3
4 4 40 0.25 4
5 1 40 0.25 5
6 5 40 0.5 1
7 11 40 0.5 2
8 13 40 0.5 3
9 8 40 0.5 4
10 3 40 0.5 5
11 9 40 0.75 1
12 12 40 0.75 2
13 14 40 0.75 3
14 11 40 0.75 4
15 5 40 0.75 5
16 6 40 1 1
17 11 40 1 2
18 12 40 1 3
19 7 40 1 4
20 4 40 1 5
Assume \[ Y_{ijk} = \mu + \tau_i + \gamma_j + (\tau\gamma)_{ij} + C_k + \varepsilon_{ijk}, \] for \(i = 1,\dots,a\), \(j = 1,\dots,b\), and \(k = 1,\dots,c\), where
Define the cell means as \[ \mu_{ij} = \mu + \tau_i + \gamma_j + (\tau\gamma)_{ij}, \quad i = 1,\dots,a, \quad j = 1,\dots,b. \]
| Sum of squares | Symbol | Formula |
|---|---|---|
| Total | \(\operatorname{SS}_{\operatorname{Tot}}\) | \(\sum_{i=1}^a\sum_{j=1}^b\sum_{k=1}^c(Y_{ijk} - \bar Y_{...})^2\) |
| A | \(\operatorname{SS}_{\operatorname{A}}\) | \(bc\sum_{i=1}^a (\bar Y_{i..} - \bar Y_{...})^2\) |
| B | \(\operatorname{SS}_{\operatorname{B}}\) | \(ac\sum_{j=1}^b (\bar Y_{.j.} - \bar Y_{...})^2\) |
| AB | \(\operatorname{SS}_{\operatorname{AB}}\) | \(c\sum_{i=1}^a\sum_{j=1}^b (Y_{ij.} - (\bar Y_{i..} + \bar Y_{.j.} - \bar Y_{...}))^2\) |
| C | \(\operatorname{SS}_{\operatorname{C}}\) | \(ab\sum_{k=1}^c (\bar Y_{..k} - \bar Y_{...})^2\) |
| Error | \(\operatorname{SS}_{\operatorname{Error}}\) | \(\operatorname{SS}_{\operatorname{Tot}}- (\operatorname{SS}_{\operatorname{A}}+ \operatorname{SS}_{\operatorname{B}}+ \operatorname{SS}_{\operatorname{AB}}+ \operatorname{SS}_{\operatorname{C}})\) |
| Source | Df | SS | MS | F value |
|---|---|---|---|---|
| A | \(a-1\) | \(\operatorname{SS}_{\operatorname{A}}\) | \(\operatorname{MS}_{\operatorname{A}}\) | \(F_{\operatorname{A}}= \operatorname{MS}_{\operatorname{A}}/ \operatorname{MS}_{\operatorname{Error}}\) |
| B | \(b-1\) | \(\operatorname{SS}_{\operatorname{B}}\) | \(\operatorname{MS}_{\operatorname{B}}\) | \(F_{\operatorname{B}}= \operatorname{MS}_{\operatorname{B}}/ \operatorname{MS}_{\operatorname{Error}}\) |
| AB | \((a-1)(b-1)\) | \(\operatorname{SS}_{\operatorname{AB}}\) | \(\operatorname{MS}_{\operatorname{AB}}\) | \(F_{\operatorname{AB}}= \operatorname{MS}_{\operatorname{B}}/ \operatorname{MS}_{\operatorname{Error}}\) |
| C | \(c-1\) | \(\operatorname{SS}_{\operatorname{C}}\) | \(\operatorname{MS}_{\operatorname{C}}\) | \(F_{\operatorname{C}}= \operatorname{MS}_{\operatorname{C}}/ \operatorname{MS}_{\operatorname{Error}}\) |
| Error | \((ab-1)(c-1)\) | \(\operatorname{SS}_{\operatorname{Error}}\) | \(\operatorname{MS}_{\operatorname{Error}}\) | |
