You are watching: Which of the following statistics are unbiased estimators of population parameters?
An estimator T is called unbiased estimator for populace parameter 0 if Ε(T )-θ,νθ. The correct answers room as follows: B. Sample variance supplied to estimate populace variance. C. Sample proportion supplied to estimate populace proportion. D. Sample mean used to estimate populace mean. The proof for the exactly answers space as follows: allow X,,X, ,.., X, be an independent random sample from a population with mean u and variance o such the E(X,)= µ, i=1,2,..,n and also Var (X,) = o',i=1,2,..., n. To prove sample average is unbiased estimator of population mean, continue as follows: The sample can be composed as follows: Σχ. I=1 |X+X, +...+X, Now, we require to show that E(X)= µ. Therefore, take expectation because that the sample mean X together follows: x -x, E(X)= E |X, + X, +...+X, =
Now, to display that sample variance is the unbiased estimator of population variance, that is, E(s³)= o². Start with L.H.S. And also proceed as follows: E(3)--Σ-1 - X)² E(s?)= E n-12(X,- F)² п-1 i=1 Σ(χ+X _2 Σ) i=1 Ε ΣΧ+nX-2XΣ. -1 i=1 i=1 ·EE(x;)+nX² – 2nx? п-1 i=1 Σ(x)-n п-1 i=1 ΣΕ(x)-Ε ( I') i=1 Now, that is known that, Var (X,) = E(x} )-
Now substitute E(X²) and E(X;)in E(s²) together follows: E(-?+ p8) -n E(s*)= п-1 in i=1
Now to display that the sample ratio is the unbiased estimator of population proportion. Let sample proportion be denoted by p and is offered as p=÷, where x is the variety of successes in the sample and n is the sample size. Allow the population proportion it is in denoted by P. Then, E(p)= E %3D -E(00
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