Mean Absolute Deviation Use the same population of {2 min, 3 min, 8 min} from Exercise 1. Show that when samples of size 2 are randomly selected with replacement, the samples have mean absolute deviations that do not center about the value of the mean absolute deviation of the population. What does this indicate about a sample mean absolute deviation being used as an estimator of the mean absolute deviation of a population?

Why Divide by n— 1? Let a population consist of the values 2 min, 3 min, 8 min. (These are departure delay times taken from American Arlines flights from New York’s JFK airport to Los Angeles. See Data.) Assume that samples of two values are randomly selected with replacementfrom this population. (That is, a selected value is replaced before the second selection is made.)

a. Find the variance σ2 of the population {2 min, 3 min, 8 min}.

b. After listing the nine different possible samples of two values selected with replacement, find the sample variance s2 (which includes division by n − 1) for each of them; then find the mean of the nine sample variances s2.

c. For each of the nine different possible samples of two values selected with replacement, find the variance by treating each sample as if it is a population (using the formula for population variance, which includes division by n), then find the mean of those nine population variances.

d. Which approach results in values that are better estimates of σ2: part (b) or part (c)? Why? When computing variances of samples, should you use division by n or n – 1 ?

e. The preceding parts show that s2 is an unbiased estimator of σ2. Is s an unbiased estimator of σ ? Explain.

Solution 46BB

From the given information, the population of delay times consists of 2 minutes, 3 minutes, and 8 minutes.

The drawn samples of size 2 by with replacement from the given population of size 3 are with respective sample means 2, 2.5, 5, 2.5, 3, 5.5, 5, 5.5, and 8.

We know that, the sample absolute mean deviation found by

Therefore,...