Stats 2 (S2) OCR (not MEI) 22nd June 2012

I hope they don't throw in any questions which require me to recall all those rules about permutations and combinations tomorrow, like they did in Jan 12... It's been 2.5 years since I did S1, and I do not remember that shizz at all!

Anyhow, good luck for tomorrow guys!

Reply 24

Original post by CraigKirk

I hope they don't throw in any questions which require me to recall all those rules about permutations and combinations tomorrow, like they did in Jan 12... It's been 2.5 years since I did S1, and I do not remember that shizz at all!

Anyhow, good luck for tomorrow guys!

Probably not because it's on jan 12. I did s1 in January and I didn't get them then :P

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Reply 25

Bwob

On the Jan 2009 paper, on question 8 part ii)
Can someone explain why you had to do 1/2n, when it came to the continuity correction?!?
I added 0.5 to the value, which is wrong :s-smilie:

I get why you have to do a correction, but not why it's 1/2n :/

Reply 26

Original post by Bwob

On the Jan 2009 paper, on question 8 part ii)
Can someone explain why you had to do 1/2n, when it came to the continuity correction?!?
I added 0.5 to the value, which is wrong :s-smilie:

I get why you have to do a correction, but not why it's 1/2n :/

(I can't see the paper at the moment, but this is from what I know)
The continuity correction is always 1/2n, but usually the sample size is 1, so it is 1/2.

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Reply 27

Original post by Bwob

On the Jan 2009 paper, on question 8 part ii)
Can someone explain why you had to do 1/2n, when it came to the continuity correction?!?
I added 0.5 to the value, which is wrong :s-smilie:

I get why you have to do a correction, but not why it's 1/2n :/

This isn't regarding that particular question, but this worked example should help you to understand.

Forty students each throw a fair dice twelve times. Each student counted the number of sixes scored in their twelve throws. Their lecturer then calculated the mean number of sixed scored per student. Find the probability that this mean was over 2.2.

First, since the students each throw a set number of times, with success being a 6 and failure being any other number, this distribution can be modelled by
X

~

B(12,\frac{1}{6} )

Now, to approximate this distribution using the normal distribution,

\mu = np

and

\sigma ^2 = npq

Therefore, using central limit theorem,

\bar X

~

N(12 \times \frac{1}{6},12 \times \frac{1}{6} \times \frac{5}{6} \times \frac{1}{40} ) = N(2, \frac{1}{24} )

(The

\frac {1}{40}

arises due to the 40 observations and because we're interested in the distribution of

\bar X

as opposed to

X

)

\bar X

represents the mean of 40 binomial variables, so it can be written in terms of

X_1, X_2, \ldots, X_{40}

as

\bar X = \frac {1}{40} (X_1+X_2+ \ldots+ X_{40})

You want to find

P(\bar X > 2.2)

.

All of the observations of

X

that make

\bar X

can be written in terms of

T

, where

T=X_1+X_2+ \ldots + X_{40}

P(\bar X > 2.2)=P(T>40 \times 2.2)=P(T>88)

. This is because there are forty values of

T

, so

\Sigma T = 2.2 \times 40 = 88

.

Since

T

is the total number of sixes gained in 480 throws,

T

~

B(480, \frac{1}{6})

.

To find

P(T>88)

, a continuity correction must be applied. So

P(V>88.5)

, where

V

is the appropriate normal approximation.

In terms of

\bar X

, the 0.5 added as the continuity correction for

V

is

\frac{1}{176}

of

88

. You need to have the same fraction of

2.2

added to that to find

P(\bar X > x)

with a continuity correction. This fraction of 2.2 is

\frac{1}{80}

, so you want

P(\bar X > 2.2+ \frac {1}{80})

.

Therefore, when applied to the mean, of a set of

n

variables, the continuity correction is

\frac {1}{2n}

, rather than

\frac {1}{2}

, as the correction

\frac {1}{2n}

happens to carry for all distributions of means.

That's hard to follow toward the end; if you don't understand it, just note that if you're dealing with the distributions of means (i.e.

\bar X or \bar Y or \bar W \ldots

), always use the continuity correction

\frac {1}{2n}

.

