Recent content by J Flanders

  1. J

    Find MLE for f(y/x) = (x + 1)y^x, 0 < y < 1 and x > -1

    OK, I just want to check this. I get: ln[((x+1)^n)*(y1^x)*(y2^x)*...*(yn^x)] = nln(x+1) + x[ln(y1) + ln(y2) + ... + ln(yn) I take the derivative with respect to x and set it equal to zero: n/(x+1) + ln(y1) + ln(y2) + ... + y(n) and with algebra this implies x (the estimator) = [-n...
  2. J

    Find MLE for f(y/x) = (x + 1)y^x, 0 < y < 1 and x > -1

    So I get: ln[((x+1)^n)*(y1^x)*(y2^x)*...*(yn^x)], but I don't see how you maximize this. I would imagine you take the derivative and set it equal to zero, but I cannot solve for x. What is the maximum value for x? Thanks for the help from before.
  3. J

    Find MLE for f(y/x) = (x + 1)y^x, 0 < y < 1 and x > -1

    This is my question: Find the Maximum Likelihood Estimator for f(y / x) = (x + 1)y^x, 0 < y < 1 and x > -1 OR 0, elsewhere. I think this is how you get started, but I get confused. I'm not sure how to continue. The likelihood function defined as the joint density of Y1, Y2, ..., Yn...
  4. J

    Chi-squared analysis and budgeting weekly repair costs

    I cannot find any tables of Gamma distributions. How would I find Gamma(5, 20)? Is there a way to turn it into a chi-squared distribution? Thanks for the help from before.
  5. J

    Chi-squared analysis and budgeting weekly repair costs

    * I posted this in the Coursework section, but I wasn't sure if it would be answered there. * Here's my question: A plant supervisor is interested in budgeting weekly repair costs for a certain type of machine. Records over the past years indicate that these repair costs have an exponential...
  6. J

    Solving Exponential Distribution Problem: Find c for P(Y1+Y2+Y3+Y4+Y5>c)=0.05

    Here's my question: A plant supervisor is interested in budgeting weekly repair costs for a certain type of machine. Records over the past years indicate that these repair costs have an exponential distribution with mean 20 for each machine studied. Let Y1, Y2, Y3, Y4, Y5 denote the repair...
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