P1.T2.209 T - statistic and confidence interval

[Dr. Jayanthi Sankaran]

Hi David,

In the following question 209.1:

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.

We don't really need the lookup table or a calculator: the t - statistic tells us that the observed sample mean is only 1.08 standard deviations (standard errors) away from the hypothesized population mean.

A two-tailed 90% confidence interval implies 1.64 standard errors, so this (72.8%) is much less confident than even 90%

FWIW: From the 'Cumulative Probabilities of the Standard Normal Distribution Table'

Z(1.08) = 0.8599
And so, 2*(1 - .8599) = 28.02%
Area under curve = 100 - 28.02 = 71.98% ~ 72.0%
A very trivial question - how do you get an area of 72.8%, instead?

My question is (1) how do you determine the p-value of 27.8%?

Thanks
Jayanthi

 

[David Harper CFA FRM]

Hi @Jayanthi Sankaran

It looks like I would have use =T.DIST.2T( 5%/sqrt((15%*85%/60), df = 59) = 28.25% which is not the same. So maybe I used rounded inputs, it's not too far off. I'd normally look into it, but it's very high either way (i.e., very insignificant) and nearby enough that I'm not going to worry about it. Thanks!

 

[AVasa7074]

Hi,

Can anyone please explain the reasoning behind SE calculation as SQRT(15%*85%/60)?
15% is the observed default rate.

Thanks!

 

[Clay Carter]

Hi,

Can anyone please explain the reasoning behind SE calculation as SQRT(15%*85%/60)?
15% is the observed default rate.

Thanks!

@AVasa7074 If 15% defaulted that means that 85% did not default. Here we have a total population so our denominator is just simply N. Hope this helps!