MAY11 Math HL paper 3

[Snoopchicken]

Question 4? Oh, I got the SL is 17% I think.

They give you the SL is defined by x=<22 and x>=38. So, what you do, is find the Poisson distribution (where the mean becomes 30 now, since you are taking 10 samples), and add the critical regions together:

P (X=<22) = poissoncdf(30,22)

P (x>=38) = 1 - poissoncdf(30,37)

SL = P(X=<22) + P(X>=38) which is approximately 17.0%.

And yes, I am positive that 96.5% is correct, and your t-value sounds familiar to me

Edited May 13, 2011 by Snoopchicken

[Nishad Potdar]

Question 4? Oh, I got the SL is 17% I think.

They give you the SL is defined by x=<22 and x>=38. So, what you do, is find the Poisson distribution (where the mean becomes 30 now, since you are taking 10 samples), and add the critical regions together:

P (X=<22) = poissoncdf(30,22)

P (x>=38) = 1 - poissoncdf(30,37)

SL = P(X=<22) + P(X>=38) which is approximately 17.0%.

And yes, I am positive that 96.5% is correct, and your t-value sounds familiar to me

Ouch I used the Normal approximation. X~Po(3)=X~N(3,3)... Do you think they won't reward this method? Well, I did solve part (b) using Poisson Dist only. How did you do part (b)? Did you use the acceptance region as 22<x<38 or 22<=x<=38? I used the latter. I think I'm wrong

[Snoopchicken]

I'm not too sure if they'd reward that method. I'm sure they'll give something - I mean, the examiners have hearts, right?

Yeah, the latter is wrong Sorry. It's 22<x>38, and thus, to calculate the Type II error:

Po(25) (since H1 is that the mean is 2.5)

P(22<x<38) = poissoncdf(25,37) - poissoncdf(25,22)

[JayC]

@Nishad, yeah, I got 96.6% And in the beginning of question 4 (part a), they tell you to use the significant level of 1%. But that's a whole other story - that's a one tailed test.

Part b was pretty easy, just not conventional. First, you find the difference between the 2 values they give in the confidence interval. This is equal to 2tSn-1/rad(n), and you have everything except "t", so obviously you find "t" from this. After you get that value of "t", you should recognize that the CI is 2 tailed and do "tcdf" on the calculator from "-t" to "+t", and degrees of freedom 9. It ends up being like 0.965... and I remember I rounded the 3rd significant digit up.

It's right BTW, I was able to check using GDC (I performed T-Interval)

BTW, JayC, for part B in question 5, the answer is 6, correct? You try to find the value of Y which gives the highest probability in the NB distribution, and thus, you look for the integer value of Y which yields the value closest to 0 in f'(x)/f(x) (function is at max when derivative=0).

I believe so, but I was so rushed and stupid enough to round it to 7. But I did write 6.xxx first, so I hope I get one mark for method or something here.

[Nishad Potdar]

I'm not too sure if they'd reward that method. I'm sure they'll give something - I mean, the examiners have hearts, right?

Yeah, the latter is wrong Sorry. It's 22<x>38, and thus, to calculate the Type II error:

Po(25) (since H1 is that the mean is 2.5)

P(22<x<38) = poissoncdf(25,37) - poissoncdf(25,22)

Oops! But I should get some marks (2-3 maybe/10) :/

[Snoopchicken]

Yeah definitely, for the Type II, they would give you probably 1 point for writing what a type II error was, 1 point for finding the new mean of the poisson distribution, and 1 point for method (with incorrect values though), so don't worry about it