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Suppose that the random variable \(X\) has a log normal distribution with parameters \(\theta=1.5\) and \(\omega=0.8\). A sample of size \(n=15\) is drawn from this distribution. Find the standard error of the sample median of this distribution with the bootstrap method using \(n_{n}=200\) bootstrap samples.

Short Answer

Expert verified
The standard error is the standard deviation of the bootstrap medians.

Step by step solution

01

Understand the Log-Normal Distribution

The random variable \(X\) is log-normally distributed, which means that \( \ln(X) \sim \mathcal{N}(\theta, \omega^2) \). So, it's important to understand that when \(X\) is distributed this way, the natural logarithm of \(X\), \(\ln(X)\), follows a normal distribution with mean \(\theta = 1.5\) and variance \(\omega^2 = 0.64\).
02

Draw the Original Sample

Draw a sample of size \(n=15\) from the log-normal distribution. Use the parameters \(\theta=1.5\) and \(\omega=0.8\) to generate the original sample data \(X_1, X_2, \ldots, X_{15}\). This is the basis for all the bootstrap samples.
03

Generate Bootstrap Samples

Take a bootstrap sample by randomly drawing \(15\) observations with replacement from the original sample. This process should be repeated \(n_{n} = 200\) times to create 200 bootstrap samples, \(X_1^*, X_2^*, \ldots, X_{15}^*\).
04

Calculate the Median for Each Bootstrap Sample

For each of the 200 bootstrap samples, calculate the sample median. Let the median of the \(i^{th}\) bootstrap sample be called \(\text{median}_i^*\) for \(i = 1, 2, \ldots, 200\).
05

Determine the Bootstrap Distribution of the Medians

Compile all 200 medians obtained from the bootstrap samples to form the bootstrap distribution of medians. Analyze how the median varies among the bootstrap samples.
06

Compute the Standard Error of the Sample Median

Calculate the standard deviation of the bootstrap distribution of sample medians. This standard deviation is the bootstrap estimate of the standard error of the sample median.

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Key Concepts

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

Log-Normal Distribution
A log-normal distribution describes a situation where a variable grows multiplicatively, meaning it's affected by random percent changes. The log-normal distribution is unique because if you take the natural logarithm of a log-normal variable, it converts into a normal distribution. In simpler terms, if you have a log-normal variable, like our random variable \(X\), then \(\ln(X)\) is approximately normal with specific parameters.
  • For our exercise, we have parameters \(\theta = 1.5\) and \(\omega = 0.8\), meaning that if \(\ln(X)\) is normally distributed with mean \(\theta\) and variance \(\omega^2 = 0.64\).
  • This is powerful because many natural phenomena behave in ways that are better modeled by log-normal distributions, particularly when values can’t be negative and they vary over orders of magnitude.
Understanding the log-normal distribution helps in correctly simulating random samples and analyzing their statistical properties.
Random Sampling
Random sampling is fundamental for gathering a representative subset from a larger population. It ensures each member of the population has an equal chance of being included. This methodology is crucial for making sound inferences.

In our exercise, we drew a sample of size \(n = 15\) from a log-normal distribution. Randomness in sampling guarantees that the sample reflects the population without bias, an essential aspect when applying further statistical methods like the bootstrap.
  • The process involves selecting samples in such a way that every possible sample configuration has an equal probability.
  • This guarantees the generalizability of the conclusions drawn from the sample back to the entire population.
When using random sampling, biases are minimized, maintaining the integrity of subsequent analysis.
Standard Error
The standard error is crucial when evaluating the reliability and precision of a sample statistic. It represents the standard deviation of the sampling distribution of a statistic, often used with parameters like the mean or median.

In our context, we employ the standard error to comprehend the variability of the sample median obtained from repeated sampling.
  • The concept is particularly significant when applying the bootstrap method, which estimates the standard error through resampling, providing a practical approach even when traditional assumptions are hard to check.
  • Our problem required computing the standard error by assessing the standard deviation of 200 bootstrap sample medians. This technique effectively conveys the expected variability in the median.
Calculating the standard error gives insight into the dispersion and reliability of the statistical estimates derived from the data.
Sample Median
The sample median is the middle value of a dataset when it is ordered from smallest to largest, providing a measure of central tendency. The median is less affected by skewed data or outliers than the mean, making it a robust statistic.

Given its effectiveness, the sample median is a pivotal statistic in describing the central position of a sample. Each bootstrap sample of our exercise involved calculating this median to construct a distribution from which variability could be analyzed.
  • With 200 bootstrap samples, the median of each reflects how the sample median might behave if we sampled the population many times.
  • By investigating this bootstrap distribution of medians, we obtain a clearer picture of the expected variability and central tendency of the sample.
The sample median’s resilience to extreme values makes it a valuable statistic in exploratory data analysis and inferential statistics.

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Most popular questions from this chapter

An electric utility has placed special meters on 10 houses in a subdivision that measures the energy consumed (demand) at each hour of the day. The company is interested in the energy demand at one specific hour - the hour at which the system experiences the peak consumption. The data from these 10 meters are as follows (in \(\mathrm{KW}): 23.1,15.6,17.4,20.1\), \(19.8,26.4,25.1,20.5,21.9,\) and \(28.7 .\) If \(\mu\) is the true mean peak demand for the 10 houses in this group of houses having the special meters, estimate \(\mu\). Now suppose that the utility wants to estimate the demand at the peak hour for all 5000 houses in this subdivision. Let \(\theta\) be this quantity. Estimate \(\theta\) using the data given. Estimate the proportion of houses in the subdivision that demand at least \(20 \mathrm{KW}\) at the hour of system peak.

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample of size \(n\) from a population with mean \(\mu\) and variance \(\sigma^{2}\) (a) Show that \(\bar{X}^{2}\) is a biased estimator for \(\mu^{2}\) (b) Find the amount of bias in this estimator. (c) What happens to the bias as the sample size \(n\) increases?

The compressive strength of concrete is normally distributed with \(\mu=2500\) psi and \(\sigma=50\) psi. Find the probability that a random sample of \(n=5\) specimens will have a sample mean diameter that falls in the interval from 2499 psi to 2510 psi.

Suppose that \(\Theta_{1}, \Theta_{2},\) and \(\Theta_{3}\) are estimators of \(\theta .\) We know that \(E\left(\Theta_{1}\right)=E\left(\Theta_{2}\right)=0, E\left(\Theta_{3}\right) \neq \theta, V\left(\Theta_{1}\right)=12, V\left(\Theta_{2}\right)=10\) and \(E\left(\Theta_{3}-\theta\right)^{2}=6 .\) Compare these three estimators. Which do you prefer? Why?

Left right arrow A random sample of 36 observations has been drawn from a normal distribution with mean 50 and standard deviation 12. Find the probability that the sample mean is in the interval \(47 \leq X \leq 53\). Is the assumption of normality important? Why?

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