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Statistical model for Complete Randomized design

$y_{ij} = \mu + \tau_i + \epsilon_{ij}$

where, $i$ denotes treatment and $j$ denotes observation.

$i=1,2,...,k\quad and \quad j=1,2,..., n_i$

$y_{ij}$ be a random variable that represents the response obtained on the $jth$ observation of the $ith$ treatment.

$\mu$ is the overall mean of the response $y_{ij}$

$\tau_i$ is the effect on the response of $ith$ treatment.

$\mu_i = \mu + \tau_i$

here $\mu_i$ denotes the true response of the $ith$ treatment.

and $\epsilon_{ij}$ is the random error term represent the sources of nuisance variation that is, variation due to factors other than treatments.

the assumption is $\epsilon_{ij}\sim^{iid} N(0,\sigma^2)$

By Least Square Estimation Procedure we estimate $\mu\quad and\quad\tau_i$.

  • Why is it also important to estimate $\sigma^2$? If we do not estimate it , what will be the effect?

Any help including reference will be appreciated.

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2 Answers 2

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The usual unbiased estimate of $\sigma^2$ is the error mean square: $$MSE = \sum_i\sum_j (y_{ij}-\bar y_{i*})^2/(n_T-k), \quad n_T=\sum_i n_i$$ If you don't estimate $\sigma^2$, i.e. if you don't compute $MSE$, then:

  • no analysis of variance table is available, since in such tables there are:

    • the treatment sum of squares: $$SSTR=\sum_i n_i(\bar y_{i*}-\bar y_{**})^2$$
    • the explained variance, the treatment mean square: $$MSTR=SSTR/(k-1)$$
    • the error (within treatment) variability, the error sum of squares: $$SSE=\sum_i\sum_j(y_{ij}-\bar{y}_{i*})^2$$
    • the unexplained (or residual) variance, the error mean square: $$MSE=SSE/(n_T-k)$$
    • the overall variability, the total sum of squares: $$SSTO=SSTR+SSE$$
  • no $F$-test is possible, because an $F$ statistic is defined as $$F^*=\frac{MSTR}{MSE}$$
  • no standard error can be computed, therefore no $t$ statistic, no p-value and no confidence interval, since the standard error is defines as $$s=\sqrt{\frac{MSE}{n_i}}$$

In brief, you don't know if you can trust in your results.

If you are looking for a reference, I'd suggest Kutner et al., Applied Linear Statistical Models, Chapter 16.

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  • $\begingroup$ If i don't estimate MSE, can't i compute SSTR in ANOVA? $\endgroup$
    – time
    Commented May 23, 2014 at 11:23
  • $\begingroup$ You can, but SSTR is the variability explained by your model ($SSTR/(k-1)$ is the explained variance) and if you do not compare it with the residual variability you can't say if it is "high" (good model) or "low" (poor model). $\endgroup$
    – Sergio
    Commented May 23, 2014 at 11:48
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It is important for making inferences about the how large the treatment effect is, and whether it exists at all. There are usually two hurdles to concluding that a treatment has a causal effect from data, identification and estimation.

Identification: When there is an association between treatment and outcome, does that mean administering the treatment will affect the outcome? With the assumptions you make here (randomized trial) that's the case.

Estimation: Is there actually an association or does it merely look like there is but it's just noise (here, noise would be variation due to factors other than the treatment). This is why you need to estimate $\sigma^2$.

Your least squares estimation procedure will give you point estimates but they might reflect just noise. To get a sense of how different they could have been had we done the experiment again, we need to estimate their standard deviation. This involves estimating $\sigma^2$.

If we don't estimate it and just look at point estimates, we may we fooled by randomness and conclude the treatment has an effect when actually it doesn't. (or conclude it doesn't have an effect when it does have a small one which was drowned by the noise)

For a reference, any introduction to statistical inference covers this topic. See here for how the formulas of the standard deviation of the point estimates include estimates of $\sigma^2$.

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