Question

Let three random samples of sizes n₁ = 20, n₂ = 10, and n₃ = 8 be taken from a population with mean µ and variance o². Let S₂, S₃, and S₃ be the sample

variances. Show that S² = (2052 + 1052 + 853)/38 is an unbiased estimator of o². (Hint: For a population with a variance of o², the sample variance s² is an unbiased estimator.)

Answer 1

S² = (20S₁² + 10S₂² + 8S₃²) / 38 is an unbiased estimator of σ².

Let's denote the pooled variance estimator as S_pooled:

Sample Sample Size Sample Variance

1 n_1 = 20 S_1^2

2 n_2 = 10 S_2^2

3 n_3 = 8 S_3^2

Then, the pooled variance estimator S_pooled is calculated as:

S_pooled = (Σ(n_i - 1)S_i^2) / (Σn_i - k)

where:

n_i is the sample size for sample i

S_i^2 is the sample variance for sample i

k is the number of samples

In this case, we have:

S_pooled = (20S_1^2 + 10S_2^2 + 8S_3^2) / 38

To show that S_pooled is an unbiased estimator of σ², we can use the fact that the expected value of the sample variance s² is equal to the population variance σ².

This can be shown using the following identity:

E(s²) = (n - 1)σ² / n

where:

n is the sample size

Substituting n_i for n, we can write the expected value of S_i^2 as:

E(S_i^2) = (n_i - 1)σ² / n_i

Then, the expected value of S_pooled is:

E(S_pooled) = ΣE(S_i^2) / k

Substituting the expected value of S_i^2, we get:

E(S_pooled) = Σ((n_i - 1)σ² / n_i) / k

Simplifying, we get:

E(S_pooled) = (σ² / k) Σ(n_i - 1)

Since Σ(n_i - 1) = N - k, where N is the total sample size, we can write:

E(S_pooled) = (σ² / k) (N - k)

Substituting N = n_1 + n_2 + n_3 = 38 and k = 3, we get:

E(S_pooled) = (σ² / 3) (38 - 3) = 11σ²

Therefore, S_pooled is an unbiased estimator of σ².