Instruction 19-4
an Agronomist Wants to Compare the Crop Yield

Instruction 19-4
An agronomist wants to compare the crop yield of three varieties of chickpea seeds. She plants all three varieties of the seeds on each of five different patches of fields. She then measures the crop yield in bushels per acre. Treating this as a randomised block design, the results are presented in the table that follows.
$$\begin{array} { | c | c | c | c | } \hline \text { Fields } & \text { Smith } & \text { Walsh } & \text { Trevor } \\\hline 1 & 11.1 & 19.0 & 14.6 \\\hline 2 & 13.5 & 18.0 & 15.7 \\\hline 3 & 15.3 & 19.8 & 16.8 \\\hline 4 & 14.6 & 19.6 & 16.7 \\\hline 5 & 9.8 & 16.6 & 15.2 \\\hline\end{array}$$ Below is the Minitab output of the Friedman rank test:
Friectman Test Yedd versus Varieties Felds
Friedmantest for Yield by Varieties blocked by Fields $S = 10.00 D F = 2 P = 0.007$

$\begin{array} { l l l l } \text { Varieties } & N & \text { EstMedian } & \text { Sum of Ranks } \\ & & & \\ \text { Smith } & 5 & 13.500 & 5.0 \\ \text { Trevor } & 5 & 15.667 & 10.0 \\ \text { Walsh } & 5 & 18.533 & 15.0 \end{array}$

$\text { Grand median }=15.900$

-Referring to Instruction 19-4,the null hypothesis for the Friedman rank test is

A) M_Smith = M_Walsh = M_Trevor.
B) H₀: M _{Field 1} = M _{Field 2} = M_{Field 3} = M_{Field 4} = M_{Field 5.}
C) H₀: $\mu$ _{Field 1} = $\mu$ _{Field 2} = $\mu$ _{Field 3} = $\mu$ _{Field 4} = $\mu$ _{Field 5.}
D) H₀: $\mu$ _Smith = $\mu$ _Walsh = $\mu$ _Trevor.