a. The given data is is of individual type
b. to find parameters by rank regression method that is by least square method
X is considered from different distribution and following random samples are taken
y weibull normal lognormal exponential gamma extreme
46 0.28101 0.72053 5.49715 2.18920 0.47069 0.52240
64 0.91160 1.48850 1.55441 0.45226 1.33067 0.04686
83 0.93759 -1.29638 3.26693 1.56018 7.02524 0.56067
105 0.64961 -0.32414 0.65800 0.17356 3.20563 -0.75814
123 0.50455 -0.64524 0.23813 1.42705 2.92275 -0.03883
150 1.79579 -0.68728 1.87687 0.56735 4.47738 -3.70976
Scatter plots of X with Y
regession model and least square estimates of model
1)
Coefficients
Term Coef SE Coef T-Value P-Value VIF
Constant 55.4 27.2 2.03 0.112
weibull 47.0 28.0 1.68 0.168 1.00
Regression Equation
2)
Coefficients
Term Coef SE Coef T-Value P-Value VIF
Constant 92.3 13.8 6.69 0.003
normal -23.5 14.5 -1.62 0.181 1.00
Regression Equation
y = 92.3 - 23.5 normal
3)
Coefficients
Term Coef SE Coef T-Value P-Value VIF
Constant 123.4 21.2 5.82 0.004
lognormal -12.93 7.55 -1.71 0.162 1.00
Regression Equation
y = 123.4 - 12.93 lognormal
4)
Coefficients
Term Coef SE Coef T-Value P-Value VIF
Constant 118.8 28.1 4.23 0.013
exponential -22.3 22.0 -1.02 0.367 1.00
Regression Equation
y = 118.8 - 22.3 exponential
5)
Coefficients
Term Coef SE Coef T-Value P-Value VIF
Constant 70.4 28.3 2.49 0.068
gamma 7.64 7.31 1.05 0.355 1.00
Regression Equation
y = 70.4 + 7.64 gamma
6)
Coefficients
Term Coef SE Coef T-Value P-Value VIF
Constant 84.5 11.5 7.35 0.002
extreme -18.88 7.29 -2.59 0.061 1.00
Regression Equation
y = 84.5 - 18.88 extreme
conclusion : as per the scatter we can conclude that data is good fitted with extreme value distribution
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