Miller Chapter 5 Pg79 of Study Notes

[INDRAJIT]

Hi David,

I am a bit confused as to how the Critical value of 1.972 is arrived at for the below example. B'coz my understanding is that if the sample size is more than 30 we can use normal distribution table if the t-table is given only upto 30 dfs. In that case 95% 2-sided is area 0.475 on each side of the mean which corresponds to 1.96. As in the given example below if the sample size is 200 then in the FRM exam can we hope that the t-table will contain the values upto 200 dfs?

Define and construct a confidence interval.
Th
e confidence interval uses the product of
[standard error
х
critical “lookup” t].
In the Stock
& Watson example, the confidence interval is given by 22.64 +/
-
(1.28)(1.96) because 1.28 is
the standard error and 1.96 is the critical t (critical Z) value ass
ociated with 95% two
-
tailed
confidence:
Sample Mean
$22.64
Sample Std Deviation
$18.14
Sample size (n)
200
Standard Error
1.28
Confidence
95%
Critical t
1.972
Lower limit
$20.11
Upper limit
$25.17

 

[David Harper CFA FRM]

Hi @INDRAJIT Great observation (I have tasked a revision to page 79 to clarify and awarded you a star toward the weekly drawing).
You are correct: in an actual exam situation, when the sample is large (i.e., n > 30), the two-sided 95% critical Z of 1.96 will be used, just as you suggest.

Technically, the example is not incorrect to use 1.97 per a two-sided critical t of 1.96 (per a realistic situation, we do not know the population's variance/sd, such that we consume a d.f. computing the sample's standard deviation as its proxy). In this way, using a critical t for any sample size is technically correct.

However, as you say--and further as the FRM always does in practice (see almost any sample paper. In fact, GARP sometimes has sometimes made the mistake of using a Z value when the sample size is not specified)--when the sample size gets large, the values converge. So, per convention, when n > 30, we use a critical Z value; which has the further advantage of not requiring a lookup table as most candidates will be familiar with two-sided Z value of 1.96 (@ 95%) and 2.58 (@ 99%). Thanks for your point!