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    (Original post by AliWilson96)
    Should've been 1.74 from memory?..
    Then did you use (mean of sample-mean)/ (sd/root n)?
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    (Original post by emma_1111)
    Yeah I got 1.74
    **** I did that first then crossed it out. I used population mean to work out standard deviation for last question

    by using sample mean to work out standard deviation i got critical value = 2.577 (something like that) then crossed it out as it looked weirdly big. I then used population mean to owrk out standard deviation and my critical value was something like 1.4 and critical was around 1.645 which looked ok, so i went with that
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    (Original post by Osmium)
    Yes. My point is that the textbook uses p and correlation. Have you noticed that?
    I hardly used the textbook, basically taught myself from markschemes and they have always had worded hypos for spearman and p=0 or pgreaterthan0 etc for product
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    I got around 62k for mean of second fan o_O anyone remember the question>?
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    Any opinions on grade boundaries for this paper? Personally I made some silly mistakes but the paper in general was pretty solid and simple. Hopefully it stays around 60-61 for an A.
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    (Original post by emma_1111)
    Then did you use (mean of sample-mean)/ (sd/root n)?
    Almost... (Sample mean - population mean) / (sd/root n)
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    (Original post by Bealzibub)
    I got around 62k for mean of second fan o_O anyone remember the question>?
    I don't remember the full details but it involved a second normal distribution where Y was distributed with an unknown standard deviation and mean. I got 6666.67 for sigma and something like 57-58k for the mean.
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    (Original post by Duskstar)
    My answers....

    (idk what is up with latex...)

    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    1.

    - \, \mathrm{graph}

    - \, \mathrm{doesnt fall within (rough) ellipse}

    \mathrm{so not bivariate normal}

    - \, d^2 = 366, r_s = -0.6636

    - \, \mathrm{not significant}

    - \, \mathrm{1% significance means 1% chance of}

    \mathrm{a false conclusion}

    - \, \mathrm{no assumptions for rank correlation}


    2.

- \, \mathrm{'random' \, means \, each \, gene \, has \, an \, equal}

\mathrm{chance \, of \, mutation}

\mathrm{'independent' \, means \, one \, gene \, mutating \, does}

\mathrm{not \, affect \, the \, probability \, of \, another \, mutating}

- \, X \sim B(20, 0.012), \, P(X=1) = 0.1908

- \, \mathrm{n \, is \, large, \, p \, is \, small}

- \, 0.0446

- \, 0.9826

- \, 0.0210 \, \mathrm{(continuity \, correction??)}

    3.

- \, 0.4722

- \, 95.02% \, \mathrm{therefore valid}

- \, h = 40093

- \, \mu = 58349, \, \sigma = 6515

- \, graph \, \mathrm{(X peak higher, on left.  Y 'flatter'<img src="images/smilies/wink.png" border="0" alt="" title=";)" class="inlineimg" />}

    4.

- \, e = 15.225, \, \chi^2 = 0.6831

- \, \mathrm{not \, significant}

- \, X \sim N(5.64, \, \dfrac{3.03701}{60})

\bar{x} = 6.2167, \, \sigma^2 = 3.03701

z = \dfrac{6.2167 - 5.64}{\sqrt{\dfrac{3.03701}{60}}  } = 2.5633

- \, \mathrm{significant}
    How did you get 3.03701 for variance?

    i remeber the data exactly



    If you don't use population mean to work out standard deviation but instead use sample mean ..
    then you get 1.74 like everyone else here, question is which one are we supposed to use

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    Does anyone know if I get any method marks if I forgot to do the continuity correction?
    and do I also get any method marks for using the wrong critical value in the last question.
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    (Original post by Bealzibub)
    How did you get 3.03701 for variance?

    i remeber the data exactly



    If you don't use population mean to work out standard deviation but instead use sample mean ..
    then you get 1.74 like everyone else here, question is which one are we supposed to use

    I think it's worthy to tell you that 1.7427^2 = 3.0370
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    (Original post by Duskstar)
    My answers....

    (idk what is up with latex...)

    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    1.

    - \, \mathrm{graph}

    - \, \mathrm{doesnt fall within (rough) ellipse}

    \mathrm{so not bivariate normal}

    - \, d^2 = 366, r_s = -0.6636

    - \, \mathrm{not significant}

    - \, \mathrm{1% significance means 1% chance of}

    \mathrm{a false conclusion}

    - \, \mathrm{no assumptions for rank correlation}


    2.

