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    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')}

    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}
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    What did you all get for rank correlation?
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    (Original post by Crozzer24)
    Had no idea but I've been told you must assume the data is not linear
    spearmans can be used whenever pearsons can't (and you can still use spearmans even if pearsons can be used) - so wether its linear or non linear, or come from a bivariate normal distribution or not. so I put down no assumptions
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    Pretty solid paper imo
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    (Original post by gemdarkstone)
    spearmans can be used whenever pearsons can't (and you can still use spearmans even if pearsons can be used) - so wether its linear or non linear, or come from a bivariate normal distribution or not. so I put down no assumptions
    I put no assumptions too so hope you're right
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    How many marks do you think I'd lose for getting the variance wrong on the last question?
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    How were we meant to do the last one? I worked out the sd by doing the variance then square rooting and then placing into equation for the mean?
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    (Original post by emma_1111)
    How were we meant to do the last one? I worked out the sd by doing the variance then square rooting and then placing into equation for the mean?
    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

    I need someone to help me with latex lol...

    But you work out the variance and mean, then the distribution is the accepted population mean, the variance is using the variance of the sample (since you have no other choice) divided by the sample size, so X \sim N(5.64, \, \dfrac{3.03701}{60})

    Then find the z-value for the sample mean and compare it with the critical value.
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    (Original post by Crozzer24)
    I put no assumptions too so hope you're right
    Spearman's rank can't be used with correlations which increase and then decrease (ie peak, so for example temperature vs enzyme action) that's why I put monotonic, but maybe I'm looking into it too much
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    great paper!! Think I got the independent definition wrong though... mixed it up with nuclear decay in physics lol

    Wasn't sure about the spearman's rank 1 marker - I put no assumptions.
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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}
    Got pretty much all the same but got 6513 for sigma, 58350 for mean and 40094 for h
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    Balls. I was checking through my paper and found I did test statistic wrong and i fixed it and it felt great.

    Checking through other questions. Felt good.

    Teacher said 1 min left and i had to check couple more so I was on that is 95% valid question.

    Phi should be 0.11 something (what I wrote down). Checked calculator, and it sais 1.1 something so i crosed all and rewrote. Thought I saved 3 marks..

    Turns out, I divided by 48000, instead of 4800. My actual answer was correct. 3 marks gone. My heart sank. Figuratively and possibly literally.

    I think it was a bit better than last years paper which had A as 65 (WTF?!).
    And I need about 87 ums in this for A Overall in maths.. ffs

    *facepalm*

    Posted from TSR Mobile
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    I got 6666.67 for my Sigma, and around 56-58k for my mean, would I still get some marks because I don't know what I did wrong...
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    (Original post by Crozzer24)
    Got pretty much all the same but got 6513 for sigma, 58350 for mean and 40094 for h
    I was using my calculator for NDF values so that's probably due to a loss in precision.
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    Agree with everything except I got 0.0211 instead of 0.0210 but im sure it doesn't matter, and I think for mu I may have gotten 58350.2 and 40094 for h or something but everything else seems right

    Posted from TSR Mobile
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    And did anyone get 1.743 as the standard deviation for the last question? Would I lose a mark for accidentally writing X~N (5.64 , 1.743) without the square lol
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    (Original post by Crozzer24)
    Got pretty much all the same but got 6513 for sigma, 58350 for mean and 40094 for h
    Agree to both of yours


    Posted from TSR Mobile
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    for the spearmans rank hypothesis test do you write 'in the underlying population'???
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    from 2014:

    For population of young adults seen at least once. Do not allow underlying population.
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    http://www.mei.org.uk/files/pdf/spearmanrcc.pdf

    "words to make the point that the test is for association in the population by adding thewords “in the underlying population” to the end of the statement about the null hypothesis,making it something like0: There is no association between the variables in the underlying population."

    it says here in notes from MEI themselves to write 'underlying population', then why does it condone it in the mark scheme ?
 
 
 
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