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Math Help - Regression

  1. #1
    Newbie
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    Quick help needed

    Hi,

    May i request to help me in providing solution for this question quickly.

    Q: Following are marks of 11 students in two papers (out of 100)

    P1 : 88 45 55 56 58 60 65 68 70 75 85
    P2 : 92 56 50 44 60 36 64 65 97 74 99

    Find the coefficient of correlation and the lines of regression.


    Thanks & Regards,
    Lalit Chugh
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by lalitchugh
    Hi,

    May i request to help me in providing solution for this question quickly.

    Q: Following are marks of 11 students in two papers (out of 100)

    P1 : 88 45 55 56 58 60 65 68 70 75 85
    P2 : 92 56 50 44 60 36 64 65 97 74 99

    Find the coefficient of correlation and the lines of regression.


    Thanks & Regards,
    Lalit Chugh
    I'm a bit concerned about this question. It reads as though you are
    being asked to find the regerssion line of P1 on P2 or vice versa, but
    such a standard regression line is not necessarily appropriate to this data.

    For a regression line of the form y=mx+b m and b
    are calculated using the equations:

    <br />
m=\frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2}<br />

    <br />
b=\frac{\sum y -m \sum x}{n}<br />

    Now we have:

    \sum P1=725,\ \sum P2=737 ,\ \sum P1^2=49493,\ \sum P2^2=53939,\ \sum P1.P2=50805.

    So if the first regression line we want is: P2=m_1P1+b_1,
    we use P1 for x and P2 for y in the normal equations. So:

    <br />
m_1=\frac{11\times 50805-725\times 737}{11\times 49493-725^2}\approx 1.305<br />

    <br />
b_1=\frac{737-1.305\times 725}{11}\approx -19.01<br />

    The second regression line we want is: P1=m_2P2+b_2,
    we use P2 for x and P1 for y in the normal equations. So:

    <br />
m_2=\frac{11\times 50805-725 \times 737}{11\times 53939-737^2}\approx 0.489<br />

    <br />
b_2=\frac{725-0.489\times 737}{11}\approx 33.1<br />

    The correlation coefficient is:

    <br />
r=\frac{n(\sum xy)-(\sum x)(\sum y)}{[n(\sum x^2)-(\sum x)^2]^{1/2}[n(\sum y^2)-(\sum y)^2]^{1/2}}<br />
,

    so:

    <br />
r=\frac{11\times 50805-725\times 737}{[11\times 49595-725^2]^{1/2}[11\times 53939-737^2]^{1/2}}\approx 0.798<br />

    RonL
    Attached Thumbnails Attached Thumbnails Regression-gash.jpg  
    Last edited by CaptainBlack; June 20th 2006 at 03:44 AM.
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  3. #3
    Newbie
    Joined
    Jun 2006
    Posts
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    Quick help needed

    Hi,

    Thanks a lot for the reply.

    I could also calculate Correlation coefficient and it came to 0.80

    Regarding, Regression, i will also try to solve and check with your response.

    I need one more help from you.

    1) Could you pls help me provide solution for my post in trignometry section. the question is

    Prove that : ((cosA + i sin A)/(sin A + i cos A))^4 = cos 8A + i sin 8A

    Thread subject is : Can someone help ASAP

    its bit urgent.

    2) Can you pls tell me how i can post my queries in the mathematical format. i mean to say, i am using word to explain my question and then post it. but in word i can not find mathematical symbols.

    Somebody, advised me to install Latex. Not sure, for windows, how to install it. pls advice.


    Best Regards,
    Lalit Chugh
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