1. ## 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

2. 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
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 $\displaystyle y=mx+b$ $\displaystyle m$ and $\displaystyle b$
are calculated using the equations:

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

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

Now we have:

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

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

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

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

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

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

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

The correlation coefficient is:

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

so:

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

RonL

3. ## 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