Question 9.3: For the data in Table 9.1, compute MSTr, MSE, and F . Find t...
For the data in Table 9.1, compute MSTr, MSE, and F . Find the P-value for testing the null hypothesis that all the means are equal. What do you conclude?
| TABLE 9.1 Brine II hardness of welds using four different fluxes | |||||||
| Flux | Sample Values | Sample Mean | Sample Standard Deviation |
||||
| 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 |
From Example 9.2, SSTr = 743.4 and SSE= 1023.6. We have I = 4 samples and N = 20 observations in all the samples taken together. Using Equation (9.13),
\operatorname{MSTr}=\frac{SSTr}{I-1}\quad MSE=\frac{SSE}{N-I} (9.13)
\operatorname{MSTr}=\frac{743.4}{4-1}=247.8\quad MSE=\frac{1023.6}{20-4}=63.975
The value of the test statistic F is therefore
F=\frac{247.8}{63.975}=3.8734
To find the P-value, we consult the F table (Table A.6). The degrees of freedom are 4 – 1 = 3 for the numerator and 20- 4 = 16 for the denominator. Under H_0 , F has an F3.16 distribution. Looking at the F table under 3 and 16 degrees of freedom, we find that the upper 5% point is 3.24 and the upper 1% point is 5.29. Therefore the P-value is between 0.01 and 0.05 (see Figure 9.3; a computer software package gives a value of 0.029 accurate to two significant digits). It is reasonable to conclude that the population means are not all equal and thus that flux composition does affect hardness.
