Finding the standard deviation from confidence intervals

[CanvasEcho]

The textbook has the following solved exercise in chapter 6







After obtaining the 5.23% figure, I would have set it equal to Standard Error * Critical T like described in equation 6.4 of the textbook.
Since we are given n = 37, I would have expected the standard deviation to be calculated as:
12.78 = 7.55 + (CriticalT * Standard Error)
5.23 = CriticalT * Standard Error
5.23 = 2.57 * Standard Error
5.23 / 2.57 = Std.Dev / Sqrt(n)
5.23/2.57 * Sqrt(37) = Std.Dev

instead they calculated it as 5.23/2.57...
Can anyone explain if this is an error or not?

Thanks

 

[David Harper CFA FRM]

Hi @CanvasEcho Yes, that's a mess, you are correct. The margin of error (ME) = SE * t_critical where SE = sample σ/sqrt(n) such that
ME = sample σ /sqrt(n) * t_critical. Therefore,
sample σ = ME/t_critical * sqrt(n) = 5.23%/2.57 * sqrt(37) = 12.379%.

And we can easily just test this with 7.55% +/- SE*t_critical = 7.55% +/- [12.379%/sqrt(37)] * 2.57 = {2.32%, 12.78%}. Confirmed!

As further evidence of sloppy editing, notice the deviate (aka, critical value) assumed is 2.57 because the solution is using the normal approximation (justified for a large sample, barely in this case). But we actually, rightly memorize 2.58 and 1.96 (i.e., the corresponding two-tailed 95% normal deviate) because the true 99.5% deviate = 2.575829304 does truly round to the highly familiar 2.58. We knows it is using the normal approximation because the exact critical t is given by T.INV.2T(1.0%, 36) = -T.INV(0.5%, 36) = 2.71948463. That's just to say that they also should be using 2.58 (because 2.57 is confusing if only because it's one of the four two-decimal deviates we certainly memorize: 1.65, 2.33, 1.96 and 2.58) such that sample σ = ME/t_critical * sqrt(n) = 5.23%/2.58 * sqrt(37) = 12.331%. Notice my SE scales by sqrt(37) but the critical value is based on 37 - 1 = 36 degrees of freedom. Thanks!

 

[CanvasEcho]

Thanks David!