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    (Original post by L'Evil Wolf)
    Statistic = If X1,X2,X3,...,Xn is a random sample of size n from some population, then a statistic T is a random variable consisting of any function of the Xi that involves no other unknown quantities. A stastic must not contain any unknown population paramaters.

    Is that okay? Or words to that effect



    Lol
    That sounds great.


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    (Original post by physicsmaths)
    That sounds great.


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    Sweet thnx
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    (Original post by L'Evil Wolf)
    Statistic = If X1,X2,X3,...,Xn is a random sample of size n from some population, then a statistic T is a random variable consisting of any function of the Xi that involves no other unknown quantities. A stastic must not contain any unknown population paramaters.

    Is that okay? Or words to that effect



    Lol
    Sounds fine to me. So anything involving mu, lambda, lowercase sigma.. isn't one
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    For the definition of a population parameter - would it be:

    a quantity or statistical measure that, for a given population, is fixed and that is used as the value of a variable in some general distribution or frequency function to make it descriptive of that population: The mean and variance of a population are population parameters.

    This is taken from Google - could not find anything else..
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    (Original post by SeanFM)
    Sounds fine to me. So anything involving mu, lambda, lowercase sigma.. isn't one
    what does lambda represent.

    sigma squared would aslso not be a statistic if used to represent variance?
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    (Original post by L'Evil Wolf)
    what does lambda represent.

    sigma squared would aslso not be a statistic if used to represent variance?
    Lambda is a rate parameter, at A-level it is commonly associated with the Poisson distribution.

    I don't think so, if the variance were unknown. If the variance were known.. I seem to remember the definition of a statistic being something that is solely calculated from the observations (i.e doesn't involve variance or anything other than observations) but that'd be in the book somewhere.
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    (Original post by SeanFM)
    Lambda is a rate parameter, at A-level it is commonly associated with the Poisson distribution.

    I don't think so, if the variance were unknown. If the variance were known.. I seem to remember the definition of a statistic being something that is solely calculated from the observations (i.e doesn't involve variance or anything other than observations) but that'd be in the book somewhere.
    yes lol ofc haha

    I see. From what is in the book sigma x /n is a statistic but for reasons earlier mentioned just mu is not a statistic. Essentially then these funny symbols, which are unknown are what we consider is a statistic or not.

    Thanks - what you said makes sense
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    Name:  Estimators q.jpg
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    Last question on this pls explain why 10 and not 10mu
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    (Original post by L'Evil Wolf)
    Name:  Estimators q.jpg
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    Last question on this pls explain why 10 and not 10mu
    What you've done is slightly messy.

    To illustrate a point, what you've got for E(Y) is correct, but what you really did was...

    Y = sum of those things which is correct.

    Since xi's are independent, identically distributed, this means that E(Y) = E(10*(x1-mu)/sd), and so using laws of expectation that gives 10*E(x1-mu)/E(sd) = 10 * (E(x1) - E(mu)) / E(sd)) = 10 * (mu - mu)/sd = 0.

    The problem is that you've changed Var(x1 - mu / sd) into Var ( var(x1) - mu / sd) which doesn't quite make sense - can you see what to do instead with your method? (It works, but it is very easy to make a mistake with it)
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    Another method is to find the distribution of one of them, and use that the sum of normally distributed things with mean mu1, mu2..mu10 and var1, var2... var 10 respectively is normally distributed with mean mu1 + mu2 ... + mu10 and variance var1 + var2... + var10.
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    (Original post by SeanFM)
    What you've done is slightly messy.

    To illustrate a point, what you've got for E(Y) is correct, but what you really did was...

    Y = sum of those things which is correct.

    Since xi's are independent, identically distributed, this means that E(Y) = E(10*(x1-mu)/sd), and so using laws of expectation that gives 10*E(x1-mu)/E(sd) = 10 * (E(x1) - E(mu)) / E(sd)) = 10 * (mu - mu)/sd = 0.

    The problem is that you've changed Var(x1 - mu / sd) into Var ( var(x1) - mu / sd) which doesn't quite make sense - can you see what to do instead with your method? (It works, but it is very easy to make a mistake with it)
    Spoiler:
    Show
    Another method is to find the distribution of one of them, and use that the sum of normally distributed things with mean mu1, mu2..mu10 and var1, var2... var 10 respectively is normally distributed with mean mu1 + mu2 ... + mu10 and variance var1 + var2... + var10.
    Thank you, this was helpful.
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    (Original post by L'Evil Wolf)
    Thank you, this was helpful.
    Do you understand how you use your method for finding Var(Y) now?
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    (Original post by SeanFM)
    Do you understand how you use your method for finding Var(Y) now?
    Sorry but I still end up with 10sigma^2/sigma giving me 10 sigma :/
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    (Original post by L'Evil Wolf)
    Sorry but I still end up with 10sigma^2/sigma giving me 10 sigma :/
    With the method you originally posted?
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    (Original post by SeanFM)
    With the method you originally posted?
    Yes similarly.

    I get that variance(mean) = 0

    It is the Var(Xi/sigma) which I don't get, is the var(Xi) not simply sigma^2.

    It obviously is sigma I suppose from the answer?
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    (Original post by L'Evil Wolf)
    Yes similarly.

    I get that variance(mean) = 0

    It is the Var(Xi/sigma) which I don't get, is the var(Xi) not simply sigma^2.

    It obviously is sigma I suppose from the answer?
    Looking at your answer - starting from the line Var(Y). The next 3 lines - all the sigma1/2s's should still be x1's /x2s etc as you should not have applied Var yet.

    Keep the xi's there until line 4, where everything else is correct, giving you 10Var((xi/sigma)-(mu/sigma)).

    This is now where you apply Variance, as it is of the form Var(aX+c), where a is 1/sigma and c is mu/sigma. And Var(aX+c) = a^2Var(x).. and the rest should follow.
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    (Original post by SeanFM)
    Looking at your answer - starting from the line Var(Y). The next 3 lines - all the sigma1/2s's should still be x1's /x2s etc as you should not have applied Var yet.

    Keep the xi's there until line 4, where everything else is correct, giving you 10Var((xi/sigma)-(mu/sigma)).

    This is now where you apply Variance, as it is of the form Var(aX+c), where a is 1/sigma and c is mu/sigma. And Var(aX+c) = a^2Var(x).. and the rest should follow.
    Thank you for doing this.

    omg how were you supposed to figure that sigma was a number/constant? Was it that they said the standard wdeviation was sigma in the question?
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    (Original post by L'Evil Wolf)
    Thank you for doing this.

    omg how were you supposed to figure that sigma was a number/constant? Was it that they said the standard wdeviation was sigma in the question?
    sigma is a constant in this situation, not a variable.

    mu/sigma is a number when you specify what mu / sigma are, eg mu = 5 sigma = 2.
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    (Original post by SeanFM)
    sigma is a constant in this situation, not a variable.

    mu/sigma is a number when you specify what mu / sigma are, eg mu = 5 sigma = 2.
    haha thank you for clearing that up for me
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    Hi, does anyone know please whether we could be asked to prove example 7 in the exam? Thanks Name:  image.jpg
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    (Original post by economicss)
    Hi, does anyone know please whether we could be asked to prove example 7 in the exam? Thanks Name:  image.jpg
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    Seems a bit bleh, but I suppose it might just be examinable but, y'know, not actually going to be asked it in the exam, I haven't really seen 'proofs' in the S3 past papers before. :dontknow:
 
 
 
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