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    (Original post by Messier31)
    Oh

    Thanks



    I'll be expecting you in about half an hour :lol:
    For any other types of papers pure/mehanics yes. But stats no haha.



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    apart from the change in 2015 where you had to do qn 5, i dont think there is much room for changing stuff up - i feel we have seen the most of what can be thrown at us
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    (Original post by Euclidean)
    (Y) is a thumbs up emoji, probably not what you think
    Lol, enlighten me, what would one think?


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    (Original post by tazza ma razza)
    apart from the change in 2015 where you had to do qn 5, i dont think there is much room for changing stuff up - i feel we have seen the most of what can be thrown at us
    The beauty of A-level statistics modules. :moon:
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    june 14 paper 1R question 1c says state the assumption necessary for a PMCC to be valid for 1 mark, dont really understand the mark scheme just says something about bivariate normal or whatever, can anyone explain?
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    calculation of test statistic is wrong, they have used the wrong ones..
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    what do we do when we need to find a z value in the table like 3.77? i assumed 3.8 and wrote a little note saying round to 3.8. is that ok?
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    (Original post by tazza ma razza)
    what do we do when we need to find a z value in the table like 3.77? i assumed 3.8 and wrote a little note saying round to 3.8. is that ok?
    round to the nearest value on the table bruh
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    For normal goodness of fit tests, if we have to estimate mean and sd, do we use unbiased estimates?
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    (Original post by tazza ma razza)
    For normal goodness of fit tests, if we have to estimate mean and sd, do we use unbiased estimates?
    Yeah, unbiased is best because it's unbiased

    Unless you're told otherwise of course
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    Name:  Q7 June 2013.jpg
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    Can anyone help me with Question 7 part b on June 2013?
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    Anyone know which papers have a question where you need to test either the normal distribution or continuous uniform distribution as a suitable model? Thanks.
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    how would you use pooling in contingency tables? like check the 2015 paper (6f), which two columns would you pool together? male or female part or the colours??
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    (Original post by Music With Rocks)
    Name:  Q7 June 2013.jpg
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    Can anyone help me with Question 7 part b on June 2013?
    \text{Standard Error of the Mean}=\frac{\sigma}{\sqrt{n}}

    Notice that

    \displaystyle \frac{\sum_{i=1}^{n} X_i}{n} \times {n} = \bar{X} \times n = \sum_{i=1}^{n} X_i

    So we can combine the samples this way as two sums of X and then calculate the value for \sigma = \sqrt{s^{2}}

    Edit: This may still not be very clear with regards to combined variance:

    \displaystyle s^{2} = \frac{\sum_{i=1}^{n} X_i^{2} - n\left(\bar{X}\right)^{2}}{n-1}

    Therefore:

    \displaystyle \sum_{i=1}^{n} X_i^{2} = n\left(\bar{X})^{2} + s^{2}(n-1)

    We can add both sums for x and x^{2} and work as normal

    Music With Rocks
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    (Original post by ubahkassim)
    how would you use pooling in contingency tables? like check the 2015 paper (6f), which two columns would you pool together? male or female part or the colours??
    I don't think you'd ever be asked that for a contingency question, they'd modify the question to avoid it unless it was a (1) marker like it is in 2016 6(f).

    All you need to explain for the mark is that you'd expect the frequencies to be below 5 so some pooling would be necessary thereby reducing the degrees of freedom
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    (Original post by Euclidean)
    \text{Standard Error of the Mean}=\frac{\sigma}{\sqrt{n}}

    Notice that

    \displaystyle \frac{\sum_{i=1}^{n} X_i}{n} \times {n} = \bar{X} \times n = \sum_{i=1}^{n} X_i

    So we can combine the samples this way as two sums of X and then calculate the value for \sigma = \sqrt{s^{2}}

    Edit: This may still not be very clear with regards to combined variance:

    \displaystyle s^{2} = \frac{\sum_{i=1}^{n} X_i^{2} - n\left(\bar{X}\right)^{2}}{n-1}

    Therefore:

    \displaystyle \sum_{i=1}^{n} X_i^{2} = n\left(\bar{X})^{2} + s^{2}(n-1)

    We can add both sums for x and x^{2} and work as normal

    Music With Rocks
    I think I kind of get it? I will make sure to memorise that formula at least

    Name:  q7 june 2013 ans.jpg
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    After that part in the answer when it says "combined sample s^2:..." why is the 178.89^2 divided by 40 rather than multiplied by 40?
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    (Original post by Music With Rocks)
    I think I kind of get it? I will make sure to memorise that formula at least

    Name:  q7 june 2013 ans.jpg
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    After that part in the answer when it says "combined sample s^2:..." why is the 178.89^2 divided by 40 rather than multiplied by 40?
    \displaystyle s^{2} = \frac{\sum_{i=1}^{n} X_i^{2} - \boxed{n\left(\bar{X}\right)^{2}  }}{n-1}

    Referring only to the boxed area now:

    \displaystyle \bar{X} = \frac{\sum_{i=1}^{n} X_i}{n}

    Therefore:

    \displaystyle \bar{X}^{2} = \frac{\left(\sum_{i=1}^{n} X_i \right)^{2}}{n^{2}}

    So \displaystyle n\left(\bar{X}\right)^{2} = \frac{\left(\sum_{i=1}^{n} X_i \right)^{2}}{n^{2}} \times n = \frac{\left(\sum_{i=1}^{n} X_i \right)^{2}}{n}

    Where:

    \displaystyle \sum_{i=1}^{n} X_i = 178.89
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    (Original post by Euclidean)
    \displaystyle s^{2} = \frac{\sum_{i=1}^{n} X_i^{2} - \boxed{n\left(\bar{X}\right)^{2}  }}{n-1}

    Referring only to the boxed area now:

    \displaystyle \bar{X} = \frac{\sum_{i=1}^{n} X_i}{n}

    Therefore:

    \displaystyle \bar{X}^{2} = \frac{\left(\sum_{i=1}^{n} X_i \right)^{2}}{n^{2}}

    So \displaystyle n\left(\bar{X}\right)^{2} = \frac{\left(\sum_{i=1}^{n} X_i \right)^{2}}{n^{2}} \times n = \frac{\left(\sum_{i=1}^{n} X_i \right)^{2}}{n}

    Where:

    \displaystyle \sum_{i=1}^{n} X_i = 178.89
    Ah I get it now, thank you very much!

    You have been a great help!
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    (Original post by Zacken)
    The CLT states that if your sample size is large enough then the sample mean is normally distributed with the mean of the population and the variance of the population divided by the sample size.

    i.e: the sample mean may be 17.2, it doesn't mean the sample mean is distributed with mean 17.2.

    In part (b), this is precisely what you are dong, you are working out the confidence interval for the mean.

    I got part a and why the mean is a + 2. I'm uncertain as to how we're suddenly using the sample mean to find a's CI in part (b). I thought you're supposed to use the population mean?
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    (Original post by paradoxequation)
    Anyone know which papers have a question where you need to test either the normal distribution or continuous uniform distribution as a suitable model? Thanks.
    2015
 
 
 
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