Standard Dev in Hypothesis Testing

[Kashif Khalid]

Hi David & Co

In the below question why is the standard dev not SQRT(60*0.15*0.85) as it seems like the binomial distribution applies? and therefore the standard error would be SQRT((60*0.15*0.85)/60)

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


The answer is 209.1 (b): No, the t-statistic is only 1.08. For a large sample, the distribution is normally approximated, such that at a 5.0% two-tailed significance, we reject if the abs(t-statistic) exceeds 1.96

The standard error = SQRT((15%*85%)/60) = 0.0460098; please note: if you used SQRT(10%*90%/60) for the standard error, that is not wrong, but also would not change the conclusion as the t-statistic is 1.29
The t-statistic = (15%-10%)/0.0460098 = 1.08. The two-sided p-value is 27.8%, but as the t statistic is well below 2.0, we cannot reject.


Thanks

 

[ShaktiRathore]

Hi,
The population std deviation is popln default rate*(1- popln default rate) here we are using tstatics and sample std deviation=sqrt(.15*(1-.15)) as proxy for popln std deviation.
Also SE for sampling distbn=popln std deviation/sqrt(sample size)=sqrt(.15*.85)/sqrt(60) so tstat=sample default rate-popln default rate/SE=10%-15%/sqrt(.15*.85/60). Yours point that we should take sqrt( 60*.15*.85 )as popln std dev but that would just give sqrt of sum variances of all companies in sample since we are interested in average variance per periid divide by 60 to get popln or sample std dev (see analogy to popln for popln n>>infinity so that would give infinite variance is not logical) Using.9*.10 shall give actual popln variance we can use it instead but due to sample size>30 i think z statistic would do equally good job.
You should know for a given probability p of variable in population std deviation is sqrtp(1-p) if no popl probability is given we use sample probability as proxy so that popln std dev is sqrt(s(s-1)) so SE =popln std dev/sqrt(n)=sqrt(s(s-1)/n).
Thanks

 

[David Harper CFA FRM]

@joehar04 This is difficult because you have correctly specified the standard deviation of the population distribution (i.e., the standard deviation of the default rate); then you just have to take the additional step of computing the standard error which is this standard deviation divided by sqrt(n); aka, sqrt(population variance divided by n). See explain at the source at https://forum.bionicturtle.com/thre...stic-and-confidence-interval.5318/#post-22826
i.e.,

Hi Joe,

• 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*(1-p)]/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]

... note the difference: the 9% variance is the variance of a single company: expects to default 10% of the time, with SQRT(9%) = 30% variance. But when we retrieve the sample average default rate from a set of n companies, the dispersion of that average is going to tighten pretty quickly; e.g., if we used half the companies, it should be very near to 10%. I hope that explains,

 

[Rolme]

So in summary this really is a Bernoulli as we're interested in the verification of the single paramter "default rate" and not a Binomial which would i.e. be looking into the probability of a certain number of defaults (i.e. repeated application of the default rate)?

thanks

 

[ShaktiRathore]

Hi
For any sample default rate is X/n where X is no of defaults in a sample is binomial variable with std deviation sqrt(np(1-p)) ,p is popln/sample default rate. These sample default rates X/n will be according to CLT will be normally distributed with mean as popln default rate(E(X/n)=np/n=p) and std deviation ,also called std error as sqrt(variance of X/n)=sqrt(p(1-p)/n).
Thanks