A variable Y is regressed against a single variable X across 24 observations. The value of the slope is 1.14, and the constant is 1.3. The mean value of X is 1.10, and the mean value of Y is 2.67. The standard deviation of the X variable is 1.10, and the standard deviation of the Y variable is 2.46. The sum of squared errors is 89.7. For an X value of 1.0, what is the 95% confidence interval for the Y value? A) -1.68 to 6.56. B) -1.83 to 6.72. C) 0.59 to 4.30.
SSE=89.7 SEE=2.01922577 SEE^2=4.077272 1/n=0.04166667 X - Xbar = 1-1.10=-0.1 n - 1 = 23 (X - Xbar)/(n - 1) = -0.1/23=-0.0043478260 sx^2 = 1.21 sf^2 = 4.077272* [1 + 0.04166667 + -0.0043478260*1.21] 4.077272* [1 + 0.04166667 - 0.00526086946] 4.077272*[1.03640580054] 4.22570835117932688 sf = 2.0556527798 tc=2.074 ycap = 1.3 + 1.14*1 = 2.44 tc*sf = 4.2634238653052 2.44-4.2634238653052 = -1.8234238653052 2.44+4.2634238653052 = 6.7034238653052 B
Swaption, looking at your answer is the reason I am letting the CFA have Quant.
How the fuck did you do that so fast? Can you give a little bit of an explanation as to what you did there?
he/she is a computer
How about using a stats package?
Shorter method 2.44 ± 2.01922577*2.074 2.44-4.18787424698 = -1.74787424698 2.44+4.18787424698 = 6.62787424698 Now select the ans that has wider abscissa spread = B
A variable Y is regressed against a single variable X across 24 observations. The value of the slope is 1.14, and the constant is 1.3. The mean value of X is 1.10, and the mean value of Y is 2.67. The standard deviation of the X variable is 1.10, and the standard deviation of the Y variable is 2.46. The sum of squared errors is 89.7. For an X value of 1.0, what is the 95% confidence interval for the Y value? A) −1.68 to 6.56. B) −1.83 to 6.72. C) 0.59 to 4.30. Y=1.3 + 1.14 X When X=1, Y=1.3+1.14(1) = 2.44 xbar=1.1, ybar=2.67, sigmax = 1.1, sigmay=2.46 SSE=89.7 n=24 see = sqrt(89.7/(24-2))=2.0192 sf^2 = see^2*(1+1/n+(xhat-xbar)^2/(n-1)*sigmax^2)) = 4.0772 * (1+1/24+.1^2/23*1.1^2) = 4.2485 sf = 2.0612 tc=2.074 2.44 +/- 2.074*2.0612 = 2.44 +/- 4.2749 = -1.8349, 6.7149 B.
@swaption: I know we can use the SEE as an approx for sf in doing these questions, what I am curious about however is weather or not sf>SEE always holds mathematically, i.e. when we find the CI do we always just go the to the next closest WIDER interval in the answer choice. I am trying to find a helpful yet easy rule of thumb so I don’t have to memorize Sf^2 for the exam and this would seem to be as simple as it gets.
YES - this was a rule confirmed last year by Joey. So I remember it. Go WIDER always and it always falls to an correct ans.
Dude thanks so much, very helpful! I love it when I learn something genuinely useful from this site that may be directly applicable and makes life simpler (cus usually its minute details that just make things harder).
Dreary Wrote: ------------------------------------------------------- > How about using a stats package? That’s how I feel about the entire quant section
After all of these explanations I still have absolutely no idea what is going on. Am I alone?
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calc. Y at the given X. = 2.44 Based on # of observation calc. t-crit. = 2.074 you know SEE. Use the Tcrit * SEE get your conf. interval based on that - then move out wider. lower on the low side, higher on the high side. (Bcos sf > see). now choose your answer.
so am I
where did you get X to be 2.44, it says X=1
MT327 Wrote: ------------------------------------------------------- > After all of these explanations I still have > absolutely no idea what is going on. Am I alone? i have no idea either dude, still crushing quant in Q-bank, but no idea what theyre doing here
You are solving for the Y interval not the X, so you have to plug the given X value into the regression in order to get your Y hat value.
I solved this question already I remember in another thread which means CPK123 probably as too 