n=120 Intercept: 6.017 Slope coefficient: 0.09251 Mean X = 91.0983 standard error = 0.81325 Now they want me to build a 5% confidence interval for an X value of 90. Ehhh? Help?

I get 12.729 - 15.956 If its close to this i’ll write out the steps.

Y = 6.017 + 0.09251*90 = 14.3429 From T-Table => t(120-2-1, 0.025) = 1.98 CI = 14.3429 ± 1.98*0.81325 CI = 14.3429 ± 1.610235 is this what was expected ??? A CI of 12.732665 -------- 15.953135 EDIT: Nib, you too quick!

You probably would have beat me if I wrote out the solution as you did. That is exactly how I got the answer.

I was looking to us T - 2 vs what you guys had as T-2-1. ( 120-2-1 ). Why use 117 vs 118. I got that part of it wrong.

GM —> DoF = n-k-1

Dinesh thanks. I messed up and was thinking of DF on a t-test.

You have a table that goes beyond 100 df? I know what I want for Christmas.

I actually have that table tattooed on my leg.

I need to upgrade. I had to search for some online t-table and all I got was a ghetto one from a UCLA professor’s website.

The CFA book says the answer is 12.780 < Y < 16.014 They first solve for variance with a formula that I have never seen before (outcome is 0.66694) and the square root of that is 0.8167 which is what they use for the interval. I also used standard error but obviously CFAI wants me to calculate the standard deviation instead which I have no idea of what they’re doing and where it comes from.

Wouldn’t df=118? There is only 1 independent variable. So k=1, so df=n-(k+1)=120-2=118. Am I missing something?

Hey guys how did you know how many slope coeff were in the question? I hvaen’t looked at quant for a long time… If I knew how many indep var there were then I could calc n-k-1, but I don’t here. Explain you quant wizards.

is this a standard error of the forecast question? then the predicted y interval is (predicted y based on x=90)+/- (t-crit)*(st. errror of forecast)

Dimes27 Wrote: ------------------------------------------------------- > is this a standard error of the forecast > question? > > then the predicted y interval is (predicted y > based on x=90)+/- (t-crit)*(st. errror of > forecast) No. That would be the S.E. around the Y value.

There is 1 Intercept (obviously) and 1 Slope coefficient (so 1 indep variable) n =120 k =1 Hence DoF = n-k-1 = 120-1-1 = 118

I still don’t understand how they get to a variance number with this data. And why can’t I use the standard error (as you guys did) and have to go for the standard deviation? I’m really puzzled and since it’s a q from the CFA book… I WANT TO UNDERSTAND IT!

Where’s the question?

Question 12 (a and b) on page 261/262 of the CFA book. Answer is in the back of the book and doesn’t help me one bit.

dinesh.sundrani Wrote: ------------------------------------------------------- > There is 1 Intercept (obviously) and 1 Slope > coefficient (so 1 indep variable) > > n =120 > k =1 > > Hence DoF = n-k-1 = 120-1-1 = 118 Okay. I didn’t know if the full question had more independant vars.