In a F test, if we have to test significance at 5% and 2.5%, and we reject null at 2.5%, then does it mean that we will also reject at 5%?
Got it…If null is getting rejected at 97.5%, then it surely is getting rejected at 95%…
Yup. You can see this in other perspective.
The lower the threshold (alpha), the narrower the confidence interval (CI) for the true value of the parameter to be within the interval. If the calculated statistic (F, T, Z, etc) rejects the null (you are inside the CI), then you are also inside a wider CI.
Got it…Have you also studied the Black and Scholes. I am not quite getting the intuition, What to do?
You meant, lower the alpha, wider the CI… 
Do you want intuition, or do you want understanding?
Oops, you got me. I think you need no help ; )
Both magician…Pls help me here…
Pls help me in Black and Scholes…
It is mainly about volatility of stock prices, if I remember well.
CFAI EOC 6th Questn, page 316-Suppose that you deleted several of the observations that had small residual values. If you re-estimated the regression equation using this reduced sample, what would likely happen to the standard error of the estimate and the R-squared?
Standard Error of the Estimate R-Squared A Decrease Decrease B Decrease Increase C** Increase **Decrease
Answer is C. Is it because the observations had small residual values? Had the residual values been higher, it would have been the other way round, ie. it SEE would decrease and R squared would increase?
Or is it because, as n decreases, SEE increases. And as SEE increases, SSE increases, which further increases Total variation. Hence, R2 decreases?
It is mainly because the lower the residual value for an observation, the lower SEE of the whole model and higher R2 (lower error of estimate implies better fit)
If we retire the best errors (the lowest ones), then the SEE of the model will be higher, therefore worse fit / lower R2.
To make this more intuitive, recall the graphic where you have the observations dispersion and an “average line” between the dispersion. This line is linear model. The distance between any observation point and the linear model is an observation error. Now, take out the observations with the lowest distance from the linear model (the lowest errors). Yup, the sum of errors will be higher (ie. SEE), therefore lower R2.
Also, the standard deviation of errors will increase.
Thank you so much Harrogath…
It is mainly because the lower the residual value for an observation, the lower SEE of the whole model and higher R2 (lower error of estimate implies better fit)
If we retire the best errors (the lowest ones), then the SEE of the model will be higher, therefore worse fit / lower R2.
To make this more intuitive, recall the graphic where you have the observations dispersion and an “average line” between the dispersion. This line is linear model. The distance between any observation point and the linear model is an observation error. Now, take out the observations with the lowest distance from the linear model (the lowest errors). Yup, the sum of errors will be higher (ie. SEE), therefore lower R2.
Also, the standard deviation of errors will increase.
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Black and Scholes - It is about pricing a call option…And we have that formula where we use probabilities…ND1, ND2…My question was on ND1…which is the delta of the call option…
It is mainly about volatility of stock prices, if I remember well.
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Pg 324,CFAI-Solution of 12th problem, EOC-For a regression with one independent variable, the t-value (and significance) for the slope coefficient should equal the t-value (and significance) of the correlation coefficient.
Why???
I would not try to go deeper in this demonstration unless you are really interested in advanced statistics. For the scope of the L2 exam, we rely on the formulas provided.
In this case, the book says that the T-statistic for the slope coefficient (…1) of a single variable regression (1 indep variable) should be equal to the T-statistic of the correlation coefficient (…2) between the dependent variable and the independent variable.
The calculation of (1) is:
T(slope) = ( bx - 0 ) / SEx
where: SEx = Standard Error of x = SD(x) / n0.5 The SEx is always given in an ANOVA table.
The calculation of (2) is exactly the same as (1):
T(correlation coefficient r) = ( r - 0 ) / SEr
where Standard Error of r is = [(1 - r2) / (n-2)]0.5 …The book lacks to specify this formula because it is too much deep in the subject of advanced statistics.
So replacing the SEr in the T® we get the same T-statistic the book shows off:
T® = ( r - 0 ) / [(1 - r2) / (n-2)]0.5 = r * (n-2)0.5 / (1 - r2)0.5
I know, we are at the same initial point… why those T-statistics are the same value. The demonstration relies on the calculation of the SEr, which is not an easy task and probably not of your business for now, unless you are really interested in those kind of demonstrations. You can check these links for help:
https://www.jstor.org/stable/2277400?seq=1#page_scan_tab_contents
I recommend you to stick to the CFAI books and no more research for now. Train for the exam instead.
BR.
Noted with thanks…
I have a doubt when we calculate the percent decline in dependent variable given percent increase in independent variable (say 2%). In the equation, in some problems, 2 is used while in some, 0.02 is used. Pls advise.
I have the same concern. However, it is usually mentioned in the problem. They will either say in percent or in decimals. This is what I have been seeing in the EOC questions in the multiple regression chapter. Hopefully it is the same thing in the exam.