Question

Suppose we are interested in analyzing the market share and customer loyalty for Murphy's Foodliner and Ashley's Supermarket, the only two grocery stores in a small town. We focus on the sequence of shopping trips of one customer and assume that the customer makes one shopping trip each week to either Murphy's Foodliner or Ashley's Supermarket, but not both. Suppose that, as part of a market research study, we collect data from 100 shoppers over a 10-week period. Suppose further that these data show each customer's weekly shopping trip pattern in terms of the sequence of visits to Murphy's and Ashley's. In reviewing the data, suppose that we find that of all customers who shopped at Murphy's in a given week, 70% shopped at Murphy's the following week while 30% switched to Ashley's. Suppose that similar data for the customers who shopped at Ashley's in a given week show that 60% shopped at Ashley's the following week while 40% switched to Murphy's. Probabilities based on these data are shown in the table below. Current Weekly Next Weekly Shopping Period Shopping Period Murphy's Foodliner Ashley's Supermarket Murphy's Foodliner 0.70 0.30 Ashley's Supermarket 0.40 0.60 Suppose that we are considering the Markov process associated with the shopping trips of one customer, but we do not know where the customer shopped during the last week. Thus, we might assume a 0.5 probability that the customer shopped at Murphy's and a 0.5 probability that the customer shopped at Ashley's at period 0; that is (0)= 0.5 and (0)= 0.5. Given these initial state probabilities, develop a table showing the probability of each state in future periods. What do you observe about the long-run probabilities of each state? If required, round your answers to four decimal places. Do not round your intermediate calculations. State Probability 0 1 2 3 4 5 6 7 8 9 10 (n) (n) Probabilities are approaching = and =

Answer 1

Civen ue mght assumepoblbly stred a might assume pobabity ar Ashley's al pid kitePobabi T, (6)-05 From the above Vectos oarht gen eba bas cs CamScanher

Probabili 0 0.5 0.5 1 0.55 0.45 2 4 10 0.622 0.388 0.5850.60950.6267 0.6382 0.6471 0.6529 0.65710.6599 0.4150.39050.3734 0.3613 0.3219 0.3129 0.34290.3401

pabily aprs app9o0 2 3 091099 1521 Cs Scanned with CamScanner