Recent content by anonymousk

This assignment is translated from another language, so some words be not be completely correct Homework Statement (Annual market rate might be yield to maturity?) Face value = 1000 Obligation----Condition----Payment profile----Maturity------Annual Coupon Rate-----Annual market rate...

Haha! "consequence of a variance for a probability." and " It will have the same order of magnitude as N.". I feel like I'm sitting in a rowboat in the middle of the Atlantic ocean when reading this, haha! I'm sorry. I've been skimming through your posts too often to count. I agree...

I believe you're referring to this one ##(N/Y)p(1-p)## And I did (N/Y)p(1−p) = (32100/3)*0,02866(1-0,02866) = 297,87. That's how I understand the formula, which was wrong, hmm. Question c) goes Explain that the number of bankruptcies in the C-sector during 2014 can be assumed to be...

Ohhhh, I got it.. So a 95% confidence interval tells us, that we are 95% confident that the true population proportion is between these 2 numbers (2.76% to 2.97%). It's the range/uncertainty of our estimator \widehat{P}=2,86% just unsure how to calculate the CI for expected number...

I'm so confused about this assignment. To my defense, I'm not this bad at the other courses (micro economics etc). It's just statistics I just can't seem to be able to wrap my mind around. I still haven't done the rest of the assignment, c), d) and e), but I think I've got the confidence...

Well, in my case I have: Mean, variance and n for the credit scores of active businesses (X) and mean, variance and n for the credit scores of bankrupt business (Y) I have to calculate the mean and variance for them put together (without adding them on top of each other. Says we can't assume...

Okay, this leaves me with 32100*0,02866(1-0,02866) = 893,6192 Sorry, but I don't feel like I'm getting any closer anymore at this point, even with these good hints you throw at me, since I've been sitting with this assignment for a whole week and still not finished it. I just can't see...

(N/Y)p(1−p) = (32100/3)*0,02866(1-0,02866) = 297,87. These are the same results as before, so not sure where I went wrong. I think I need to sleep on it, since it's 5am here, hehe. I appreciate your help and patience with me! :)

0,02866(1-0,02866)/(3*32100) = 0,000000289 This is the variance of the estimator of P. Again a really low number. Thinking of giving up on this assignment, because it simply doesn't catch. 32100*0,02866(1-0,02866)/3 = 297.. This number makes sense .. so the standard deviation from the mean...

\widehat{P} I've got right. 0,02866 Estimate for the expected number bankruptcies is the 3 year average, \overline{X}=920 32100*0,02866(1-0,02866) = 893,6192/3 = 297,87 This is where I'm confused. I divided 893,6192 with 321002 = 0,00000087/3 = 0,00000029 (which is the variance of...

Aah .. so: ((sum of x)*(sum of Y)) / ((mean of X)*(mean of Y))? (I have a similar assignment, but seems I did it wrong)

Hmm.. I don't know. I don't understand why the variance for PC is a higher number than the variance for the expected number bankruptcies during 2014 either. Edit: Im also unsure of the numbers I calculated now. The two variances I found 297,91 for PC and the variance for the...

I appreciate that. It's quite frustrating using these formulas not knowing what's happening behind it all. Still struggling to understand this all, but feel like I'm getting there.(slowly) Just to be clear: Y in this case will be the 3 years the data is calculated from? (2009, 2010 and...

Been looking at this for 2 days now, and simply cannot get what you're trying to lure me towards here (combining those equations.) However, I looked into Standard Error of the Mean What I'm sitting with at the moment is Variance of the estimate PC: σ/√n = 29,9/√3 = 17,26 = std dev σ2=17,262=...

Okay, I've now slept on it. Is the estimate the same in these both?? Estimate for PC being (920*3)/(32100*3) = 0,02866 And estimate for the expected number of bankruptcies during 2014 being 920/32100= 0,02866, but it's the variance where they differ. The variance for PC is...