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    Please help with part iii, where did they get 0.2 from? Name:  image.png
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    And if you have any additional tips to getting better at statistics other than practise practise practise, please share it 😭 The statistic paper mark scheme doesn't help me at all, especially when random numbers just come up, they should at least mention when they are using a table value 😭
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    (Original post by Christina Tiana)
    Please help with part iii, where did they get 0.2 from? Name:  image.png
Views: 154
Size:  111.9 KB

    And if you have any additional tips to getting better at statistics other than practise practise practise, please share it 😭 The statistic paper mark scheme doesn't help me at all, especially when random numbers just come up, they should at least mention when they are using a table value 😭
    1 day in 5 = 1/5 = 0.2
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    (Original post by Zacken)
    1 day in 5 = 1/5 = 0.2
    Thank you,

    I'm not gonna survive s2 👋🏼
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    (Original post by Zacken)
    1 day in 5 = 1/5 = 0.2
    Can you explain the calculation they did? I now know how they got the value, but I don't understand it, so 0.14154 is the probability of type 2 error on that day, why would I times that by the probability of type 2 error not occurring?(0.2923) and I add on the probablility of the type 2 error not occurring on that day (0.85846) times the probability of type 1 error(0.0481), why?????
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    Someone help me understand this please?
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    (Original post by Christina Tiana)
    Someone help me understand this please?
    I must admit that the first explanation given seems pretty obscure. I'll have a think about it!

    But for the moment, do you see how the second explanation works (the one labelled "OR" - I think that one is much clearer. You get to it by drawing a tree diagram. The first branch is a split for p = 0.5 or 0.6 (with probabilities 0.2 and 0.8 respectively) then the second split is Reject/Do not Reject H0 (with probabilities that you can read off the answers you've already got) and the third branch split is again Reject/Do not Reject H0 - but this time you must remember that p is adjusted to 0.6 if the null hypothesis is rejected first time around in the 0.5 branch.
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    (Original post by Christina Tiana)
    Can you explain the calculation they did? I now know how they got the value, but I don't understand it, so 0.14154 is the probability of type 2 error on that day, why would I times that by the probability of type 2 error not occurring?(0.2923) and I add on the probablility of the type 2 error not occurring on that day (0.85846) times the probability of type 1 error(0.0481), why?????
    What they are giving here is an argument based on conditioning; so they are aiming to use something of the form:

    \displaystyle P(A) = P(A|B)P(B) + P(A|\bar{B})P(\bar{B})

    So the question, what event are they conditioning on, that is, what is B. If you disentangle it, what they are trying to compute is P(Reject 2nd H0) by conditioning on the value of p at the time of the second test. So they have

    P(Reject 2nd H0) = P(Reject 2nd H0) | p=0.5) P(p=0.5)
    + P(Reject 2nd H0 | p=0.6) P(p=0.6)

    Their first calculation is to find P(p=0.5) at the time of the second test. The only way that this can happen is if a type II error has occurred on the days that p=0.5 to start with. Hence P(p=0.5) = 0.2 x 0.7077 = 0.14154. But now all the other terms fall out of what we have already. P(p=0.6) = 1 - P(p=0.5) = 0.85846. We know that P(Reject 2nd H0 | p=0.6) = 0.0481 and we can work out that P(Reject 2nd H0) | p=0.5) = 1 P(Type II error) = 1 - 0.7077 = 0.2923.
 
 
 
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