1. Checking on Neem trees that were planted many years ago along a country road, a forestry official obtained the following arrangement of healthy H and diseased D trees : H H H H D D D H H H H H H H D D H H D D D D. Test at the 0'05 level of significance whether this arrangement may be regarded as random. (Given the tabulated value of test statistics 1.96)
[December 2007 Question Paper]
Ans:
The hypothesis to be tested here is
H0: The arrangement of trees is random.
H1: The arrangement of trees is not random
n1= number of healthy trees; n2= number of diseased trees
n= total number of trees.
n=n1+n2.
r= number of runs( Note: A run is a sequence of identical symbols or elements which are followed and processed by different types of symbols or elements or by no symbols on either side.)
Hence n=22, n1=13, n2=9, r=6.
Mean= E(r)= [(2*n1*n2)/ (n1+n2)]+1 and
Standard Deviation= S.D.(r)= sqrt[ 2*n1*n2[(2*n1*n2)-n1-n2]/ ((sqr(n1+n2)(n1+n2-1))]
E(r)= [(2*13*9)/(13+9)]+1= 11.64
S.D.(r)= sqrt[ 2*13*9[(2*13*9)-13-9]/ ((sqr(13+9)(13+9-1))]
=sqrt(234(234-22))/(sqr(22)(21)= sqrt( 49608/10164)=2.209
The null hypothesis is tested using the test statistics:
|Z|= |r-E(r)|/S.D.(r)
Therefore |Z| = |6-11.64|/2.209= 2.55
The tabulated value of test statistics 1.96. The calculated value of Z is 2.55 which is greater than the tabulated value. Hence, the hypothesis H0 is rejected.
Thus the arrangement of trees is not random.
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