- \, \mathrm{'random' \, means \, each \, gene \, has \, an \, equal}

\mathrm{chance \, of \, mutation}

\mathrm{'independent' \, means \, one \, gene \, mutating \, does}

\mathrm{not \, affect \, the \, probability \, of \, another \, mutating}

- \, X \sim B(20, 0.012), \, P(X=1) = 0.1908

- \, \mathrm{n \, is \, large, \, p \, is \, small}

- \, 0.0446

- \, 0.9826

- \, 0.0210 \, \mathrm{(continuity \, correction??)}

    3.

- \, 0.4722

- \, 95.02% \, \mathrm{therefore valid}

- \, h = 40093

- \, \mu = 58349, \, \sigma = 6515

- \, graph \, \mathrm{(X peak higher, on left.  Y 'flatter'<img src="images/smilies/wink.png" border="0" alt="" title=";)" class="inlineimg" />}

    4.

- \, e = 15.225, \, \chi^2 = 0.6831

- \, \mathrm{not \, significant}

- \, X \sim N(5.64, \, \dfrac{3.03701}{60})

\bar{x} = 6.2167, \, \sigma^2 = 3.03701

z = \dfrac{6.2167 - 5.64}{\sqrt{\dfrac{3.03701}{60}}  } = 2.5633

- \, \mathrm{significant}
    For the meaning of random, could you also have said "The event that a gene mutates cannot be predetermined"?
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    (Original post by Bealzibub)
    How did you get 3.03701 for variance?

    i remeber the data exactly



    If you don't use population mean to work out standard deviation but instead use sample mean ..
    then you get 1.74 like everyone else here, question is which one are we supposed to use

    You should always use the sample mean when working out the sd, they normally say "assume the standard deviation is the same as the population" if they dont make you work it out
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    I have 4 queries, regarding these worded answers. Would appreciate it anyone could say if they agree/ disagree? I hate the worded ones.
    1) "Define 1% significance level"
    I put: "A 1% significance level means that the probability of rejecting the null hypothesis when it is in fact true is 1%/0.01"
    2) "Assumptions of underlying distribution for Spearmans test"
    I put something like: "There is none, as it is calculated using ranks and not the actual values, so it is not concerned with the actual underlying distribution."
    3) "Define random for the mutations of genes"
    I put something like: "Mutations being random means that their occurences are unpredictable."
    4) "Define independent for the mutations of genes"
    I put something like: "Mutations being independant means that the occurence of one mutations does not effect the probability of another mutation occuring."

    They're all a little weird to me, not sure?

    Other than that I think I got all the marks, but I'll be very upset if I got these wrong.
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    (Original post by AliWilson96)
    You should always use the sample mean when working out the sd, they normally say "assume the standard deviation is the same as the population" if they dont make you work it out
    how many marks will i lose if i did everything else correctly? it was 11mark question .. because of this mistake end conclusion was also wrong -,-
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    (Original post by ComputerMaths97)
    I have 4 queries, regarding these worded answers. Would appreciate it anyone could say if they agree/ disagree? I hate the worded ones.
    1) "Define 1% significance level"
    I put: "A 1% significance level means that the probability of rejecting the null hypothesis when it is in fact true is 1%/0.01"
    2) "Assumptions of underlying distribution for Spearmans test"
    I put something like: "There is none, as it is calculated using ranks and not the actual values, so it is not concerned with the actual underlying distribution."
    3) "Define random for the mutations of genes"
    I put something like: "Mutations being random means that their occurences are unpredictable."
    4) "Define independent for the mutations of genes"
    I put something like: "Mutations being independant means that the occurence of one mutations does not effect the probability of another mutation occuring."

    They're all a little weird to me, not sure?

    Other than that I think I got all the marks, but I'll be very upset if I got these wrong.
    1,3,4 are pretty much exact same answers me, i'm pretty sure they are model answers ... i got them from mark scheme
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    (Original post by Studious_Student)
    For the meaning of random, could you also have said "The event that a gene mutates cannot be predetermined"?
    Yeah I thought the point was that random means unpredictable, at least that's what it is in the June 2011 mark scheme
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    (Original post by Bealzibub)
    1,3,4 are pretty much exact same answers me, i'm pretty sure they are model answers ... i got them from mark scheme
    Ah sweet thanks bro! That fills me with a lot of confidence I thought that too just worried about how picky they'll be.

    Hopefully I got at least 70/72, then I should be good. How do you feel about how the exam went?
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    I used "insignificant" instead of "not significant". Would I still get marks for the conclusion? Because apparently "insignificant" is not correct :O
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    Which one are people talking about a continuity corrections?

    I only remember is for the 650 one, where it was P(x >/= 650) = P(Z > (649.5-600)/s.d)
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    (Original post by ComputerMaths97)
    I think it's worthy to tell you that 1.7427^2 = 3.0370
    how many marks will i lose if i did everything else correctly? it was 11mark question .
 
 
 
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