Q209.1

[Jo_]

Hi David,

I had a question when working through the chapters on hypothesis testing. When i take the study notes page 76, question 209.1 mentions a question where 9 companies on a random sample of 60 defaulted (15%). historical default rate is 10% and question probes to test if this is truly the case or if we can reject it.

The suggested answer creates the test statistic by (15%-10%)/SE, where SE is sqrt(15%*85%/60) = 0,046098

Could you please clarify the rationale behind the SE in this case? When solving this exercise, i was going from the assumption that we have a Bernoulli distribution at hand here, so the stdev = sqrt (n*p*q) = sqrt(60*15%*85%) = 2,7658. divided by sqrt (n) to come to SE gives 0,357, significantly different than the 0?046 which you obtained. Then again, this is the assumed stdev from the underlying distribution and not the sample stdev, so i guess this methodology isn't correct for that reason. Still leaves me wondering where your formula comes from though?

Jo

 

[David Harper CFA FRM]

Hi @JoeriG Yes, good question, but I think we've got some good discussion in the source question (I think eventually we'll add links from the notes to the source, like we do in the PQ, so I understand why it's not easy to find this one) @ https://forum.bionicturtle.com/thre...stic-and-confidence-interval.5318/#post-22826 e.g.,

• as this is a binomial, the variance of the number of defaults is (indeed) = n*p*(p-1)
• but this is default rate; e.g., 9 defaults/60 = 15% default rate
• so [n*p*(p-1)]/n = p*(1-p) is the variance of the default rate ( same as variance of the Bernoulli, hence the importance of random sample = i.i.d)
• But this p*(1-p) is the variance, in % terms, of the really the population distribution; e.g., if p = 10% is actually true, then the variance = 9.0%
• Okay but we want here the sample average and per CLT it is the population variance/n = p*(1-p)/n, such that the standard error = SQRT[p*(1-p)/n]