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Consider a Weibull distribution with shape parameter 1.5 and scale parameter \(2.0 .\) Generate a graph of the probability distribution. Does it look very much like a normal distribution? Construct a table similar to Table \(7-1\) by drawing 20 random samples of size \(n=10\) from this distribution. Compute the sample average from each sample and construct a normal probability plot of the sample averages. Do the sample averages seem to be normally distributed?

Short Answer

Expert verified
The Weibull distribution does not look like a normal distribution. However, the sample averages appear normally distributed in a normal probability plot.

Step by step solution

01

Understand the Weibull distribution

The Weibull distribution is a continuous probability distribution. It is defined by two parameters: a shape parameter denoted by \(k\) (in this case \(1.5\)) and a scale parameter denoted by \(\lambda\) (here \(2.0\)). The probability density function (PDF) of a Weibull distribution is given by: \[ f(x; k, \lambda) = \frac{k}{\lambda} \left( \frac{x}{\lambda} \right)^{k-1} e^{-(x/\lambda)^k} \] where \(x \geq 0\).
02

Graph the Weibull distribution

To graph the Weibull distribution, use plotting software or a graphing calculator capable of handling probability distributions. Plot the function f(x) using the PDF for values of \(x\) ranging from 0 to roughly 5 or 6, to cover most of the area under the curve for this distribution.
03

Compare with Normal Distribution

Observe the shape of the Weibull distribution graph. A normal distribution is symmetric and bell-shaped. A Weibull distribution with a shape parameter \(k > 1\) tends to be right-skewed, particularly when \(k\) is significantly larger than 1, hence it does not perfectly resemble a normal distribution.
04

Draw Random Samples

Use statistical software or a random number generator programmed to understand the Weibull distribution. Generate 20 random samples, where each sample consists of 10 values drawn from the Weibull distribution with parameters \(k = 1.5\) and \(\lambda = 2.0\).
05

Calculate Sample Averages

For each of the 20 samples, calculate the sample mean by summing the values of the sample and dividing by 10. These sample means represent the sampling distribution.
06

Construct a Normal Probability Plot

Using statistical software or graphing tools, construct a normal probability plot of the 20 sample means. A normal probability plot plots the cumulative probabilities of a dataset against a theoretical normal distribution. If the points form a straight line, the dataset is approximately normally distributed.
07

Analyze Sample Averages' Distribution

Examine the normal probability plot. If the plotted points are approximately linear (i.e., form a straight line), then the sample averages can be considered approximately normally distributed by the Central Limit Theorem.

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Key Concepts

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

Probability Density Function (PDF)
The probability density function (PDF) is a crucial part of the Weibull distribution, which is a continuous probability distribution often used to model life data, reliabilities, and survival analysis. The PDF provides the probabilities of all possible values that a random variable can take on, meaning it depicts how probabilities are distributed across different values in a continuous space.
In terms of the Weibull distribution, the PDF is expressed by the formula:\[ f(x; k, \lambda) = \frac{k}{\lambda} \left( \frac{x}{\lambda} \right)^{k-1} e^{-(x/\lambda)^k} \]Here:
  • \(x\) is the random variable.
  • \(k\) is the shape parameter, influencing if the distribution is right or left-skewed.
  • \(\lambda\) is the scale parameter, affecting the stretching or compressing of the distribution graph along the x-axis.
The Weibull distribution's PDF for shape parameter 1.5 and scale parameter 2.0 is therefore skewed right and doesn't mimic the bell shape of a normal distribution entirely due to these parameters, which makes it valuable yet distinguishable from other distributions.
Normal Probability Plot
A normal probability plot is a graphical tool used to determine if a dataset approximately follows a normal distribution. It's particularly valuable when assessing the nature of sample averages, as seen in statistical analyses of distributions like the Weibull.
The process of creating a normal probability plot typically involves:
  • Plotting the sorted sample data (in this case, the sample averages) on one axis.
  • Plotting the expected quantiles from a standard normal distribution on the other axis.
When the sample data follow a normal distribution, the points plotted on this graph form a straight line. For the exercise at hand, sample averages from the Weibull distribution were assessed using a normal probability plot. If these points align in a straight-line trend, the distribution of averages can be considered approximately normal, confirming the application of the Central Limit Theorem.
Central Limit Theorem
The Central Limit Theorem (CLT) is one of the fundamental principles in statistics. It states that the distribution of sample means approaches a normal distribution as the sample size becomes larger, regardless of the original distribution of the population. This principle is crucial when working with random samples from distributions like the Weibull.
Here's how CLT plays out in the given exercise:
  • 20 samples, each of size 10, were drawn from a Weibull distribution with specific parameters.
  • The sample means were computed to explore this derived sampling distribution.
  • Under the CLT, even if the original Weibull distribution is not normal, the distribution of these sample means tends to approximate a normal distribution.
By analyzing the normal probability plot of the sample means, one can visually confirm the application of the CLT. This analysis helps comprehend the tendency of the sample means to align closely with normal distribution properties, showcasing yet again the robustness of the Central Limit Theorem in practical, statistical scenarios.

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Most popular questions from this chapter

An exponential distribution is known to have a mean of \(10 .\) You want to find the standard error of the median of this distribution if a random sample of size 8 is drawn. Use the bootstrap method to find the standard error, using \(n_{a}=100\) bootstrap samples.

Data on the oxide thickness of semiconductor wafers are as follows: 425,431,416,419,421,436,418,410 , \(431,433,423,426,410,435,436,428,411,426,409,437,\) 422,428,413,416 (a) Calculate a point estimate of the mean oxide thickness for all wafers in the population. (b) Calculate a point estimate of the standard deviation of oxide thickness for all wafers in the population. (c) Calculate the standard error of the point estimate from part (a). (d) Calculate a point estimate of the median oxide thickness for all wafers in the population. (e) Calculate a point estimate of the proportion of wafers in the population that have oxide thickness of more than 430 angstroms.

Two different plasma etchers in a semiconductor factory have the same mean etch rate \(\mu\). However, machine 1 is newer than machine 2 and consequently has smaller variability in etch rate. We know that the variance of etch rate for matching 1 is \(\sigma_{1}^{2}\), and for machine 2 , it is \(\sigma_{2}^{2}=a \sigma_{1}^{2}\). Suppose that we have \(n_{1}\) independent observations on etch rate from machine 1 and \(n_{2}\) independent observations on etch rate from machine \(2 .\) (a) Show that \(\hat{\mu}=\alpha \bar{X}_{1}+(1-\alpha) \bar{X}_{2}\) is an unbiased estimator of \(\mu\) for any value of \(\alpha\) between zero and one. (b) Find the standard error of the point estimate of \(\mu\) in part (a). (c) What value of \(\alpha\) would minimize the standard error of the point estimate of \(\mu\) ?

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.

Researchers in the Hopkins Forest (see Exercise 7.16 ) also count the number of maple trees (genus acer) in plots throughout the forest. The following is a histogram of the number of live maples in 1002 plots sampled over the past 20 years. The average number of maples per plot was 19.86 trees with a standard deviation of 23.65 trees. (a) If we took the mean of a sample of eight plots, what would be the standard error of the mean? (b) Using the central limit theorem, what is the probability that the mean of the eight would be within 1 standard error of the mean? (c) Why might you think that the probability that you calculated in (b) might not be very accurate?

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