Actuarial Outpost #3 Power of the Test
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#1
11-02-2005, 10:22 AM
 ActingActuary Member Join Date: Jan 2005 Posts: 37
#3 Power of the Test

Did anyoen happen to answer this question? If so, what was your answer?
#2
11-02-2005, 10:24 AM
 azaremba Member Join Date: May 2005 Location: Chicago Posts: 95

I am pretty sure I got 61 so the 60-80 range. I actually got confused between two ways... both got me to that range which was a huge sign... anyone else agree with this?
#3
11-02-2005, 10:28 AM
 sticks1839 Member CAS AAA Join Date: Jul 2005 Posts: 1,138

I think I got 59, just below the 60-80 range. I'm not sure if I found the correct C value for the test though because of it being a poisson.
#4
11-02-2005, 10:28 AM
 ActingActuary Member Join Date: Jan 2005 Posts: 37

I'm not saying this is right, but this is how I did it.
I determined that the critical region was X>2, by finding that Pr (X>2|lambda = 1) < .1 (X has to be an integer becasue it's a single observation from a discrete distribution)

Then Power = 1 - Pr (Type II error) then Pr (Type II error) = (Pr H0 not rejected | H0 false) = Pr (X<2 | lambda = 2) = .672, but then the power would be 1-.672-.323

OR you could just say Power = Probability H0 is rejected given that it's false which is
Pr (X>2 | lambda =2 ) =.323

Last edited by ActingActuary; 11-02-2005 at 10:31 AM..
#5
11-02-2005, 10:33 AM
 sticks1839 Member CAS AAA Join Date: Jul 2005 Posts: 1,138

I did it the same way but because the question said significance level of "at least" 10%, I made the critical region was X>1. The rest was the same and that gave me the .59 or so answer.
#6
11-02-2005, 10:34 AM
 azaremba Member Join Date: May 2005 Location: Chicago Posts: 95

I was in the notion that what you solved is the Power, and then taking 1- the .672 value would be the type 2 error... that is the way I learned it from Mahler. It looks like we did the same method, we just used the opposite values for Power.
#7
11-02-2005, 10:35 AM
 ActingActuary Member Join Date: Jan 2005 Posts: 37

Quote:
 Originally Posted by sgibson18 I did it the same way but because the question said significance level of "at least" 10%, I made the critical region was X>1. The rest was the same and that gave me the .59 or so answer.

I have the exam in front of me, and it doesn't say "at least 10%" it says "10%" . I'm not sure if that changes your opinion, but I thougth I would tell you.
#8
11-02-2005, 10:52 AM
 MrsFrog Member Join Date: Jul 2004 Studying for NOT Cas 7 Favorite beer: On Tap Posts: 4,150

I got .59, but I thought the critical region was X>= 2. I don't have the test in front of me, so I can't tell you why I did this
#9
11-02-2005, 08:26 PM
 Turkey Club Member CAS Join Date: Oct 2005 Studying for CAS Exam 9 Favorite beer: Corona Posts: 367

Quote:
 Originally Posted by MrsFrog I got .59, but I thought the critical region was X>= 2. I don't have the test in front of me, so I can't tell you why I did this
I got 76%, so an answer of D: 60 to 80%

I got the same .323 answer but thought the power was the other way: 1- .323 or soemthing.

Actually I got a Z a little over 5 which leads to a power right around .700. Were you guys using an integer as your K? I had my K set to 3 since that was the next integer higher than 2.282.

I wasn't sure if that was what we were supposed to do or not

Last edited by Turkey Club; 11-02-2005 at 08:48 PM..
#10
11-02-2005, 08:55 PM
 ActingActuary Member Join Date: Jan 2005 Posts: 37

Quote:
 Originally Posted by Turkey Club I got 76%, so an answer of D: 60 to 80% I got the same .323 answer but thought the power was the other way: 1- .323 or soemthing. Actually I got a Z a little over 5 which leads to a power right around .700. Were you guys using an integer as your K? I had my K set to 3 since that was the next integer higher than 2.282. I wasn't sure if that was what we were supposed to do or not
Yes I used an integer, but I had .323 (not 1-.323). Using K >=3 or K>2 (same thing because we are dealing with integers)

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