Please assist with ABCD. Show all workings.
A gauge length of 1mm was utilized to measure the strength of single carbon fibers, and impregnated 1000-carbon fiber tows were measured in GPa, yielding the following dataset:
2.247,2.64,2.908,3.099,3.126,3.245,3.328,3.355,3.383,3.572,3.581,3.681,3.726 3.727,3.728,3.783,3.785,3.786,3.896,3.912,3.964,4.05,4.063,4.082, 4.111, 4.118, 4.141,4.246,4.251,4.262,4.326,4.402,4.457,4.466,4.519,4.542,4.555,4.614,4.632, 4.634,4.636,4.678,4.698,4.738,4.832,4.924,5.043,5.099,5.134,5.359,5.473,5.571, 5.684,5.721,5.998,6.06.
To find the maximum likelihood estimate for the parameter eta iteratively, use the mean of the data as an initial value eta ^((1)) and employ successive approximations for a distribution with the density function:
f(x;eta )=(2)/(eta )xexp(-(x^(2))/(eta )),0<=x<infty ;eta >0.
(a) (4 points) Newton-Raphson algorithm:
1. A gauge length of 1 mm was utilized to measure the strength of single carbon fibers and impregnated 1000-carbon fiber tows were measured in GPa, yielding the following dataset:
2.247, 2.64, 2.908, 3.099, 3.126, 3.245, 3.328, 3.355, 3.383, 3.572, 3.581, 3.681, 3.726 3.727, 3.728, 3.783. 3.785, 3.786, 3.896, 3.912, 3.964, 4.05, 4.063, 4.082, 4.111, 4.118 4.141, 4.246, 4.251, 4.262, 4.326, 4.402, 4.457, 4.466, 4.519, 4.542, 4.555, 4.614, 4.632 4.634, 4.636, 4.678, 4.698, 4.738, 4.832, 4.924, 5.043, 5.099, 5.134, 5.359, 5.473, 5.571 5.684, 5.721, 5.998, 6.06.
To find the maximum likelihood estimate for the parameter iteratively, use the mean of the data as an initial value (i) and employ successive approximations for a distribution with the density function:
f(x;3)
0<x<oo;3>0.
Um-1 3(m)=3(m-1) Um-1)
a) (4 points) Newton-Raphson algorithm:
U(m-1) 3(m 3(m-1+ m-1)
(b) (3 points) Method of scoring :
(c) (1 point) Compare the two results in (1) and (2)
(d) (2 points) Construct the 95% confidence interval for . Comment on your result.
