Question 9.6: The following output (from MTNITAB) presents the ANOVA table...
The following output (from MTNITAB) presents the ANOVA table for the weld data in Table 9.1 (in Section 9.1). Which pairs of fluxes, if any, can be concluded, at the 5% level, to differ in their effect on hardness?
One-way ANOVA : A. B. C. D
\begin{array}{lrrrrr}\text{Source}& \text{OF}& \text{SS}& \text{MS}& F & P \\\text{Factor}& 3 & 743.40 & 247.800 & 3.87 & 0.029 \\\text{Error}& 16 & 1023.60 & 63.975 & & \\\text{Total}& 19 & 1767.00 & & &\end{array}s = 7. 998 R-Sq = 42 . 07% R-Sq(adj) 31 . 21%
| TABLE 9.1 Brine II hardness of welds using four different fluxes | |||||||
| Flux | Sample Values | Sample Mean | Sample Standard Deviation |
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| A | 250 | 264 | 256 | 260 | 239 | 253.8 | 9.7570 |
| B | 263 | 254 | 267 | 265 | 267 | 263.2 | 5.4037 |
| C | 257 | 279 | 269 | 273 | 277 | 271.0 | 8.7178 |
| D | 253 | 258 | 262 | 264 | 273 | 262.0 | 7.4498 |
There are J = 4 levels, with J = 5 observations at each level, for a total of N = 20 observations in all. To test at level α = 0.05, we consult the Studcntized range table (Table A.7) to find q_{ 4, 16 , 05} = 4.05.
The value of MSE is 63.975. Therefore q_{l, N-I, \alpha}\sqrt{MSE / J}=4.05 \sqrt{63.975 / 5}= 14.49 . The four sample means (from Table 9.1) arc as follows:
\begin{array}{l|cccc}\text{Flux}& \text{A}& \text{B}& \text{C}& \text{D}\\\hline \text{Mean hardness}& 253.8 & 263.2 & 271.0 & 262.0\end{array}
There is only one pair of sample means, 271.0 and 253.8, whose difference is greater than 14.49. We therefore conclude that welds produced with flux A have a different mean hardness than welds produced with flux C. None of the other differences are significant at the 5% level.