Instruction 19-6
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} { | l | l | l | l | } \hline Fi e l d s & \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:

Friedman Test: Yield versus Varieties, Fields
Friedman test for Yield by Varieties blocked by Fields

\mathrm { S } = 10.00 \mathrm { DF } = 2 \mathrm { P } = 0.007

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

Grand median

= 15.900

-Referring to Instruction 19-6,the null hypothesis for the Friedman rank test is
A)H₀: 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: ?_{Field 1} = ?_{Field 2}= ?_{Field 3}= ?_{Field 4}= ?_{Field 5.}
D)H₀: ?_Smith = ?_Walsh= ?_Trevor.

If the Sample Sizes in Each Group Are Larger Than