t-test and p value approach

[Adzi]

Define and interpret the p value

2011 Quantitative Analysis Study Notes:
On page 30, Test t statistic = 2.06 and p value = 4.09%.
On page 33, “P/E ratios of 28 NYSE companies” example, Test t statistic = 2.65 and p value = 1.3%.
I still can’t figure out the way you computed p values of 4.09 % and 1.3%, respectively. Could you please explain?

I know that there are three possible tests when we deal with a t-test and p value or z-test and p value:
* The parameter is greater than (>) the stated value (right-tailed test), or
* The parameter is less than (<) the stated value (left-tailed test), or
*The parameter is either greater than or less than (≠) the stated value (two-tailed test).

Could you, please, elaborate more on how to get the p value of each case without using Excel?

When should we use the formula provided on page 43 of the 2011 Quantitative Analysis Study Notes?
p-value = Pr(Z > t act) = 1 – Phi(t act)

Thanks a lot in advance,

 

[David Harper CFA FRM]

Hi Denis,

You cannot (to my knowledge) calculate p-value with your calculator (TI BA II+ or HP 12c). You need Excel or you can "reverse-lookup" the student's t (or normal Z) table in the statistics book.

In these cases, I computed with:
p 30: = T.DIST.2T(2.06, 199) = 4.09%
p 33: = T.DIST.2T(2.65, 27) = 1.33%

If you refer to the Student's t lookup table on page 32, here is a method many seem not to be aware of:
Using the 27 df. example, go to row where d.f. = 27 (yellow column), then scan right until you find the computed value (2.65). Note it is between 2.473 and 2.771 . The row headers are, in fact, p-values!
e.g., if the computed t were 2.771 (@ 27 df), then the two tailed p value would be 1% (0.01)
... in this case, it's between, but if i interpolate i get 1.4% which is close enough

I think the more useful points are:
1. In terms of the exam, you can't be expected to calculate the p-value
2. In terms of interpretation, the p-value is the probability that you would get the test statistic if the null hypothesis is true; e.g., page 30:
if the population mean is really 20 (i.e., null is true!), then the probability of observing a sample mean that is +/-1.28 standard deviations (errors) away from that true value is only 4.09%. In the same way, 4.09% is the unlikely area in both tails: it is randomly possible to get 1.28 sigma from true 20 because each sample is a random sample.

Re: which test: these are both typical test of sample mean, unless otherwise instructed, assume two-tail.

I hope that helps!
(btw, he's a little more advanced, but an interesting link from last Friday on p-values @ http://davegiles.blogspot.com/2011/04/may-i-show-you-my-collection-of-p.html)

David

 

[Adzi]

Hi David,

Thanks a lot for your quick answer to this question.

I used the Linear Interpolation Method to compute p-values as you advised.
Below are the answers I have got even if they are not exactly the same as the ones in the notes.

Page 30: using d.f. of 199 and Test t statistic of 2.06, I found p-value of 4.30% close to 4.09% as in the notes.
Page 33: using d.f. of 27 and Test t statistic of 2.65, I found p-value of 1.4% close to 1.3% as in the notes.

Once again thanks David,
Denis.

 

[David Harper CFA FRM]

Hi Denis,

Right, well done, interesting! If we meditate on the meaning of the p value (p value = the area under the tail(s) if the null is true; in the case, the area under the curve to the right of 2.06/2.65 if the population mean is correct), then I think we can see why the interpolation will always give a larger number. The z/t curve is non-linear; if the tail were a rectangle, then your interpolation should exactly match the result. The interpolation is, in fact, drawing a rectangle over the partial tail region, which must have less area ....

The way i think about the p value, as a probabilistic concept, this precision is instructive but not so useful in practice. Similar to the post above, where the professor notes the p value is itself a random variable (!). What is the practical difference between 95.70% and 95.91% confidence, in the context of a random sample (i.e., the next sample will give a different p value)? as a human user, i cannot treat them with realistic difference ... in this way, the interpolation works for me as approximation (with knowing overestimation) because the *meaning* of the result does not deserve high precision.

Thanks, David