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    Need help with q3 from this paper.

    http://www.ocr.org.uk/Images/62244-q...atistics-2.pdf

    I have worked out the Critical value to be 17.0925.

    The bit I dont fully understand is the type 2 error thing.

    It is P(accept Ho| H1 is true) , but I am not sure what that means tbh.

    Thanks
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    (Original post by Super199)
    Need help with q3 from this paper.

    http://www.ocr.org.uk/Images/62244-q...atistics-2.pdf

    I have worked out the Critical value to be 17.0925.

    The bit I dont fully understand is the type 2 error thing.

    It is P(accept Ho| H1 is true) , but I am not sure what that means tbh.

    Thanks
    It's a bit like conditional probability, but under the assumption that mu = 20 you've found the values for which you will reject H0, so now you're told that mu is actually 15, and with that what's the probability that you'll get those values again and so not reject H0?

    P(accept Ho| H1 is true) just means that mu = 15 but you get values which suggest that H0 is true.
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    (Original post by SeanFM)
    It's a bit like conditional probability, but under the assumption that mu = 20 you've found the values for which you will reject H0, so now you're told that mu is actually 15, and with that what's the probability that you'll get those values again and so not reject H0?

    P(accept Ho| H1 is true) just means that mu = 15 but you get values which suggest that H0 is true.
    hmm I am sort of confused. Does the critical value show us the rejection region?
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    (Original post by Super199)
    hmm I am sort of confused. Does the critical value show us the rejection region?
    As this is a one tailed test, you can find your critical region (the rejection region) by looking at one side of the critical value which you work out based on the hypotheses. (In this case it will be values that are less than the critical value - can you see why?)
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    (Original post by SeanFM)
    As this is a one tailed test, you can find your critical region (the rejection region) by looking at one side of the critical value which you work out based on the hypotheses. (In this case it will be values that are less than the critical value - can you see why?)
    Not really something I have just accepted tbh. The area under to the left of the critical value is the significance level which shows the values you deem to be unlikely?
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    (Original post by Super199)
    Not really something I have just accepted tbh. The area under to the left of the critical value is the significance level which shows the values you deem to be unlikely?
    I think you've kind of got it, but I'm not sure what you mean there.

    As a rule of thumb the critical region is where, if x is in the critical region, P(X=x) gets smaller and smaller. If you think about a binomial distribution with say n = 20 and we're testing whether p = 0.8 or less than 0.8, and we (made up) found a critical value that if x is less than 4, then we reject that p = 0.8. We see that P(X=4) > P(X=3) > P(X=2)... so the x values are getting less and less likely, and the sum of their probabilities make up the significance level, and those values are the 'unlikely' values. If we took x>4 it wouldn't really make sense.

    Most, if not all of the time, in examples with the mean you should be 'moving away' from the mean rather than going towards it (in this case the mean was 16 (np = 20*0.8) and we're moving away from it (x<4) rather than (x>4).
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    (Original post by SeanFM)
    I think you've kind of got it, but I'm not sure what you mean there.

    As a rule of thumb the critical region is where, if x is in the critical region, P(X=x) gets smaller and smaller. If you think about a binomial distribution with say n = 20 and we're testing whether p = 0.8 or less than 0.8, and we (made up) found a critical value that if x is less than 4, then we reject that p = 0.8. We see that P(X=4) > P(X=3) > P(X=2)... so the x values are getting less and less likely, and the sum of their probabilities make up the significance level, and those values are the 'unlikely' values. If we took x>4 it wouldn't really make sense.

    Most, if not all of the time, in examples with the mean you should be 'moving away' from the mean rather than going towards it (in this case the mean was 16 (np = 20*0.8) and we're moving away from it (x<4) rather than (x>4).
    got it cheers
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    Can anyone answer this??? Suppose that A, B and C are events in a sample space S, such that A∪B∪C = S and A∩B∩C= ∅ If P(A) = 0.139, P(B) = 0.268, P(A∩B)= 0.045, P(B∩C)= 0.204 and P(A∩C) = 0.055, find P(C), and P(A|C).
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    (Original post by AFraggers)
    Can anyone answer this??? Suppose that A, B and C are events in a sample space S, such that A∪B∪C = S and A∩B∩C= ∅ If P(A) = 0.139, P(B) = 0.268, P(A∩B)= 0.045, P(B∩C)= 0.204 and P(A∩C) = 0.055, find P(C), and P(A|C).
    Draw a Venn diagram.

    In the future it may be helpful to make your own thread
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    (Original post by SeanFM)
    Draw a Venn diagram.

    In the future it may be helpful to make your own thread
    Do you know how to answer this question??
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    (Original post by AFraggers)
    Do you know how to answer this question??
    I think so

    How about you, how would you start to answer this question?
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    (Original post by SeanFM)
    I think so

    How about you, how would you start to answer this question?
    Well i know that A, B and C are mutually exclusive as the intersections equal an empty set. then i dont really know where to go from there to find P(C).
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    (Original post by AFraggers)
    Well i know that A, B and C are mutually exclusive as the intersections equal an empty set. then i dont really know where to go from there to find P(C).
    Draw a Venn diagram with all of the information given, then see if you can work it out.

    Or equivalently, you can use the inclusion exclusion principle if you've come across that.
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    (Original post by SeanFM)
    Draw a Venn diagram with all of the information given, then see if you can work it out.

    Or equivalently, you can use the inclusion exclusion principle if you've come across that.
    ive drawn the intersection but im still stuck. ive looked up the principle but it still hasnt helped me
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    (Original post by AFraggers)
    ive drawn the intersection but im still stuck. ive looked up the principle but it still hasnt helped me
    Okay, you've drawn the intersection of all 3, which is good, and that should have nothing in it. You can fill in the three sets of intersections (where it's A intersect B etc)... and then look at P(A) and P(B) and see what you have to take away because of the intersections.. and that should leave you with P(C).
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    (Original post by SeanFM)
    Draw a Venn diagram with all of the information given, then see if you can work it out.

    Or equivalently, you can use the inclusion exclusion principle if you've come across that.
    I know that once I find P(C) I can easily find P(A|C)
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    (Original post by AFraggers)
    I know that once I find P(C) I can easily find P(A|C)
    Good, then you just need to find P(C) using the steps I've hinted at in my previous post.
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    (Original post by SeanFM)
    Okay, you've drawn the intersection of all 3, which is good, and that should have nothing in it. You can fill in the three sets of intersections (where it's A intersect B etc)... and then look at P(A) and P(B) and see what you have to take away because of the intersections.. and that should leave you with P(C).
    So it should look something like this...
    Attached Images
     
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    Not quite - you've got the intersection values correct but you haven't accounted for them when it comes to just P(A) and P(B).

    Remember that P(A) = P(A but not B or C) + P(A intersect B) + P(A intersect B) + P(A intersect B intersect C). If you shade in the circle A, starting with where it doesn't intersect, by the time you have shaded the circle you will have circled all of the points in that formula so all of that makes up A. Do you see what I mean?
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    (Original post by SeanFM)
    Not quite - you've got the intersection values correct but you haven't accounted for them when it comes to just P(A) and P(B).

    Remember that P(A) = P(A but not B or C) + P(A intersect B) + P(A intersect B) + P(A intersect B intersect C). If you shade in the circle A, starting with where it doesn't intersect, by the time you have shaded the circle you will have circled all of the points in that formula so all of that makes up A. Do you see what I mean?
    Yeah I kind of understand.If you were doing this question using the inclusion- exclusion formula, how would you get to your answer??
 
 
 
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