Solved

Instruction 19-7
an Agronomist Wants to Compare the Crop Yield

Question 2

Short Answer

Instruction 19-7
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.
1elds Smith  Walsh  Trevor 111.119.014.6213.518.015.7315.319.816.8414.619.616.759.816.615.2\begin{array} { | l | l | l | l | } \hline 1 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
S=10.00DF=2P=0.007\mathrm { S } = 10.00 \mathrm { DF } = 2 \mathrm { P } = 0.007
 Est  Sum of  Varieties N Median  Ranks  Smith 513.5005.0 Trevor 515.66710.0 Walsh 518.53315.0\begin{array} { l l l l } & & \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 \end{array}
Grand median = 15.900
-Referring to Instruction 19-7,what are the degrees of freedom of the Friedman rank test for the difference in the medians at a level of significance of 0.01?

Correct Answer:

verifed

Verified

Related Questions