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Chapter 5: Problem 45

Carry out a simulation experiment using a statistical computer package or other software to study the sampling distribution of \(\bar{X}\) when the population distribution is lognormal with \(E(\ln (X))=3\) and \(V(\ln (X))=1\). Consider the four sample sizes \(n=10,20,30\), and 50 , and in each case use 500 replications. For which of these sample sizes does the \(\bar{X}\) sampling distribution appear to be approximately normal?

Short Answer

Expert verified
The sampling distribution of \(\bar{X}\) appears approximately normal for \(n = 30\) and \(n = 50\).

Step by step solution

01

Define the Lognormal Distribution Parameters

The lognormal distribution is characterized by the mean and variance of the natural logarithm of your data, which are given as follows: \(E(\ln (X))=3\) and \(V(\ln (X))=1\). This means the parameters for the underlying normal distribution are \(\mu = 3\) (mean) and \(\sigma^2 = 1\) (variance).
02

Set Up the Simulation

For each sample size \(n = 10, 20, 30, 50\), we will simulate 500 samples from a lognormal distribution with parameters calculated in Step 1. Each sample consists of \(n\) observations.
03

Calculate Sample Means

For each of the 500 replications, compute the sample mean \(\bar{X}\). This is done by taking the average of each simulated sample.
04

Analyze the Sampling Distribution for Each Sample Size

Plot histograms or use statistical software to visualize the distribution of the 500 sample means for each sample size. This will allow us to visually assess the shape of the sampling distribution.
05

Assess Normal Approximation

Evaluate the shape of each distribution by looking for symmetry and bell shape characteristic of a normal distribution. Additionally, use statistical tests such as the Shapiro-Wilk test to check normality.
06

Interpret Results

Compare the results from each sample size. The Central Limit Theorem suggests that larger sample sizes should yield a sampling distribution of \(\bar{X}\) that is closer to normal.

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Key Concepts

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

Lognormal Distribution
A lognormal distribution is quite a fascinating statistical concept. It occurs when the logarithm of a variable is normally distributed. This means that if we take any variable and transform it using the natural logarithm, and this transformed variable follows a normal distribution, then the original variable follows a lognormal distribution. Lognormal distributions have some intriguing properties that make them very useful:
  • They are positively skewed, meaning they have a longer tail on the right side.
  • They are bound at zero and can't be negative, which makes them useful for modeling things like stock prices or any value that can't logically be less than zero.
  • The parameters for the normal distribution of the log of the variables, which are the mean (\(\mu\)) and variance (\(\sigma^2\)), define the lognormal distribution.
In our exercise, we are given that \(E(\ln (X)) = 3\) and \(V(\ln (X)) = 1\), which translates into the parameters \(\mu = 3\) and \(\sigma^2 = 1\) for the normal distribution. Understanding these properties is critical for interpreting our simulation results.
Central Limit Theorem
The Central Limit Theorem (CLT) is a cornerstone of statistics. It asserts that the distribution of sample means will approximate a normal distribution more closely as the sample size increases, regardless of the original distribution's shape. This is why it鈥檚 such a powerful concept:
  • The CLT allows statisticians to make inferences about populations from samples.
  • It applies even to non-normally distributed populations, which is extremely valuable in real-world data analysis.
  • The rate at which the distribution of the sample means becomes normal as the sample size increases can differ depending on the skewness and kurtosis of the data.
In the original exercise, the CLT is used to explain why the sample means (\(\bar{X}\)) from a lognormal distribution, which is a skewed distribution, become more normal-looking as the sample size increases. The exercise involves testing various sample sizes to see at what point the approximation to a normal distribution becomes apparent.
Simulation Experiment
A simulation experiment in statistics involves creating artificial data using computer software to study statistical phenomena. This technique is particularly useful when analytical solutions are complex or infeasible. Let's break down how a simulation experiment is carried out:
  • Define the parameters of your population distribution, which in our case is the lognormal distribution with \(\mu = 3\) and \(\sigma^2 = 1\).
  • Use a computer program to generate a large number of random samples from this distribution. In the exercise, it was specified to conduct 500 replications for each of the four different sample sizes.
  • Calculate the statistical measure of interest, which here is the sample mean (\(\bar{X}\)), for each replication.
  • Analyze the distribution of the sample means to evaluate the sampling distribution's shape.
Simulation experiments provide empirical insights into statistical laws such as the Central Limit Theorem and can clarify the properties of complex distributions like the lognormal.
Shapiro-Wilk Test
The Shapiro-Wilk test is a widely used statistical test for normality. Its main aim is to determine whether a sample comes from a normally distributed population. Here's what you need to know about it:
  • The test calculates a W statistic, which assesses how much a set of data deviates from a normal distribution.
  • A small W value might indicate a significant departure from normality.
  • The test provides a p-value; a p-value less than a chosen significance level (commonly 0.05) means rejecting the hypothesis that the data are from a normal distribution.
  • It is powerful and suitable for small to medium-sized samples.
In the context of the exercise, the Shapiro-Wilk test helps to determine at which sample size the sampling distribution of \(\bar{X}\) starts to resemble a normal distribution. Utilizing such statistical tests in conjunction with visual analysis (like histograms) strengthens conclusions about the normality of the data.

