Question
1 1 111.5 1 2 97.7 1 3 126.1 2 1 94.4 2 2 70.5 2...
| 1 | 1 | 111.5 |
| 1 | 2 | 97.7 |
| 1 | 3 | 126.1 |
| 2 | 1 | 94.4 |
| 2 | 2 | 70.5 |
| 2 | 3 | 93.1 |
| 3 | 1 | 73.9 |
| 3 | 2 | 56.2 |
| 3 | 3 | 84.6 |
| 26. What is the estimate of the average (pooled) variance for this experiment (ie. error mean square)? | Wrong | |
| 27. What are the error degrees of freedom for the pooled variance? | Wrong | |
| 28. What is the least significant difference (α = 0.05) for comparing shading treatments in this experiment? |
Homework Answers
Answer #1
by using given data
2 - way ANOVA
Q. 26: The estimate of the
average (pooled) variance ;
MSE / df = 17.63777778 / 4 = > 4.409444445 ~ 4.41
Q. 27 : The degrees of freedom for the pooled variance = 4
Q.28: The least significant difference (a = 0.05) for comparing shading treatments in this experiment ;
LSD =
2.78*
= 6.16
NOTE: if you have any doubts please free to ask in comment section
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