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

Question 12

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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.
 Fields  Smith  Walsh  Trevor 111.119.014.6213.518.015.7315.319.816.8414.619.616.759.816.615.2\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=10.00DF=2P=0.007S = 10.00 D F = 2 P = 0.007

 Varieties N EstMedian  Sum of Ranks  Smith 513.5005.0 Trevor 515.66710.0 Walsh 518.53315.0\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}

 Grand median =15.900\text { Grand median }=15.900

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


A) MSmith = MWalsh = MTrevor.
B) H0: M Field 1 = M Field 2 = MField 3 = MField 4 = MField 5.
C) H0: μ\mu Field 1 = μ\mu Field 2 = μ\mu Field 3 = μ\mu Field 4 = μ\mu Field 5.
D) H0: μ\mu Smith = μ\mu Walsh = μ\mu Trevor.

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