Also, this worked example is from a book, which happens to ignore the fact that the normal approximation at the start of the calculation is not a very good approximation, since

np=2\not > 5

. The rest of it works though, so don't let that put you off.

Reply 28

Killjoy-

8

Original post by CraigKirk

I hope they don't throw in any questions which require me to recall all those rules about permutations and combinations tomorrow, like they did in Jan 12... It's been 2.5 years since I did S1, and I do not remember that shizz at all!

Anyhow, good luck for tomorrow guys!

Lol, I wrote all the probabilities out one by one for that one!

Reply 29

Original post by Killjoy-

Lol, I wrote all the probabilities out one by one for that one!

I guess that if something comes up in the exam on P/Cs and I have a LOT of spare time at the end, I may do so too. Otherwise, since I've self-taught it, I'm not gonna waste my own time writing it all out! I'd rather just drop the few marks (only two on this occasion, luckily!).

Reply 30

On June 2008 Q7)ii) I dont understand where the markscheme gets 16 from. Could you help me please anyone? :smile:

Oh the probability was >..... I was working with the wrong value!!!

(edited 11 years ago)

Reply 31

Original post by Matthew692692

On June 2008 Q7)ii) I dont understand where the markscheme gets 16 from. Could you help me please anyone? :smile:

Oh the probability was >..... I was working with the wrong value!!!

Unparseable latex formula:

P(X \geq x) <0.05[br]\[br]P(X \leq x-1) > 0.95[br]\[br]P(X \leq 15)=0.9513>0.95[br]\[br]x-1=15 \Rightarrow x = 16

Reply 32

Original post by CraigKirk

Unparseable latex formula:

P(X \geq x) <0.05[br]\[br]P(X \leq x-1) > 0.95[br]\[br]P(X \leq 15)=0.9513>0.95[br]\[br]x-1=15 \Rightarrow x = 16

Thank you!! It makes much more sense now :-)

Reply 33

How do you calculate the critical region? What's the formula? My revision book doesn't have it, and I can't find the page from my textbook

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Reply 34

Original post by Matthew692692

How do you calculate the critical region? What's the formula? My revision book doesn't have it, and I can't find the page from my textbook

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(\pm) z =\displaystyle \frac {\bar x - \mu}{\frac{\sigma}{\sqrt n}} \Rightarrow \bar x = (\pm) z \frac{\sigma}{\sqrt n} + \mu

The

\pm

signs depend on which critical region you're after. If it's a critical region above the mean, use +. If it's below the mean, use -. If it's a two-tail test, use

\pm

.

Reply 35

Original post by CraigKirk

(\pm) z =\displaystyle \frac {\bar x - \mu}{\frac{\sigma}{\sqrt n}} \Rightarrow \bar x = (\pm) z \frac{\sigma}{\sqrt n} + \mu

The

\pm

signs depend on which critical region you're after. If it's a critical region above the mean, use +. If it's below the mean, use -. If it's a two-tail test, use

\pm

.

Thank you!! Now to learn it for the exam!!!

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Reply 36

Sgt.Incontro

20

Can someone explain type II errors? I really don't get it.

Cheers.

Reply 37

OCR S2 Revision Sheet.pdf89.3KB

Original post by Sgt.Incontro

Can someone explain type II errors? I really don't get it.

Cheers.

A type II error is when a false Ho is accepted. i.e. you shouldn't have come to that conclusion, and H1 should actually have been accepted.
To calculate a Type II error, you use the same values you did for the hypothesis test, but instead you use the new value for

\mu

I'm not sure how clearly I've explained it, I'm not very good at explaining things, but i hope it makes sense...

Reply 38

Sgt.Incontro

20

Original post by Matthew692692

A type II error is when a false Ho is accepted. i.e. you shouldn't have come to that conclusion, and H1 should actually have been accepted.
To calculate a Type II error, you use the same values you did for the hypothesis test, but instead you use the new value for

\mu

I'm not sure how clearly I've explained it, I'm not very good at explaining things, but i hope it makes sense...

Thanks mate, I understand it now.

The only thing I don't understand now are calculating critical regions for hypothesis tests.

Good luck mate.

ALSO: For you and anyone else, this might also come in useful.

Reply 39