Regression

**lalitchugh** · June 19th 2006, 09:37 AM

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

**CaptainBlack** · June 19th 2006, 11:09 PM

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:

$ m=\frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2} $

$ b=\frac{\sum y -m \sum x}{n} $

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:

$ m_1=\frac{11\times 50805-725\times 737}{11\times 49493-725^2}\approx 1.305 $

$ b_1=\frac{737-1.305\times 725}{11}\approx -19.01 $

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:

$ m_2=\frac{11\times 50805-725 \times 737}{11\times 53939-737^2}\approx 0.489 $

$ b_2=\frac{725-0.489\times 737}{11}\approx 33.1 $

The correlation coefficient is:

$ 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}} $ ,

so:

$ 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 $

RonL

**lalitchugh** · June 20th 2006, 01:07 AM

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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