Question

An article presents a study that investigated the effect of varying the type of fertilizer on the height of certain Mediterranean woody tree species. In one experiment, three samples, each consisting of ten trees, were grown with three different fertilizers. One, the control group, was grown with a standard fertilizer. Another was grown with a fertilizer containing only half the nutrients of the standard fertilizer. The third was grown with the standard fertilizer to which a commercial slow-release fertilizer had been added. Following are the heights of the trees after one year.

Height Fertilizer Control 17.9 12.2 14.9 13.8 26.1 15.4 20.3 16.9 20.8 14.8 Deficient 7.5 74 13.8 116 115 17 132 129 176 9.5 Slow-releasel 19.8 20.3 16.1 179 124 12.5 174 19.9 27.3 14.4
a) If you want to construct a hypothesis to test for heights of the trees, state your null and alternative hypothesis.
b) Is the experiment a factorial balanced CR design?
c) What is the degree of freedom of treatment sum of square and error sum of square? If SSTrt and SSE is given: SSTrt=192.374 SSE=439.389, what is the value of your test statistics?
d) Compute the SST?
e) We get a p-value of 0.007 and we have significance level 0.05 for this case, state your conclusion of the test.

Answer 1

Step-by-step explanation:

a.

The experiment under study from the article involves a single factor fertilizer type, at three levels.

Hence, this is a completely randomized design with p=3 treatments.

Assuming :

`\mu_ 1 ; \mu_2 \ and \ \mu_3` are mean heights for three types of fertilizer. Therefore, the null and alternative hypotheses are:

Null hypothesis:

`H_o : \mu_ 1 = \mu_2 \ = \ \mu_3` ( i.e all type of population mean heights for three different fertilizers f are equal )

Alternative hypothesis:

`H_1 :\mu_ 1 \\eq \mu_2 \ \\eq \ \mu_3` ( at least one of the fertilizer is not equal )

b.

This is a completely randomized design as p=3 treatment means are compared, where treatments are randomly assigned to the experimental units.

c.

`SSTr(df)=p-1 \\ \\ =3-1 \\ \\=2, \\ \\ df(SSE)=n-p \\ \\ =30-3 \\ \\ =27 \\ \\ MSTr=SSTr/df(SSTr) \\ \\=192.374/2 \\ \\ =96.187; MSE=SSE/df(SSE) \\ \\ =439.389/27 \\ \\ =16.2737 \\ \\ F=MSTr/MSE \\ \\ =96.187/16.2737 \\ \\ =5.91`

d.

`SST=SSTr+SSE \\ \\ =192.374+439.389\\ \\ =631.763`

e.

Given that the p-value is 0.007 and F = 5.91

Also;

p-value 0.007 is less than the level of significance 0.05.

Thus ; we reject the null hypothesis and conclude that there is difference in mean of all types of the fertilizers.