# 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 800 early replacement buyers yields a mean number of dealers visited equal to 3.3, and assume that the population standard deviation equals .71. Calculate a 99 percent confidence interval for the population mean number of dealers visited by all early replacement buyers.

b. Suppose that a random sample of 500 late replacement buyers yields a mean number of dealers visited equal to 4.3, and assume that the population standard deviation equals .66. Calculate a 99 percent confidence interval for the population mean number of dealers visited by all late replacement buyers.

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

a. Suppose that a random sample of 800 early replacement buyers yields a mean number of dealers visited equal to 3.3, and assume that the population standard deviation equals .71. Calculate a 99 percent confidence interval for the population mean number of dealers visited by all early replacement buyers.

b. Suppose that a random sample of 500 late replacement buyers yields a mean number of dealers visited equal to 4.3, and assume that the population standard deviation equals .66. Calculate a 99 percent confidence interval for the population mean number of dealers visited by all late replacement buyers.

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

## Answer to relevant Questions

Explain how each of the following changes as the number of degrees of freedom describing a t curve increases: a. The standard deviation of the t curve. b. The points tα and tα/2. Whole Foods is an all- natural grocery chain that has 50,000 square foot stores, up from the industry average of 34,000 square feet. Sales per square foot of supermarkets average just under $ 400 per square foot, as reported ...Consider a population having a standard deviation equal to 10. We wish to estimate the mean of this population. a. How large a random sample is needed to construct a 95 percent confidence interval for the mean of this ...Suppose we are using the sample size formula in the box on page 313 to find the sample size needed to make the margin of error in a confidence interval for p equal to E. In each of the following situations, explain what ...Consumer Reports (January 2005) indicates that profit margins on extended warranties are much greater than on the purchase of most products. In this exercise we consider a major electronics retailer that wishes to increase ...Post your question

0