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# Edexcel S3 - Wednesday 25th May AM 2016 watch

1. (Original post by Zacken)
Nope, then you work out v with the new number of rows and columns.
Oh so if one of the 'squares' of the table have frequencies<5, the entire column is merged with the one on the left?
2. (Original post by thesmallman)
Oh so if one of the 'squares' of the table have frequencies<5, the entire column is merged with the one on the left?
Yes. That's what I said on the previous page?
3. Zacken sorry for bothering you again, but:

Attachment 537927537929
Attachment 537927537929537900

I'm talking about part b of the question. I used interpolation and got 0.8418 (to 4dp), and they don't seem to be accepting that in the mark scheme. Why? How else would I do it? Btw I attached the mark scheme and part of the normal distribution table.
Attached Images

4. (Original post by gagafacea1)
Zacken sorry for bothering you again, but:

Attachment 537927537929
Attachment 537927537929537900

I'm talking about part b of the question. I used interpolation and got 0.8418 (to 4dp), and they don't seem to be accepting that in the mark scheme. Why? How else would I do it. Btw I attached the mark scheme and part of the normal distribution table.
Convert it to P( C - A > 0 ) = 0.2 and then use the table (percentage point table on the page after the normal distrition table/below it) where it gives P(Z > 0.8416) = 0.2 so you can ake it from there.

[sorry for the terse reply, currently in bed and on my phone]
5. (Original post by zacken)
convert it to p( c - a > 0 ) = 0.2 and then use the table (percentage point table on the page after the normal distrition table/below it) where it gives p(z > 0.8416) = 0.2 so you can ake it from there.

[sorry for the terse reply, currently in bed and on my phone]
yess!!! Thank you!!!
6. How the hell is part A done
7. (Original post by ArafH)
How the hell is part A done
From the interval, you should be able to spot the 'midpoint' (for want of a better word, the xbar if you like) as that 90% interval is xbar +/- sd/sqrt(n) * z(0.95) where z(0.95) is the appropriate z value. So that interval is symmetric, so this tells you that ...

Then, as you know what xbar is, you can work what the interval is going to be (n is unknown but you can use what I've said above to leave it as an unknown and still work out the confidence interval).
8. (Original post by SeanFM)
From the interval, you should be able to spot the 'midpoint' (for want of a better word, the xbar if you like) as that 90% interval is xbar +/- sd/sqrt(n) * z(0.95) where z(0.95) is the appropriate z value. So that interval is symmetric, so this tells you that ...

Then, as you know what xbar is, you can work what the interval is going to be (n is unknown but you can use what I've said above to leave it as an unknown and still work out the confidence interval).
Thanks soooo much Sean mate
9. One last,
I don't want to lose accuracy marks so what do i round all my fina answers to?
In mechanics it is 3sf so what about Stats?

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10. (Original post by physicsmaths)
One last,
I don't want to lose accuracy marks so what do i round all my fina answers to?
In mechanics it is 3sf so what about Stats?

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usually 2dp for values for means/variances in unbiased estimates. 4dp for probability and 3dp for test statistics. Just give it to the same accuracy in the question or 3sf generally
11. When do we use two tailed testing apart from confidence intervals?

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12. Sorry, last minute panic, the central limit theorem means we assume the population mean or sample mean is normally distributed please? Thanks
13. (Original post by 260498)
When do we use two tailed testing apart from confidence intervals?

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Whether the qn asks for it lol w/o being a smart ass
14. (Original post by tazza ma razza)
Whether the qn asks for it lol w/o being a smart ass
Oh, thank god I was going through a last minute panic. Thank you

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15. (Original post by economicss)
Sorry, last minute panic, the central limit theorem means we assume the population mean or sample mean is normally distributed please? Thanks
the latter
16. (Original post by Zacken)
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...]

Yep! Love your handwriting, by the way! And thank you for helping out tonight, really saved me from having to answer tons of questions.
Never thought id hear that from anybody xD, Thank you! & anytime, I noticed you do get drowned usually!
17. (Original post by 260498)
When do we use two tailed testing apart from confidence intervals?

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Sometimes it says test "whether or not" something is a certain value. This means H0 = That value, and H1 is not that value, so you conduct a two tailed test because a "negative" answer or "positive" answer both counts as not the answer you want.

Hope that helps!
18. When it says what issignificance of CLT,
I am not sure that saying 'assume' is correct,
Can I say, 'relevance of CLT is that since samples are large it assures that these sample means are normally distributed'

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19. (Original post by lai812matthew)
the latter
Thanks!!
20. (Original post by physicsmaths)
When it says what issignificance of CLT,
I am not sure that saying 'assume' is correct,
Can I say, 'relevance of CLT is that since samples are large it assures that these sample means are normally distributed'

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I'm pretty sure the best wording is along the lines of "as the samples are large, we are able to assume that the sample means are distributed approximately normally".

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