# Question

In the chapter, we discussed briefly the rationale for using n –1 in the calculation of s, a sample standard deviation intended to serve as an estimate of σ, the population standard deviation. We suggested that if we use n rather than n – 1 in the calculation, we would produce estimates of σ that (on average) tend to be too low. The correcting effect of dividing by n –1 rather than by n can be seen clearly in the variance calculation. (Remember, variance is just the square of the standard deviation.) To demonstrate, suppose you have a population consisting of three values:

A = 10, B = 20 and C = 30.

a. Compute the population variance, σ2 .

b. Show all possible samples of size two that can be selected from this population, sampling with replacement. (You should produce nine samples.) Also calculate the mean of each sample.

c. Compute the variance for each of the samples in part b using n = 2 in the denominator of your calculation.

d. Compute the variance for each of the samples in part b using n – 1 = 1 in the denominator of your calculation.

e. Compute the average for the nine sample variances that you produced in part c. Compute the average for the nine sample variances that you produced in part d.

Compare the result in the two cases to the value of σ2. You should see that for the set of nine variances calculated with a denominator of n = 2, the average sample variance will be smaller than σ2; for the set of nine variances calculated with a denominator of n – 1 = 1, the average sample variance will match σ2 precisely.

A = 10, B = 20 and C = 30.

a. Compute the population variance, σ2 .

b. Show all possible samples of size two that can be selected from this population, sampling with replacement. (You should produce nine samples.) Also calculate the mean of each sample.

c. Compute the variance for each of the samples in part b using n = 2 in the denominator of your calculation.

d. Compute the variance for each of the samples in part b using n – 1 = 1 in the denominator of your calculation.

e. Compute the average for the nine sample variances that you produced in part c. Compute the average for the nine sample variances that you produced in part d.

Compare the result in the two cases to the value of σ2. You should see that for the set of nine variances calculated with a denominator of n = 2, the average sample variance will be smaller than σ2; for the set of nine variances calculated with a denominator of n – 1 = 1, the average sample variance will match σ2 precisely.

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