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Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a continuous distribution with pdf \(f(x ; \theta)\). For large \(n\), the variance of the sample median is approximately \(1 /\left\\{4 n[f(\tilde{\mu} ; \theta)]^{2}\right\\}\). If \(X_{1}, X_{2}, \ldots, X_{n}\) is a random sample from the normal distribution with known standard deviation \(\sigma\) and unknown \(\mu\), determine the efficiency of the sample median.

Short Answer

Expert verified
The sample median efficiency is approximately 63.66% compared to the sample mean for a normal distribution.

Step by step solution

01

Understanding Efficiency

Efficiency in statistics refers to the comparison of variances between two unbiased estimators. Given two estimators, if one has a smaller variance, it is considered more efficient. In this problem, we need to compare the efficiency of the sample median with the sample mean, which is known to be the most efficient estimator for the mean of a normal distribution.
02

Identify the Variances

For the sample mean of a normal distribution, the variance is \(\frac{\sigma^2}{n}\). For the sample median, based on the problem statement, this is approximately \(\frac{1}{4n[f(\tilde{\mu}; \theta)]^2}\). For a normal distribution, we know \(f(\tilde{\mu}; \theta)\) corresponds to the normal density at its mean which is given by \(\frac{1}{\sqrt{2\pi}\sigma}\).
03

Calculate the Variance of the Sample Median

Substitute the normal density into the variance formula for the median. This gives: \[\frac{1}{4n\left(\frac{1}{\sqrt{2\pi}\sigma}\right)^2} = \frac{\sigma^2\pi}{2n}.\]Thus, the variance of the sample median for a normal distribution is approximately \(\frac{\sigma^2\pi}{2n}\).
04

Compute the Efficiency of the Sample Median

The efficiency of an estimator is the ratio of the variances of the sample mean and the sample median. Compute the efficiency using: \[\frac{Variance\ of\ Sample\ Mean}{Variance\ of\ Sample\ Median} = \frac{\frac{\sigma^2}{n}}{\frac{\sigma^2\pi}{2n}}\]Simplifying this, we find: \[Efficiency = \frac{2}{\pi}.\]
05

Interpret the Efficiency Value

The efficiency \(\frac{2}{\pi}\) indicates that the sample median is roughly 63.66% as efficient as the sample mean for estimating the mean of a normal distribution when variance is considered. This means the sample mean has a lower variance, making it a superior estimator in terms of efficiency.

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Key Concepts

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

Sample Median
The sample median is a statistical measure that identifies the middle value in an ordered data set. If there is an odd number of observations, the median is precisely the center value. If there is an even number, it is the average of the two central values.

The sample median is particularly useful when dealing with skewed distributions or outliers since it is robust to extreme values. This characteristic makes it a reliable measure of central tendency in such cases since it doesn't get disproportionately affected by anomalies as the sample mean does.
  • Order the data set from smallest to largest.
  • Identify the middle value.
  • If even number, compute the average of the two middle values.
Normal Distribution
The normal distribution, also known as the Gaussian distribution, is a continuous probability distribution characterized by its bell-shaped curve. It is defined by two parameters: the mean (\(\mu\)) and the standard deviation (\(\sigma\)). This distribution describes data that clusters around a mean and occurs frequently in nature and statistics.

A feature of the normal distribution is that it is symmetric around its mean. This symmetry implies that the left and right halves of its histogram are mirror images. Moreover, the spread of the data is determined by its standard deviation — larger values indicate a wider spread. Understanding the normal distribution is crucial because many statistical methods are based on this model due to the Central Limit Theorem, which states that averages of samples of observations of random variables independently drawn from the same distribution converge in distribution to the normal as the number of observations goes to infinity.
  • Symmetrical bell curve.
  • Defined by mean and standard deviation.
  • Natural data tendency.
Estimator Variance
Estimator variance refers to the variability of an estimate in a statistical model. It represents how much the estimate would vary if we repeated sampling from the same population with the same sample size. Lower variance indicates that the estimate is more reliable in reproducing the true parameter value.

In our scenario, understanding the estimator variance is essential when comparing the performance of different statistical estimators, such as the sample mean and sample median, for the same parameter. The sample mean typically has a smaller variance than the sample median, indicating its higher efficiency when data are normally distributed. However, the variance can vary significantly depending on the estimator used and the characteristics of the data.
  • Variability of a statistical estimate.
  • Importance for estimator reliability.
  • Smaller variance indicates higher consistency.
Sample Mean Efficiency
Sample mean efficiency is a measure used in statistics to compare the performance of the sample mean as an estimator against other possible estimators. It is determined by the ratio of the variances of two estimators, with the sample mean often serving as the benchmark since it is the Best Linear Unbiased Estimator (BLUE) for normally distributed data.

In the context of this problem, calculating the efficiency involves comparing the variance of the sample mean to that of the sample median. For normal distributions, the sample mean is considered most efficient because it yields the smallest variance. As seen in the provided solution, the efficiency of the sample median relative to the sample mean is \(\frac{2}{\pi}\), or approximately 63.66%. This suggests that while the sample median can be useful, it isn't as efficient as the sample mean for estimating the mean of a normal distribution due to its higher variance.
  • Comparison measure of variances.
  • Sample mean is usually the most efficient.
  • Efficiency ratio indicates precision advantage.

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Most popular questions from this chapter

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