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 45BB

(a)

We know that the mean of the population by the following formula:

Therefore, the mean of the population is 4.33 minutes.

We know that population variance found by the following formula:

Therefore, the variance of the given population is.

(b)

We know that, sample of size taken from the population of size by with replacement in ways. From the given information, therefore, the possible number of samples of size 2 from a population of size 3 in ways.

These 9 samples and their respective means and sample variances are listed in the following table:

... Sample | Sample mean | Sample Variance |