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An engineer wishes to estimate the mean yield of chemical process based on the yield measurements $X_1$, $X_2$, $X_3$ from three independent runs of an experiment with variance $\sigma^2$.

Consider the following two estimators of the mean yield $\theta$:

$$\dfrac{X_1+X_2+X_3}{3} \quad\text{and}\quad \dfrac{X_1+2X_2+X_3}{4}.$$

Which is a better estimator? Justify with the appropriate statistical procedures.

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closed as off-topic by heropup, Meta, Shailesh, Leucippus, choco_addicted Apr 3 at 3:13

This question appears to be off-topic. The users who voted to close gave this specific reason:

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We prefer to use caps for random variables. Note that $$E\left(\frac{X_1+X_2+X_3}{3}\right)=\frac{1}{3}\left(E(X_1)+E(X_2)+E(X_3)\right)=\mu,$$ where $\mu$ is the common mean of the $X_i$.

Similarly, $$E\left(\frac{X_1+2X_2+X_3}{4}\right)=\frac{1}{4}\left(E(X_1)+2E(X_2)+E(X_3)\right)=\mu.$$

So our two estimators are both unbiased estimators of $\mu$.

Since the $X_i$ are independent, we have $$\text{Var}\left(\frac{X_1+X_2+X_3}{3}\right)=\frac{1}{9}\left(\text{Var}(X_1)+\text{Var}(X_2)+\text{Var}(X_3)\right)=\frac{\sigma^2}{3}.$$

Also, $$\text{Var}\left(\frac{X_1+2X_2+X_3}{4}\right)=\frac{1}{16}\left(\text{Var}(X_1)+2^2\text{Var}(X_2)+\text{Var}(X_3)\right)=\frac{6\sigma^2}{16}.$$

The variance of the second estimator is greater than the variance of the first. Thus the first estimator is "better" than the second.

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