R8.P1.T2.Miller_v3 - 209.1

[nilz]

209.1. Nine (9) companies among a random sample of 60 companies defaulted. The companies

were each in the same highly speculative credit rating category: statistically, they represent a

random sample from the population of CCC-rated companies. The rating agency contends that

the historical (population) default rate for this category is 10.0%, in contrast to the 15.0%

default rate observed in the sample. Is there statistical evidence, with any high confidence,

that the true default rate is different than 10.0%; i.e., if the null hypothesis is that the true

default rate is 10.0%, can we reject the null?

a) No, the t-statistic is 0.39

b) No, the t-statistic is 1.08

c) Yes, the t-statistic is 1.74

d) Yes, the t-statistic is 23.53

Question - As per explanation the Standard error is calculated in 2 ways-

n=60

standard error = SQRT(15%*85%/60) = 0.046098 OR
standard error = SQRT(10%*90%/60)=0.046098

I did not understand the term ( 15%*85% or 10%*90% in numerator ) . how does that define the std deviation?

The Standard error formula is
standard error = Variance / Sqrt(n) or
standard error = SQRT(std deviation / n)

Please explain.

Thanks
Nilesh

 

[Alex_1]

Hi Nilesh, for more detail you can refer to the following thread:
https://forum.bionicturtle.com/threads/p1-t2-209-t-statistic-and-confidence-interval.5318/
and from there to the answer by David: https://forum.bionicturtle.com/thre...stic-and-confidence-interval.5318/#post-22826.
Best regards, Alex

 

[David Harper CFA FRM]

Thank you @Alex_1 for sharing a direct pointer

 

[filip313]

Hi,

I red the explanation in the thread, but I still can't understand the trick of the Standard error in 209.1.
I do not understand how do you go from n*p(1-p) to p(1-p)/n ? Why do you divide by n^2?
Could someone please point me in the right direction?

Many Thanks

 

[Alex_1]

Hi,

I red the explanation in the thread, but I still can't understand the trick of the Standard error in 209.1.
I do not understand how do you go from n*p(1-p) to p(1-p)/n ? Why do you divide by n^2?
Could someone please point me in the right direction?

Many Thanks

Hi @filip313 , I also had some difficulties at the beginning, but I found David's explanation in the thread helpful. Let me try to "rephrase" it: in a binomial distribution the variance of the number of defaults is n*p*(1-p). The question here is however related to the default rate. Since the default rate is the number of defaults divided by the total number of companies (= n) this means that the variance of the default rate is [n*p*(1-p)]/n, which results to p*(1-p). What he have obtained up to this point is the variance of the population distribution, but we are looking for the variance of the sample distribution, hence we have to divide p*(1-p) by the sample size (= n). In my opinion one doesn't have to think of these steps as "dividing the general binomial variance formula by n^2", but rather going through the steps mentioned before (explained actually by David and just rephrased by myself). I hope this helps?

Best regards,
Alex

 

[filip313]

Hi Alex,

Thank you for the explanation. I think i understand now.
I was a bit confused at the beginning because of the default rate and also because on my text the formula for the standard error is actually
Std Error = Std Dev/SQRT(n).


Thanks,
filip