Calculating Simultaneous Confidence Intervals

Question

$$\begin{array}{|c|c|}\hline \text{Abuse} & \text{Neglect} & \text{Nonabuse Control} \\ \hline \overline{y}_1 = 81.06 & \overline{y}_2 = 78.56 & \overline{y}_3 = 87.81\\ \hline s_1 = 17.05 & s_2 = 15.43 & s_3 = 14.36\\ \hline n_1 = 32 & n_2 = 16 & n_3 = 16\\\hline\end{array}$$

Calculate simultaneous confidence intervals for the differences.

So I think this is the formula I should be using:

$$\overline{y}_{j}-\overline{y}_{j'}\pm t_{\alpha/2m}\cdot s\sqrt{\frac{1}{n_j}+\frac{1}{n_{j'}}}$$

I think the degrees of freedom would be $n-k=64-3=61$, $m =3$ since there are are $3\choose{2}$$=3$ ways to choose $2$ out of the $3$ groups.

However, how could I calculate $s$? I am reading that $s=\sqrt{\text{MSE}}$ but it looks like I need individual data values to calculate the MSE. Should I use

$$s_{\text{pooled}}=\sqrt{\frac{(n_j-1)s_j^2+(n_{j'}-1)s_{j'}^2}{n_j+n_{j'}-2}}$$

If this is all correct, here is one of the three calculations I would need to make, just to see if I'm doing this right:

$$\begin{align*} \overline{y}_{1}-\overline{y}_{2}\pm t_{\alpha/2m}\cdot s\sqrt{\frac{1}{n_1}+\frac{1}{n_{2}}} &= 81.06-78.56\pm t_{0.05/6}\cdot s_{\text{pooled}}\sqrt{\frac{1}{32}+\frac{1}{16}}\\\\ &= 2.5 \pm 2.73\cdot s_{\text{pooled}}\sqrt{\frac{1}{32}+\frac{1}{16}} \\\\ &= 2.5 \pm 13.26 \\\\ &= (-10.76, 15.76) \\\\ \end{align*}$$

Attempt 2

$$s^2=\frac{(32-1)(17.05^2)+(16-1)(15.43^2)+(16-1)(14.36^2)}{32+16+16-3}\Rightarrow s=16.03$$

$$t_{61, .05/3}=2.46$$

All together,

$$\begin{align*} \overline{y}_{1}-\overline{y}_{2}\pm t_{61,.05/3}\cdot s\sqrt{\frac{1}{n_1}+\frac{1}{n_{2}}} &= 81.06-78.56\pm t_{61,0.05/3}\cdot s\sqrt{\frac{1}{32}+\frac{1}{16}}\\\\ &= 2.5 \pm 2.46\cdot s\sqrt{\frac{1}{32}+\frac{1}{16}} \\\\ &= 2.5 \pm 12.074 \\\\ &= (-9.57, 14.57) \\\\ \end{align*}$$

Answer 1

Comment continued: With a bit of fussing I made fake normal data to mimic your sample sizes, means, and standard deviations.

Then did the Bonferroni-adjusted pairwise.t.test procedure in R. The output is the three P-values.

It will be interesting to know whether the R procedure is (as I suspect) the same as the one you are doing, and whether your P-values match those from the R procedure.

mean(ab); sd(ab)
[1] 89.06
[1] 17.05
mean(ne); sd(ne)
[1] 78.56
[1] 15.43
mean(nc); sd(nc)
[1] 87.81
[1] 14.36

y = c(ab, ne, nc)
gp = as.factor(c(rep(1,32), rep(2,16), rep(3,16)))

pairwise.t.test(y, gp, p.adj="bonferroni")

        Pairwise comparisons using t tests with pooled SD 

data:  y and gp 

  1    2   
  2 0.11 -   
  3 1.00 0.32

P value adjustment method: bonferroni 

Thus there are no significant differences among the group means according to the Bonferroni method. This raises the question whether the main F-test for the one-way ANOVA rejected $H_0: \mu_{ab} = \mu_{ne} = \mu_{nc}.$ As shown below it does not, at the 5% level. According to the usual methods of analysis, that failure to reject would mean that you should not be looking at multiple-comparison methods.

summary(aov(y ~ gp))
            Df Sum Sq Mean Sq F value Pr(>F)  
gp           2   1237   618.4   2.406 0.0987 .
Residuals   61  15676   257.0                 
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Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1