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Suppose we have a random sample of size \(2 n\) from a population denoted by \(X,\) and \(E(X)=\mu\) and \(V(X)=\sigma^{2} .\) Let $$\bar{X}_{1}=\frac{1}{2 n} \sum_{i=1}^{2 n} X_{i} \quad \text { and } \quad \bar{X}_{2}=\frac{1}{n} \sum_{i=1}^{n} X_{i} $$be two estimators of \(\mu\). Which is the better estimator of \(\mu\) ? Explain your choice.

Short Answer

Expert verified
\( \bar{X}_1 \) is the better estimator because it has a smaller variance.

Step by step solution

01

Understand the Estimators

We have two estimators, \( \bar{X}_1 \) which is the average of a sample of size \( 2n \), and \( \bar{X}_2 \) which is the average of a sample of size \( n \). Our goal is to find out which of these two is a better estimator of the population mean \( \mu \).
02

Check Unbiasedness of Estimators

Both estimators are unbiased if their expected values are equal to \( \mu \).\[ E(\bar{X}_1) = \mu \quad \text{and} \quad E(\bar{X}_2) = \mu \] since both \( \bar{X}_1 \) and \( \bar{X}_2 \) are sample means. Thus, both are unbiased estimators of \( \mu \).
03

Calculate the Variance of the Estimators

Calculate the variance of \( \bar{X}_1 \) and \( \bar{X}_2 \): \[ V(\bar{X}_1) = \frac{\sigma^2}{2n} \quad \text{and} \quad V(\bar{X}_2) = \frac{\sigma^2}{n} \] \( \bar{X}_1 \) has lower variance than \( \bar{X}_2 \) because \( 2n > n \).
04

Compare the Variances

A smaller variance indicates a more precise estimator. Since \( V(\bar{X}_1) = \frac{\sigma^2}{2n} < \frac{\sigma^2}{n} = V(\bar{X}_2) \), \( \bar{X}_1 \) is more precise or consistent than \( \bar{X}_2 \).
05

Conclusion on the Better Estimator

Given that both estimators are unbiased, we choose the one with the lower variance as the better estimator for \( \mu \). Therefore, \( \bar{X}_1 \) is the better estimator since it has a smaller variance.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Unbiased Estimators
In statistical analysis, an unbiased estimator is crucial because it ensures that, on average, the estimator equals the parameter it is estimating. In simpler terms, the expected value of an unbiased estimator matches the true value of the parameter. In our exercise, we have two estimators, \( \bar{X}_1 \) and \( \bar{X}_2 \), aiming to estimate the population mean \( \mu \).

The concept of unbiasedness in this context means:
  • The average calculated using the sample size of \( 2n \) for \( \bar{X}_1 \).
  • The average calculated using the sample size of \( n \) for \( \bar{X}_2 \).
For both, the expected value of the sample mean equals \( \mu \), confirming that both \( \bar{X}_1 \) and \( \bar{X}_2 \) are unbiased estimators. This quality is critical because it determines the legitimacy of these estimators in accurately reflecting the true population mean.
Variance Analysis
Variance measures the dispersion of a set of data points. In the context of estimators, variance denotes how much the estimations will fluctuate when sampling is repeated.

For our exercise, we know:
  • The variance of \( \bar{X}_1 \) is \( \frac{\sigma^2}{2n} \).
  • The variance of \( \bar{X}_2 \) is \( \frac{\sigma^2}{n} \).
These variances highlight the difference in precision between the two estimators. Since the variance of \( \bar{X}_1 \) is less than that of \( \bar{X}_2 \), it signifies that \( \bar{X}_1 \) provides more consistent and reliable estimates of \( \mu \). Lower variance means that the estimates are less spread out and thus closer to \( \mu \) on average, imparting \( \bar{X}_1 \) with a higher level of precision.
Population Mean Estimation
Estimating the population mean \( \mu \) involves using sample data to deduce the average value of the entire population.

Here's a brief overview:
  • \( \bar{X}_1 \) and \( \bar{X}_2 \) are two different approaches with sample sizes \( 2n \) and \( n \), respectively.
  • A larger sample size, as used by \( \bar{X}_1 \), generally results in a more accurate estimation of \( \mu \).
  • This is because a larger sample size typically captures more variability in the data, leading to better representation.
In the estimation of the population mean, it is not only the unbiased nature of an estimator that counts but also its variance. In our case, since both \( \bar{X}_1 \) and \( \bar{X}_2 \) are unbiased and \( \bar{X}_1 \) has a lower variance, \( \bar{X}_1 \) is the preferred choice, making it a better estimator of \( \mu \). Understanding these principles is essential for effectively interpreting statistical data and making informed decisions based on analyses such as these.

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