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Most popular questions from this chapter

A binary communication channel transmits a sequence of "bits" ( 0 s and \(1 \mathrm{~s})\). Suppose that for any particular bit transmitted, there is a \(10 \%\) chance of a transmission error (a 0 becoming a 1 or a 1 becoming a 0 ). Assume that bit errors occur independently of one another. a. Consider transmitting 1000 bits. What is the approximate probability that at most 125 transmission errors occur? b. Suppose the same 1000-bit message is sent two different times independently of one another. What is the approximate probability that the number of errors in the first transmission is within 50 of the number of errors in the second?

Annie and Alvie have agreed to meet between 5:00 P.M. and 6:00 P.M. for dinner at a local health-food restaurant. Let \(X=\) Annie's arrival time and \(Y=\) Alvie's arrival time. Suppose \(X\) and \(Y\) are independent with each uniformly distributed on the interval \([5,6]\). a. What is the joint pdf of \(X\) and \(Y\) ? b. What is the probability that they both arrive between \(5: 15\) and \(5: 45 ?\) c. If the first one to arrive will wait only \(10 \mathrm{~min}\) before leaving to eat elsewhere, what is the probability that they have dinner at the health- food restaurant? [Hint: The event of interest is \(A=\left\\{(x, y):|x-y| \leq \frac{1}{6}\right\\}\).

A student has a class that is supposed to end at 9:00 A.M. and another that is supposed to begin at 9:10 A.M. Suppose the actual ending time of the 9 A.M. class is a normally distributed rv \(X_{1}\) with mean \(9: 02\) and standard deviation \(1.5\) min and that the starting time of the next class is also a normally distributed rv \(X_{2}\) with mean \(9: 10\) and standard deviation \(1 \mathrm{~min}\). Suppose also that the time necessary to get from one classroom to the other is a normally distributed rv \(X_{3}\) with mean \(6 \mathrm{~min}\) and standard deviation \(1 \mathrm{~min}\). What is the probability that the student makes it to the second class before the lecture starts? (Assume independence of \(X_{1}, X_{2}\), and \(X_{3}\), which is reasonable if the student pays no attention to the finishing time of the first class.)

In cost estimation, the total cost of a project is the sum of component task costs. Each of these costs is a random variable with a probability distribution. It is customary to obtain information about the total cost distribution by adding together characteristics of the individual component cost distributions-this is called the "roll-up" procedure. For example, \(E\left(X_{1}+\cdots+X_{n}\right)=E\left(X_{1}\right)+\cdots+E\left(X_{n}\right)\), so the roll-up procedure is valid for mean cost. Suppose that there are two component tasks and that \(X_{1}\) and \(X_{2}\) are independent, normally distributed random variables. Is the roll-up procedure valid for the 75 th percentile? That is, is the 75 th percentile of the distribution of \(X_{1}+X_{2}\) the same as the sum of the 75 th percentiles of the two individual distributions? If not, what is the relationship between the percentile of the sum and the sum of percentiles? For what percentiles is the roll-up procedure valid in this case?

The breaking strength of a rivet has a mean value of 10,000 psi and a standard deviation of \(500 \mathrm{psi}\). a. What is the probability that the sample mean breaking strength for a random sample of 40 rivets is between 9900 and 10,200 ? b. If the sample size had been 15 rather than 40 , could the probability requested in part (a) be calculated from the given information?
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