Question

In an article in the Journal of Marketing, Bayus studied the differences between "early replacement buyers” and "late replacement buyers” in making consumer durable good replacement purchases. Early replacement buyers are consumers who replace a product during the early part of its lifetime, while late replacement buyers make replacement purchases late in the product’s lifetime. In particular, Bayus studied automobile replacement purchases. Consumers who traded in cars with ages of zero to three years and mileages of no more than 35,000 miles were classified as early replacement buyers. Consumers who traded in cars with ages of seven or more years and mileages of more than 73,000 miles were classified as late replacement buyers. Bayus compared the two groups of buyers with respect to demographic variables such as income, education, age, and so forth. He also compared the two groups with respect to the amount of search activity in the replacement purchase process. Variables compared included the number of dealers visited, the time spent gathering information, and the time spent visiting dealers.

(a) Suppose that a random sample of 805 early replacement buyers yields a mean number of dealers visited of x¯x¯ = 3.2, and assume that σ equals .74. Calculate a 99 percent confidence interval for the population mean number of dealers visited by early replacement buyers. (Round your answers to 3 decimal places.)

(b) Suppose that a random sample of 507 late replacement buyers yields a mean number of dealers visited of x¯x¯ = 4.7, and assume that σ equals .61. Calculate a 99 percent confidence interval for the population mean number of dealers visited by late replacement buyers. (Round your answers to 3 decimal places.)

(c) Use the confidence intervals you computed in parts a and b to compare the mean number of dealers visited by early replacement buyers with the mean number of dealers visited by late replacement buyers. How do the means compare?

Answer 1

Ans:

a)

99% confidence interval for the population mean number of dealers for early replacement

=3.2+/-2.576*(0.74/sqrt(805))

=3.2+/-0.067

=(3.133, 3.267)

b)

99% confidence interval for the population mean number of dealers for late replacement

=4.7+/-2.576*(0.61/sqrt(507))

=4.7+/-0.070

=(4.630 , 4.770)

c)

As,lower limit of CI of late replacement buyers is greater than upper limit of CI for early replacement buyers.

So,mean number of buyers for late replacement buyers is greater than mean number of buyers for early replacement buyers.