TABLE 12-19
an Agronomist Wants to Compare the Crop

TABLE 12-19
An agronomist wants to compare the crop yield of 3 varieties of chickpea seeds. She plants all 3 varieties of the seeds on each of 5 different patches of fields. She then measures the crop yield in bushels per acre. Treating this as a randomized block design, the results are presented in the table that follows.
$$\begin{array} { l l l l } \underline{\text { Fields }} & \underline{\text { Smith} } &\underline{ \text { Walsh }} & \underline{\text { Trevor} } \\1 & 11.1 & 19.0 & 14.6 \\2 & 13.5 & 18.0 & 15.7 \\3 & 15.3 & 19.8 & 16.8 \\4 & 14.6 & 19.6 & 16.7 \\5 & 9.8 & 16.6 & 15.2\end{array}$$ Below is the Minitab output of the Friedman rank test:
Friedman Test: Yield versus Varieties, Fields
$\text { Friedman test for Yield by Varieties blocked by Fields }$

$\mathrm{S}=10.00 \quad \mathrm{DF}=2 \quad \mathrm{P}=0.007$
$\begin{array}{lrccc} & & \text { Est } & \text { Sum of } & \\\text { Varieties } & \mathrm{N} & \text { Median } & \text { Ranks } \\\text { Smith } & 5 & 13.500 & 5.0 \\\text { Trevor } & 5 & 15.667 & 10.0 \\\text { Walsh } & 5 & 18.533 & 15.0 \\\\{\text { Grand median }=} & 15.900\end{array}$
-Referring to Table 12-19, what is the null hypothesis for the Friedman rank test?

A) H₀: µField1 = µField2 = µField3 = µField4 = µField5
B) H₀: MSmith = MWalsh = MTrevor
C) H₀: µSmith = µWalsh = µTrevor
D) H₀: MField1 = MField2 = MField3 = MField4 = MField5