3.4 Estimation
Under the side condition \(\sum n_i\tau _i = 0\), the least squares estimates are \[ \hat \mu = \bar y_{\cdot \cdot } , \qquad \hat \tau _i = \bar y_{i\cdot } - \bar y_{\cdot \cdot } , \qquad \hat \sigma ^2 = MSE , \] and the treatment means are estimated by \(\hat \mu _i = \hat \mu + \hat \tau _i = \bar y_{i\cdot }\), whatever side condition is used. Because \(\bar y_{i\cdot } \sim N(\mu _i, \sigma ^2/n_i)\) independently of \(MSE\), and \((N - t)MSE/\sigma ^2 \sim \chi ^2_{N-t}\), \[ \frac {\bar y_{i\cdot } - \mu _i}{\sqrt {MSE/n_i}} \sim t_{N-t} , \] and a \(100(1 - \alpha )\%\) confidence interval for \(\mu _i\) is \[ \bar y_{i\cdot } \pm t_{\alpha /2,\,N-t}\sqrt {\frac {MSE}{n_i}} . \] Similarly, for the difference of two treatment means, \[ \bar y_{i\cdot } - \bar y_{k\cdot } \pm t_{\alpha /2,\,N-t}\sqrt {MSE\Bigl (\frac {1}{n_i} + \frac {1}{n_k}\Bigr )} . \] These intervals use the pooled estimate \(MSE\) from all \(t\) treatments, with \(N - t\) degrees of freedom, rather than the data on treatments \(i\) and \(k\) alone; this is the gain from the common-variance assumption.
Example 3.4.1. For the coating data of Example 3.0.1, the model is \[ y_{ij} = \mu + \tau _i + \varepsilon _{ij} , \qquad i = 1, \dots , 5 , \quad j = 1, \dots , 4 , \] where \(y_{ij}\) is the conductivity of the \(j\)th tube with coating \(i\), \(\mu \) is the overall mean conductivity, \(\tau _i\) is the effect of coating type \(i\), with \(\sum \tau _i = 0\), and \(\varepsilon _{ij}\) is the random error. The errors are assumed independent \(N(0, \sigma ^2)\), with the same variance for every coating; the coating effects are fixed, since these five coatings are the only ones of interest.
With \(N = 20\), \(t = 5\) and \(y_{\cdot \cdot } = 1437\), the correction factor is \(1437^2/20 = 103248.45\), and \begin {align*} SST &= 104471 - 103248.45 = 1222.55 , \\ SS_{\mathrm {Treatments}} &= \frac {331^2 + 308^2 + 287^2 + 260^2 + 251^2}{4} - 103248.45 \\ &= 1100.30 , \\ SSE &= 1222.55 - 1100.30 = 122.25 . \end {align*}
| source | SS | df | MS | \(F_0\) |
| coating types | \(1100.30\) | \(4\) | \(275.075\) | \(33.75\) |
| error | \(122.25\) | \(15\) | \(8.150\) | |
| total | \(1222.55\) | \(19\) |
Since \(F_0 = 33.75 > F_{0.05,4,15} = 3.06\) (the \(p\)-value is \(2.5 \times 10^{-7}\)), we reject \(H_0\): the coating type affects conductivity.
The estimates are \(\hat \mu = 71.85\), \(\hat \sigma ^2 = 8.15\) and \[ \hat \tau _1 = 10.90 , \quad \hat \tau _2 = 5.15 , \quad \hat \tau _3 = -0.10 , \quad \hat \tau _4 = -6.85 , \quad \hat \tau _5 = -9.10 . \] With \(t_{0.025,15} = 2.131\) and \(\sqrt {MSE/4} = 1.427\), the \(95\%\) interval for the mean conductivity with coating \(1\) is \(82.75 \pm 3.04\), that is \((79.71,\ 85.79)\), and for the difference between coatings \(1\) and \(2\) it is \[ 5.75 \pm 2.131\sqrt {8.15\Bigl (\frac 14 + \frac 14\Bigr )} = 5.75 \pm 4.30 , \] that is, \((1.45,\ 10.05)\).
Remark. The intervals above are for comparisons chosen in advance. If the pair to compare is chosen after seeing the data, for instance the largest and smallest means, or if many comparisons are made, the confidence level of each interval no longer describes the whole collection. Methods that control the error rate over a family of comparisons are the subject of the section on treatment contrasts.
Questions on this section
Stuck on something here? Ask below and it stays attached to this topic.