Let X1, . . . , X n, be a random sample from a continuous distribution with the probability density function 3(36)2, 6Sx6+1, 0, otherwise fX(~T;9) ={ Here, 6 is an unknown parameter. Assume that the sample size n = 10 and the observed data are 1.46,1.72,154,175,177,1.15,1.60,1.76,1.62,1.57 (a) Find the maximum likelihood estimator of 6, 6ML, and show that W = 6ML 6 is a pivot. Use this pivot to construct a twosided 90% condence interval for 6 using the observed data. (b) Find a method of moments estimator of 6, 6MM. Using approximate normality of 6 M M, construct a twosided 90% condence interval for 6 using the observed data. > x= C(1.46, 1.72, 1.54, 1.75, 1.77, 1.15, 1.60, 1.76, 1.62, 1.57) > meaan) [1] 1.594 > vaer) [1; @.@3533778