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# Hypothesis about the difference between two population means.

#### Elizabeth_Babalola

##### Member
Hi @David Harper CFA FRM

I am having a hard time understanding the calculations here:

My confusion lies in how 0.78 and 1.02 was gotten.

Kindly assist.

Edited by Nicole to note: This is referencing QA-6 (Chapter 6) study notes starting on page 28.

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#### David Harper CFA FRM

##### David Harper CFA FRM
Staff member
Subscriber
HI @Elizabeth_Babalola 0.78 = 0.398 * NORM.S.INV(97.5%) = 0.398 * 1.96 = 0.78. And 1.02 = 0.398 * NORM.S.INV(99.5%) = 0.398 * 2.58 = 1.88, where 1.96 is the 2-tailed 95.0% confident standard normal deviate/quantile and 2.58 is the 2-tailed 99.0% confident standard normal deviate. I admit that my label N^(-1) could be confusing because it might implied 1-tailed (I haven't pulled up the associated note text ...). Thanks,

#### Elizabeth_Babalola

##### Member
Thanks @David Harper CFA FRM

The explanation is quite clear. If $\sigma\;(diff)$ implies the difference between both standard deviations, the i guess it should be $\sqrt{4\;}$ - $\sqrt1$

#### David Harper CFA FRM

##### David Harper CFA FRM
Staff member
Subscriber
@Elizabeth_Babalola Okay, right, that's fair ... I just pulled up the text (thank you @Nicole Seaman for giving the exact reference, that's why I didn't look before because without a pointer it just takes too long to find, but your pointer makes it quicker) ....

σ (diff) = sqrt[(2^2 + 1^2 - 2*2*1*0.4)/24] where 2^2 is variance of first fund and 1^2 is variance of second fund. So it is the denominator of the test statistic (i.e., test of two sample means) which is a standard error. A standard error is standard deviation, is why I labelled it such, but I see it can be improved. It's really the standard error of the difference between sample means.

btw, page 29 ends with truncated sentence (cc @Nicole Seaman tagged to wrike)

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#### Elizabeth_Babalola

##### Member
Many thanks David.

All clear now. The point on referencing is also noted.

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