| Total | \(abc-1\) | \(\operatorname{SS}_{\operatorname{Tot}}\) |
Analysis of Variance Table
Response: resp
Df Sum Sq Mean Sq F value Pr(>F)
noise 1 65 65.0 28.38 1.1e-05 ***
shock 3 219 73.1 31.90 3.6e-09 ***
subj 4 362 90.5 39.48 4.0e-11 ***
noise:shock 3 13 4.2 1.84 0.16
Residuals 28 64 2.3
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
y <- skin$resp
y... <- predict(lm(resp ~ 1,data = skin))
yi.. <- predict(lm(resp ~ noise,data=skin))
y.j. <- predict(lm(resp ~ shock,data=skin))
y..k <- predict(lm(resp ~ subj,data=skin))
yij. <- predict(lm(resp ~ noise + shock + noise:shock,data=skin))
SSA <- sum((yi.. - y...)^2)
SSB <- sum((y.j. - y...)^2)
SSC <- sum((y..k - y...)^2)
SSAB <- sum((yij. - (yi.. + y.j. - y...))^2)
SST <- sum((y - y...)^2)
SSE <- SST - (SSA + SSB + SSC + SSAB)
a <- 2
b <- 4
c <- 5
MSA <- SSA/(a-1)
MSB <- SSB/(b-1)
MSC <- SSC/(c-1)
MSAB <- SSAB/((a-1)*(b-1))
MSE <- SSE/((a*b-1)*(c-1))
FA <- MSA / MSE
FB <- MSB / MSE
FAB <- MSAB / MSE
FC <- MSC / MSE
pA <- 1 - pf(FA,a-1,(a*b-1)*(c-1))
pB <- 1 - pf(FB,b-1,(a*b-1)*(c-1))
pAB <- 1 - pf(FAB,(a-1)*(b-1),(a*b-1)*(c-1))
pC <- 1 - pf(FC,c-1,(a*b-1)*(c-1))| Source | Df | Expected mean square |
|---|---|---|
| A | \(a-1\) | \(bc \theta_A^2 + \sigma_\varepsilon^2\) |
| B | \(b-1\) | \(ac \theta_B^2 + \sigma_\varepsilon^2\) |
| AB | \((a-1)(b-1)\) | \(c \theta_{AB}^2 + \sigma_\varepsilon^2\) |
| C | \(c-1\) | \(ab\sigma_C^2 + \sigma_\varepsilon^2\) |
| Error | \((ab-1)(c-1)\) | \(\sigma_\varepsilon^2\) |
In the above
\[ \dot \sigma_C^2 = \frac{\operatorname{MS}_{\operatorname{C}}- \operatorname{MS}_{\operatorname{Error}}}{ab} \quad \text{ and }\quad \displaystyle \dot \sigma_{\varepsilon}^2 = \operatorname{MS}_{\operatorname{Error}}. \]
Obtain REML estimators of \(\sigma_C^2\) and \(\sigma_{\varepsilon}^2\) on the skin response data.
Linear mixed model fit by REML ['lmerModLmerTest']
Formula: resp ~ noise + shock + noise:shock + (1 | subj)
Data: skin
REML criterion at convergence: 145
Random effects:
Groups Name Std.Dev.
subj (Intercept) 3.32
Residual 1.51
Number of obs: 40, groups: subj, 5
Fixed Effects:
(Intercept) noise80 shock0.5 shock0.75
4.8 2.0 3.2 5.4
shock1 noise80:shock0.5 noise80:shock0.75 noise80:shock1
3.2 -0.6 2.4 0.4
Obtain MoM estimates for \(\sigma_C^2\) and \(\sigma_{\varepsilon}^2\) on the skin response data.
We have \(\dot \sigma_C = 3.320\) and \(\dot \sigma_{\varepsilon}= 1.514\).