Edexcel S3 - Wednesday 25th May AM 2016

Pretty sure the textbook says the proof isn't examinable.

Its not letting me upload it so i put it on imgur.

When you write Sum of O² / E , You have to write every number your adding together, basically showing how your adding all your observed frequenices squared divided by your expected frequencies and how your taking away the original Total and then how you got the Chi-squared value. Then compared the Chi-squared value with the critical value from the table showing if its in the critical region or not. Then stating your conclusion.

It is June 2014 Question 6.

I'm not sure why a proof like that would only be 1 mark?

Putting it up here for you, hope you don't mind:

What's this total error and total mean error i've seen people talking about?!

Thanks!

What's this total error and total mean error i've seen people talking about?!

I thought he was asking about proving that $s^2$ was an unbiased estimator of $\sigma^2$ which isn't examinable. (un)biased estimators for other functions are most definitely examinable though, yes.

I thought he was asking about proving that $s^2$ was an unbiased estimator of $\sigma^2$ which isn't examinable. (un)biased estimators for other functions are most definitely examinable though, yes.

Nothing. Ignore it, it's coming from nowhere.

It's because $U_1$ and $\bar{U}$ are not independent. The formula you quoted are only for independent random variables $X,Y$. They are not independent because they come from the same sample, but the important thing here is that they are not independent.

Why isn't it examinable? It's a fairly straight forward proof...

HAHA! I reckon its people doing IAL, @Zacken is that right? ( I know nothing about IAL )

Cool, I see that now. But then isn't the method to working out the variance as 4sigma^2/5 using the same formula?

Is the mark scheme attached not using Var(F) = Var(U1-Ubar) = Var(U1) + Var(Ubar) ??
What is different about it?

It is using the same formula because it's written $\bar{U}$ in terms of $U_1, U_2, \cdots, U_5$ all of which are independent of one another.

It's fairly algebraically dense, see page 29 of the textbook. Anyhow, lots of proofs are straighforward, doesn't mean they're examinable.



Nope, IAL and UK have the exact same specification. Total error doesn't mean anything in terms of S3-level stuff.

Thank you very much, it makes sense to me now.

One last question, why are Ubar and U1 not independent to each other, but U1, U2, ..., U5 are?

Ohhh, someone really shook the whole forum up!

Well, think of it, two things being independent means knowing something about one thing gives you no information about the other. And the other way around for dependent.

So, if we know something about U1, that gives us a bit of information about Ubar since we calculate Ubar using U1 (and other stuff), but the key thing is that knowing U1 or U2 or U3 or U4 or U5 gives us information about Ubar.

So they are not independent.

However, knowing something about U1 gives you no information whatsoever about U2 or U3 or anything, hence U1 and U2 are independent, same for U2 and U3 and U1 and U3 and etc...

Remember that just because the things are all independent to one another (U1 independent to U2 to U3 to U4, etc...) doesn't mean that the sum of those things (Ubar) is also independent of them.

[Lost the will to LaTeX...]