MSE : correlation regression

[sweta]

the regressive coefficients of regression equations are 0.5 and 0.7.
 then the correlation coefficient is?
1. 0.55
2.  0.59
3. 0.65
4. 0.63

 q2.the regression coefficients of regression eqns  are -a and -b where a and b are positiv fractional values, then the correlation coefficient is
a) (a+b)/2
b) (a-b)/2
c) -c (fractional value)
d)  c (fractional value)

the minimum value of f(x)=x^x is at
a) x=1/e
b) x=-1/e
c) x=e
d) x= -e

[Chinni18]

For the first question:

Correlation coefficient is basically the square root product of the regression coefficients
So correlation coefficient = (0.7*0.5)^1/2
                                  = (0.35)^1/2
                                  = 0.57
Option b

[Chinni18]

For Q.3
The minimum value of f(x)=x^x is at 1/e
Let f(x) = y
y = x^x
=> log y = x* log x
=> (1/y) (dy/dx) = logx + 1
dy/dx = (x^x) (log x + 1 )
For minimum, put dy/dx = 0
=> log x + 1 = 0
=> x = e^ -1 = 1/e
If you substitute this value in the second derivative of y, you'll get > 0, therefore it is indeed a point of minimum
Option a

[sweta]

thanks a lot!!

@chinni18.

[sweta]

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[sweta]

@chinni 18
can u try dis out
consider the  data mean: X= 65, Y= 67 , std deviation (sigma) of x= 2.5 and that of y=3.5
and the correlation coefficient is r=0.8, then the value of y at x= 70 is
1. 62.6
2. 82.6
3.72.6
4. 70.6

[Chinni18]

Sweta, the answer is (c) 72.6
You have to first find the regression equation of Y on X (no need to waste time finding X on Y, because it's not been asked)
Y-Ybar = r*(Sy/Sx) (X – Xbar)
Y – 67 = (0.8)(3.5/2.5)(X-65)
Y = 1.12 x – 5.8
Substitute X=70 and you get the value of Y

[sweta]

thank you!