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 807 early replacement buyers yields a mean number of dealers visited of x¯x¯ = 3.3, and assume that σ equals .79. Calculate a 99 percent confidence interval for the population mean number of dealers visited by early replacement buyers.
The 99 percent confidence interval is [, ].

Answer 1

a) The 99 percent confidence interval for the population mean number of dealers visited by early replacement buyers is approximately
`\([3.15, 3.45]\).`

Step-by-step explanation:

To calculate the 99 percent confidence interval for the population mean
`(\(\mu\))` number of dealers visited by early replacement buyers, we use the formula
`\(\bar{x} \pm Z * (\sigma)/(√(n))\), where \(\bar{x}\)` is the sample mean,
`\(Z\)` is the Z-score corresponding to the confidence level (in this case, 99 percent),
`\(\sigma\)` is the population standard deviation, and
`\(n\)` is the sample size. Given that
`\(\bar{x} = 3.3\), \(\sigma = 0.79\),`and the sample size
`\(n = 807\)` , we find the Z-score for a 99 percent confidence level and substitute the values into the formula.

The critical Z-value for a 99 percent confidence interval is approximately 2.576. Therefore, the confidence interval is calculated as follows:

`\[3.3 \pm 2.576 * (0.79)/(√(807))\]`

After performing the calculations, the 99 percent confidence interval is approximately
`\([3.15, 3.45]\)`. This interval provides a range within which we are 99 percent confident that the true population mean number of dealers visited by early replacement buyers lies.

Understanding confidence intervals is essential in statistics to quantify the uncertainty associated with sample estimates. In this case, the interval
`\([3.15, 3.45]\)` suggests that if we were to repeat this sampling process many times, 99 percent of the calculated intervals would contain the true population mean number of dealers visited by early replacement buyers.