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# S2 Question Watch

1. http://static1.squarespace.com/stati...+June+2015.pdf

Very last question, 8ii)

I have no idea how to do this, apart from draw a tree diagram with 81 branches.

MS

http://static1.squarespace.com/stati...ark+scheme.pdf
2. (Original post by 16characterlimit)
http://static1.squarespace.com/stati...+June+2015.pdf

Very last question, 8ii)

I have no idea how to do this, apart from draw a tree diagram with 81 branches.

MS

http://static1.squarespace.com/stati...ark+scheme.pdf
Do you follow the reasoning in the mark-scheme? They work out

P(Type II error | p = 0.5) = 0.6047

and

P(Type II error | p = 0.7) = 0.0933

(what happens to the p = 0.3 case?)

Then they work out P(Type II error) in one test (where p is randomly chosen), then they work out P(Type II error) in at least one of the four tests, assuming that the tests are independent.
3. (Original post by Gregorius)
Do you follow the reasoning in the mark-scheme? They work out

P(Type II error | p = 0.5) = 0.6047

and

P(Type II error | p = 0.7) = 0.0933

(what happens to the p = 0.3 case?)

Then they work out P(Type II error) in one test (where p is randomly chosen), then they work out P(Type II error) in at least one of the four tests, assuming that the tests are independent.
Yeah, I got those, its just there's so many outcomes, such as p = 0.3 then p = 0.5 then p = 0.7, and you can't do a binomial distribution or any distribution if there's not a fixed chance of occurrence
4. (Original post by 16characterlimit)
Yeah, I got those, its just there's so many outcomes, such as p = 0.3 then p = 0.5 then p = 0.7, and you can't do a binomial distribution or any distribution if there's not a fixed chance of occurrence
Once you've got P(Type II error | p = 0.5) and P(Type II error | p = 0.7), you've eliminated any need to think about distributions from there on in. You use the law of total probability to get P(Type II error), and then work out the probability of (at least) one of four independent events occurring.
5. (Original post by Gregorius)
Once you've got P(Type II error | p = 0.5) and P(Type II error | p = 0.7), you've eliminated any need to think about distributions from there on in. You use the law of total probability to get P(Type II error), and then work out the probability of (at least) one of four independent events occurring.
In that case I would work out 1 - P(NoTypeII) right?

So P(NoTypeII) could be P(p=0.3)^4 or P(p=0.5)*P(TypeIIp=0.5)^(3 or 2 or 1) or P(p=0.7)*P(TypeIIp=0.7)^(3 or 2 or 1) and so on...

I don't think I get it, because there is way to many possible things to consider.
6. (Original post by 16characterlimit)

So P(NoTypeII) could be P(p=0.3)^4 or P(p=0.5)*P(TypeIIp=0.5)^(3 or 2 or 1) or P(p=0.7)*P(TypeIIp=0.7)^(3 or 2 or 1) and so on...
No no. Use the law of total probability to get P(Type II), unconditional on anything.

P(Type II)= (1/3) P(Type II | P = 0.7) + (1/3) P(Type II | P = 0.5) + (1/3) P(Type II | P = 0.3)
7. (Original post by Gregorius)
No no. Use the law of total probability to get P(Type II), unconditional on anything.

P(Type II)= (1/3) P(Type II | P = 0.7) + (1/3) P(Type II | P = 0.5) + (1/3) P(Type II | P = 0.3)
That gives the right answer, but isn't that for a single test? The question says 4 tests are carried out, so I thought you need to consider this and all the possible subsequent events.
8. Here's my working, had to write it up because Latex isn't working nicely at the moment..

Obviously it would take too much time to work with all combinations of type (II) errors so 1-p(No type (II) error) is the way to go.
9. (Original post by Parallex)
Here's my working, had to write it up because Latex isn't working nicely at the moment..

Obviously it would take too much time to work with all combinations of type (II) errors so 1-p(No type (II) error) is the way to go.
Thank's that all makes sense now.
10. (Original post by 16characterlimit)
That gives the right answer, but isn't that for a single test? The question says 4 tests are carried out, so I thought you need to consider this and all the possible subsequent events.
There are four tests, but they are independent of each other. All the variety of possible combinations are dealt with by using the law of total probability, followed by the independence assumption of the tests.

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