Edexcel S4 - Friday 24th June 2016 AM

Thought S4 would be hard having not looked at it at all until today (too many other exams!), but it doesn't seem to bad in truth. Really helps that there's so much overlap with AQA Statistics too, as I have S4 and SS05 on the same day >.<

Yeah, it's not sooo bad though out of interest, how/why are you doing SS05?

Doing all 18 Edexcel modules on top of the 6 AQA Statistics ones. People say it's a waste of time, I say it's a challenge

Fairs I wouldn't think you're sitting all 24 in one year, but that is still a bit.. stressful

Please link the paper

https://7c9a99abd139eb14bde7da77d86ad3346f91c6f3.googledrive.com/host/0B1ZiqBksUHNYOFBYbFFYWW9iOWs/June%202013%20(R)%20QP%20-%20S4%20Edexcel.pdf
Hope it works. It's on Physics and Maths tutor.

Would anyone be able to give me a hint for this one? I don't know how to work with the sums of X^2 that come with S^2 etc.

And does anyone know why they gave me information about a chi-square variable?

@SeanFM

Well.. starting with 7, what have you tried/thought about?

I've gotta be honest, I'm not sure on this one. I've had a look at all of the definitions and stuff but can't really think of what to do.

It might have something to do with this: if it comes up in S4..

Please don't. I don't understand why they used the Critical value equation from part a) and used another one in b) (with mu=200 and z=1.6449). I got the numbers but I didn't understand what to do with them. How would I calculate P(type II error) in this specific case (not the general definition of accepting Ho when it's false).
8 I started to understand.

I've gotta be honest, I'm not sure on this one. I've had a look at all of the definitions and stuff but can't really think of what to do.

It might have something to do with this: if it comes up in S4..

Suggested solution: It is necessary to first show that S^2 is an unbiased estimator of sigma squared. By rearranging the equation above to make S^2 the subject we get: (Chi squared)(sigma squared)/(n-1), the chi squared variable can be replaced by the random variable Y.

It is known that E(Y) = V, therefore:

E(S^2) = E(Y)(sigma squared)/(n-1) = V(sigma squared)/(n-1)

In general V = degrees of freedom = n-1.
Replacing V with n-1:
E(S^2)= (n-1)(sigma squared)/(n-1)
Common terms n-1 cancel out. Hence E(S^2) = sigma squared is an unbiased estimator.

Using the same idea, Var(S^2) can be calculated to be equal to:
2(sigma)^4/(n-1). Following this, as n approaches infinity, the variance of S^2 approaches zero. It is therefore a consistent estimator.

In general, we can use the chi squared variable Y to derive the properties of S^2.

Nice one. Thanks for that

I would've thought it wasn't necessary to show that it was unbiased (i.e irrespective of whether it is unbiased or not, it can still be consistent?)

you always gotta do the bias test - otherwise it is an unsuitable estimator.

https://7c9a99abd139eb14bde7da77d86ad3346f91c6f3.googledrive.com/host/0B1ZiqBksUHNYOFBYbFFYWW9iOWs/June%202015%20QP%20-%20S4%20Edexcel.pdf

Qn 4 part c i actually don't understand where to begin - I drew out a table of probabilities but then it got me no where (unless i did it wrong...which i doubt)

Use the definition of expectation in the discrete case (where you write all the values that it can take, and multiply it by the probability) i.e $E(X) = x_1 P(X=x_1) + x_2 P(X=x_2)...$.