Question

In: Statistics and Probability

Drew knows that the urine from his dog Rosie is bad for vegetation: there are brown...

  1. Drew knows that the urine from his dog Rosie is bad for vegetation: there are brown patches all over the lawn. He hypothesised that urine not only reduces grass cover, it also affects the abundance of soil-dwelling invertebrates in the patch versus the surrounding vegetation. To investigate this hypothesis, Drew randomly chose 6 brown patches in the lawn. For each patch, he collected three soil cores: a core in the patch centre, a core just outside the patch, and a core 15cm from the patch edge. Invertebrates were counted within each core.

(iv) What kind of test is appropriate for these data?
(v) How many groups are there?
(vi) How many replicates are in each group?
(vii) What are the test statistic and degrees of freedom for the test?

Solutions

Expert Solution

Drew randomly chooses 6 brown patches in the lawn. For each patch, he collected three soil cores: a core in the patch centre, a core just outside the patch, and a core 15cm from the patch edge.

Part iv) Here, testing with RBD is appropriate. Inorder to tackle blocking of soil RBD is done. The blocking is done in such a way that intrablock variablility is removed and interblock variability is significantly present. This blocking is done by virtue of local control. Now treatments are allocted to plot in a block randomly.

Part v) Here, we consider three groups which are three soil cores: a core in the patch centre, a core just outside the patch, and a core 15cm from the patch edge.

Part vi) Here, there are 6 replicates in each group which are the 6 browm patches chosen in the lawn.

Part vii) The test statistic is F statistic that is

where, MST = mean square due to treatment that is mean square due to 6 brown patches in the lawn.

MSE = mean square due to error in the model.

Now, here t = no of treatments = 6 and r = no of blocks = 3.

So the degrees of fredom of total sum of squares = rt - 1 =17

degrees of fredom of sum of squares due to treatment = t - 1 =5

degrees of fredom of sum of squares due to block = r - 1 =2

degrees of fredom of sum of squares due to error= (r-1)(t-1)= 10

degrees og fredom of F statistic is F5,10.


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