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MS-95 Research Methodology for Management Decisions Question Papers/ Assignments
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.
[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.
Tuesday, May 5, 2009
Date sheet for June 2009 exams announced
IGNOU has published the date sheets for the term-end exams (T.E.E.) to be held in June 2009. The timetable is available at http://www.ignou.ac.in/datesheet.htm
Friday, May 1, 2009
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