hw9sol - Stat 511 Homework 9 Solution Spring 2011 1. Last...

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Unformatted text preview: Stat 511 Homework 9 Solution Spring 2011 1. Last homework, I described the New Zealand pasture fertilization study. The data are in ryegrass.txt. This week, we analyze the data. (a) Last week, you derived the E MS for variety. Some of the other E MS are: Effect E MS Block*Variety 2 s + 2 2 m Fert. 2 s + Q () Fert. Variety 2 s + Q () Residual 2 s where Q () indicates a quadratic form involving only fixed effects. Construct the ANOVA table and test the main effects of variety, fertilization, and the interaction. Table 1: ANOVA table Source df Sum Sq Mean Sq F value p-value variety 3 11221 3740 4.1532 0.042 block 3 13879 4626 variety*block 9 8105 901 fertilizer 1 138864 138864 185.56 < . 0001 variety*fertilizer 3 13758 4586 6.13 0.009 residual 12 8980 748 cor.total 31 If you use R, it really helps that these data are balanced. When the design is balanced, the SS can be obtained by fitting a sequence of fixed effects models (or using anova() after lm() to get sequential SS on the fixed effects model block + variety + block:variety + fertilizer + variety:fertilizer). Then construct the F tests yourself. If you use SAS, both the sequential and partial ANOVA tables are part of the proc mixed output. Variety*block is in the random statement. Block is either in the random or model statement. The other terms are in the model statement. (b) Pastures will be seeded with a single variety of ryegrass. Hence, land managers are especially interested in the simple effects of fertilization for each variety (i.e. the difference between heavy and average manure separately for each variety). Estimate the increase in dry matter production caused by heavy manuring Report each simple effect, its s.e....
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This note was uploaded on 02/11/2012 for the course STAT 511 taught by Professor Staff during the Spring '08 term at Iowa State.

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hw9sol - Stat 511 Homework 9 Solution Spring 2011 1. Last...

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