Question

1.

A market researcher wants to determine whether a new model of a personal computer that had been advertised on a​ late-night talk show had achieved more​ brand-name recognition among people who watched the show regularly than among people who did not. After conducting a​ survey, it was found that 16​% of all people both watched the show regularly and could correctly identify the product.​ Also, 20​% of all people regularly watched the show and 48​% of all people could correctly identify the product. Define a pair of random variables as shown below. Complete parts​ (a) through​ (c).

X	=	1 if regularly watch the show	X	=	0 otherwise
Y	=	1 if product correctly identified	Y	=	0 otherwise

a. Find the joint probability function of X and Y.

		X
		0		1
Y	0	nothing		nothing
	1	nothing		nothing

b. Find the conditional probability function of​ Y, given X=1.

​P(0​|1​)=

​(Round to three decimal places as​ needed.)

​P(1|1​)=

​(Round to three decimal places as​ needed.)

c. Find and interpret the covariance between X and Y.

​Cov(X,Y)=

​(Do not​ round.)

Interpret the covariance.

A.

The covariance indicates that there is no association between watchers of a​ late-night talk show and​ brand-name recognition.

B.

The covariance indicates that there is a positive association between watchers of a​ late-night talk show and​ brand-name recognition.

C.

The covariance indicates that there is a negative association between watchers of a​ late-night talk show and​ brand-name recognition.

Answer 1

Let us assume that there are 100 people in total, then we can comfortably generate the below-given   table

	CORRECTLY IDENTIFY	DIDN’T IDENTIFY	TOTAL
WATCH	16	4	20
DON’T WATCH	32	48	80
TOTAL	48	52	100

The values highlighted are derived on the basis of our (total 100 people) assumption.

a) Conditional probability of X, given Y=1

Y=1, correctly identified the product

If X=0 then conditional probability is (32/100) / (48/100) = .32/.48 = 0.67 (approx)

There is 67% probability of finding someone who has correctly identified but not watched the show.

If X=1, then similarly conditional probability would be = .16/.48 = 0.33 (approx)

b) Conditional probability of Y, given X=1

If Y=0, i.e. didn’t identify, then conditional probability would be = .04/.20 = 0.20

If Y=1, then conditional probability would be = .16/.20 = 0.80

c) Cov(X,Y)= E(XY) - X'Y'

X'---> expected value of X

Y'---> expected value of Y

Cov(X,Y)= (1)(1)(0.16)-(0.5)(0.5)

=0.16-0.25

=-0.09

This shows how two variables move together.

C----> there is a negative association between watchers of a​ late-night talk show and​ brand-name recognition.