Please provide steps to solutions 2 Consider the information

2. Consider the information given in Exercise 2 in Section 4.2 regarding the amount, X, of drug adminis- tered to a randomly selected laboratory rat, and number, Y, of tumors the rat develops. (a) Would you expect X and Y to be positively or nega- tively correlated? Explain your answer, and confirm it by computing the covariance. (b) Compute the linear correlation coefficient of X and Y.

Solution

Given that

X = amount of drug administered to a randomly selected lavoratory rat and

Y = number of tumors the rat developes

Here we have to find cov(X,Y) and correlation between X and Y.

The formula for covariance is,

Cov(X , Y) = E(XY) - E(X) * E(Y)

E(X) = x*p(x) = (0 * 0.4) + (1 * 0.5) + (2 * 0.1) = 0.7

E(Y) = y*p(y) = (0 * 0.963)+(1 * 0.025)+(2 * 0.012) = 0.049

E(XY) = xy * p(x,y)

= (0 * 0.388) + (0 * 0.009) + (0 * 0.003) + (0 * 0.485) + (1 * 0.01) + (2 * 0.005) + (0 * 0.09)+(2 * 0.006)+(4 * 0.004) = 0.048

Cov(X,Y) = 0.048 - 0.7 * 0.049 = 0.0137

Covariance is positive that means there is positive correlation between X and Y.

Next we find correlation between X and Y.

= Cov(X , Y) / _x * _y

²_x = x² p(x) - E(X)²

= (0² * 0.4) + (1²  * 0.5) + (2² * 0.1) - 0.7² = 0.41

_x  = sqrt(0.41) = 0.6403

²_y = y² p(y) - E(Y)²

   ⁼ (0² * 0.963) + (1² * 0.025) + (2² * 0.012) - 0.049² = 0.070599

_y = sqrt(0.070599) = 0.2657

= 0.0137 / (0.6403 * 0.070599) = 0.3031

The linear correlation between X and Y is 0.3031.

			y
	p(x,y)	0	1	2	total
	0	0.388	0.009	0.003	0.4
x	1	0.485	0.01	0.005	0.5
	2	0.09	0.006	0.004	0.1
	total	0.963	0.025	0.012	1