Statistics Homework Solutions 219

Statistics Homework Solutions 219 - 20 021417 5 275. The...

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SECTION 9.4 219 (b) Since the interaction terms are not equal to 0, ( F 4 18 5 2567, P 0 006), we cannot interpret the main effects. Therefore we compute the cell means. These are DCM (ml) PVAL 50 40 30 0.5 97.8 92.7 74.2 1.0 93.5 80.8 75.4 2.0 94.2 88.6 78.8 We conclude that a DCM level of 50 ml produces greater encapsulation efficiency than either of the other levels. If DCM = 50, the PVAL concentration does not have much effect. Note that for DCM = 50, encapsulation efficiency is maximized at the lowest PVAL concentration, but for DCM = 30 it is maximized at the highest PVAL concentration. This is the source of the significant interaction. Section 9.4 1. (a) Liming is the blocking factor, soil is the treatment factor. (b) Source DF SS MS F P Soil 3 1.178 0.39267 18.335 0.000 Block 4 5.047 1.2617 58.914 0.000 Error 12 0.257 0.021417 Total 19 6.482 (c) Yes, F 3 12 18 335, P 0. (d) q 4 12 05 4 20, MSAB 0 021417, and J 5. The 5% critical value is therefore 4
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Unformatted text preview: 20 021417 5 275. The sample means are X A 6 32, X B 6 02, X C 6 28, X D 6 70. We can therefore conclude that D differs from A, B, and C, and that A differs from B. 3. (a) Let I be the number of levels for treatment. let J be the number of levels for blocks, and let K be the number of replications. Then I 3, J 4, and K 3. The number of degrees of freedom for treatments is I 1 2. The number of degrees of freedom for blocks is J 1 3. The number of degrees of freedom for interaction is I 1 J 1 6. The number of degrees of freedom for error is IJ K 1 24. The mean squares are found by dividing the sums of squares by their respective degrees of freedom. The F statistics are found by dividing each mean square by the mean square for error. The number of degrees...
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