Skip to main content
Cross Validated

Return to Answer

added 449 characters in body
Source Link
Aniko
Aniko
  • 11.1k
  • 34
  • 35

Start edit

The variance term in this formula gives you the square of the standard error for your plot. The point to show would be the mean of the within-day means. If you don't want to assume that the within-day variability is constant, just replace $\sigma$ with $\sigma_i$ - the within-day standard deviation on day 1, so the summation would be $\sum_iSE_i^2$, where $SE_i$ is the standard error of the within-day mean.

End edit

If for a moment you assume $n_i=n$, then the variance is $\frac{\tau^2}{k} + \frac{\sigma^2}{n}$. In your application the key observation is that $n >> k$, so the first term will dominate. In essence, you can ignore the within-day replicates, and just use them as an expensive way to obtain one (day-specific) observation.

If for a moment you assume $n_i=n$, then the variance is $\frac{\tau^2}{k} + \frac{\sigma^2}{n}$. In your application the key observation is that $n >> k$, so the first term will dominate. In essence, you can ignore the within-day replicates, and just use them as an expensive way to obtain one (day-specific) observation.

Start edit

The variance term in this formula gives you the square of the standard error for your plot. The point to show would be the mean of the within-day means. If you don't want to assume that the within-day variability is constant, just replace $\sigma$ with $\sigma_i$ - the within-day standard deviation on day 1, so the summation would be $\sum_iSE_i^2$, where $SE_i$ is the standard error of the within-day mean.

End edit

If for a moment you assume $n_i=n$, then the variance is $\frac{\tau^2}{k} + \frac{\sigma^2}{n}$. In your application the key observation is that $n >> k$, so the first term will dominate. In essence, you can ignore the within-day replicates, and just use them as an expensive way to obtain one (day-specific) observation.

Source Link
Aniko
Aniko
  • 11.1k
  • 34
  • 35

For simplicity, first consider only one group. Your data can be modeled as $$Y_{ij} = \mu + \delta_i + \epsilon_{ij},$$ where $\delta_i \sim N(0, \tau^2), i=1,\ldots,k$ is the random day effect, and $\epsilon_{ij}\sim N(0, \sigma^2), j=1,\ldots, n_i$ is the random within-day replicate effect. Then you can obtain an unbiased estimate of the group average $$\hat\mu = \frac 1k \sum_{j=1}^j \bar{Y}_{i.} \sim N(\mu, \frac{\tau^2}{k} + \frac 1k \sum_{i=1}^k \frac{\sigma^2}{n_i}).$$ It is fairly simple to obtain estimates of the variance term with $\tau^2$ being the variance of the day means, and $\sigma^2$ the pooled within-day variance.

If for a moment you assume $n_i=n$, then the variance is $\frac{\tau^2}{k} + \frac{\sigma^2}{n}$. In your application the key observation is that $n >> k$, so the first term will dominate. In essence, you can ignore the within-day replicates, and just use them as an expensive way to obtain one (day-specific) observation.

I have run a "correct" mixed ANOVA analysis and a simple ANOVA on group means, and they give essentially the same result:

library(lme4)
m1 <- lmer(Meas ~ Treatment*Temp + (1|Rep), data=dat)
m1

Output:

Linear mixed model fit by REML 
Formula: Meas ~ Treatment * Temp + (1 | Rep) 
   Data: dat 
  AIC  BIC logLik deviance REMLdev
 3351 3376  -1669     3343    3339
Random effects:
 Groups   Name        Variance Std.Dev.
 Rep      (Intercept) 15.414   3.9260  
 Residual             42.000   6.4808  
Number of obs: 504, groups: Rep, 12

Fixed effects:
                 Estimate Std. Error t value
(Intercept)       16.1579     5.2369   3.085
TreatmentT2       -9.3838     7.3903  -1.270
Temp              -0.4723     0.3315  -1.425
TreatmentT2:Temp   1.0377     0.4683   2.216

And

m2 <- lm(mean ~ Treatment*Temp, data=dat2)
summary(m2)

with output

Coefficients:
                 Estimate Std. Error t value Pr(>|t|)  
(Intercept)       16.1037     5.2451   3.070   0.0153 *
TreatmentT2       -9.2846     7.4177  -1.252   0.2460  
Temp              -0.4694     0.3317  -1.415   0.1948  
TreatmentT2:Temp   1.0302     0.4691   2.196   0.0594 .
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

Residual standard error: 4.063 on 8 degrees of freedom
Multiple R-squared: 0.5955,     Adjusted R-squared: 0.4438 
F-statistic: 3.926 on 3 and 8 DF,  p-value: 0.05412