A management made a certain proposal to company\'s management representatives in

Question

A management made a certain proposal to company's management representatives in sales regions Questionnaires were sent to representative, and the results obtained were as follows: Region East Midwest West opposed 200 Not opposed. 100 120 200 250 150 Total (a) what is the probability that a questionnaire selected at random is that of a Western sales representative in favor of the proposal? (b) What is the probability that it is that of a Midwestern sales representative? 104 Chapter 2 Probability Models and Discrete Distributions (c) If a questionnaire is selected at random from the group that responded unfavorably to the what is the conditional probability that the respondent comes from the East? (d) regional district and opinion on the proposal independent? If yes, prove your assertion. If no, with the same marginal totals, specify what the numbers in the cells of the table would have been had the two factors been independent. If the regions are denoted by A1, A2, A3, and opinions are called B1, B2. respectively, is it true that P(Ain B) P(A)P(B), for all i 1, 2, 3 and j so, regions and opinions would be independent.

Explanation / Answer

a.) Probability that questionaire selected at random is that of a Western sales representative in favour of the proposal

=Number of proposals of Western sales representative in favour of the proposal / Total number of proposals

=200 / 600 = 0.333

b.)  Probability that questionaire selected at random is that of a Midwestern sales representative

= Number of proposals Midwestern sales representatives / Total number of proposals

=200(80+120) / 600 = 0.333

c.) P(Unfavourable proposal/Proposal from East representative)

= P(Unfavourable proposal from East representative) / P(Proposal from East representative)

=50 / 150 = 0.333

d.) If the regions are denoted by A1,A2,A3 and opinions are denoted by B1,B2 respectively then let us check if

P(AiBj)=P(Ai)P(Bj) for i=1,2,3 and j=1,2

thus

P(A1B1)=50/600 and P(A1)=150/600 ,P(B1)=180/600

In this case we see that P(A1B1) P(A1)P(B1)

Thus the regions and opinions are not independent.

We now form a contingency table and calculate expected cell frequencies by using the formula

Expected frequency=(row total)(coloumn total)/n

thus form the table

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