/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 173 When \(X_{1}, X_{2}, \ldots, X_{... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

When \(X_{1}, X_{2}, \ldots, X_{n}\) are independent Poisson random variables, each with parameter \(\lambda,\) and \(n\) is large, the sample mean \(\bar{X}\) has an approximate normal distribution with mean \(\lambda\) and variance \(\lambda / n .\) Therefore, $$Z=\frac{\bar{X}-\lambda}{\sqrt{\lambda / n}}$$ has approximately a standard normal distribution. Thus, we can test \(H_{0}: \lambda=\lambda_{0}\) by replacing \(\lambda\) in \(Z\) by \(\lambda_{0} .\) When \(X_{i}\) are Poisson variables, this test is preferable to the large-sample test of Section \(9-2.3,\) which would use \(S / \sqrt{n}\) in the denominator because it is designed just for the Poisson distribution. Suppose that the number of open circuits on a semiconductor wafer has a Poisson distribution. Test data for 500 wafers indicate a total of 1038 opens. Using \(\alpha=0.05,\) does this suggest that the mean number of open circuits per wafer exceeds \(2.0 ?\)

Short Answer

Expert verified
We do not reject \(H_0\); there is insufficient evidence to show the mean exceeds 2.0.

Step by step solution

01

Define Hypotheses

We need to test if the mean number of open circuits per wafer is greater than 2.0. Hence, the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_a\)) are:\[H_0: \lambda = 2.0\]\[H_a: \lambda > 2.0\]
02

Calculate Sample Mean

The sample mean \(\bar{X}\) is the total number of open circuits divided by the number of wafers, so:\[\bar{X} = \frac{1038}{500} = 2.076\]
03

Compute Test Statistic Z

Using the given test statistic formula:\[Z = \frac{\bar{X} - \lambda_0}{\sqrt{\frac{\lambda_0}{n}}}\]Substitute \(\bar{X} = 2.076\), \(\lambda_0 = 2.0\), and \(n = 500\):\[Z = \frac{2.076 - 2.0}{\sqrt{\frac{2.0}{500}}} = \frac{0.076}{0.0632455532} \approx 1.201\]
04

Determine Critical Value

For a one-tailed test at \(\alpha = 0.05\), using a standard normal distribution table, the critical value \(Z_{\alpha}\) is approximately 1.645.
05

Conclusion

Compare the computed test statistic \(Z \approx 1.201\) to the critical value \(Z_{\alpha} = 1.645\). Since \(1.201 < 1.645\), we do not reject the null hypothesis \(H_0\). There is not enough evidence at the 0.05 significance level to conclude that the mean number of open circuits exceeds 2.0.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Poisson Distribution
The Poisson Distribution is a fundamental concept in probability theory and statistics. It's often used to model the number of times an event occurs in a fixed interval of time or space. For this distribution, two parameters are important:
  • Mean (\( \lambda \)): which represents the average number of occurrences in the interval.
  • Variance (\( \lambda \)): which is the same as the mean for a Poisson distribution.
The unique property here is that both the mean and variance are equal. This distribution is typically skewed if the mean is small, but it becomes more symmetrical as the mean increases.
The Poisson model is particularly useful for count data and modeling rare events, making it ideal for our semiconductor wafer example where each wafer has a small probability of defects. As the sample size increases, like with our 500 wafers, the Poisson distribution begins to approximate a normal distribution.
Standard Normal Distribution
The Standard Normal Distribution is a special kind of normal distribution. It has a mean of zero and a standard deviation of one. This is a key concept for statisticians as it simplifies the process of hypothesis testing and data analysis.
When data is normally distributed, you can assess probabilities and critical values using the standard normal distribution, often through standardized test scores known as Z-scores. By converting various distributions to fit a standard normal, you eliminate the initial complexity and only deal with the standardized form of the data.
  • Z-score: This score tells us how many standard deviations an element is from the mean. It's calculated using:\[Z = \frac{X - \mu}{\sigma}\]with \( \mu \) being the mean and \( \sigma \) the standard deviation of the population.
This is particularly useful as we transition from a Poisson to a normal distribution model in larger samples, allowing us to perform hypothesis testing effectively, like in our semiconductor wafer example.
Test Statistic
A Test Statistic is a standardized value that is calculated from sample data during a hypothesis test. It is used to decide whether to reject the null hypothesis. In the problem above, the Z-test statistic calculated from the sample data of the wafer's open circuits is used.
The test statistic provides a means to evaluate the discrepancy between the sample data and the null hypothesis. In our example, \( Z\) was determined using:\[Z = \frac{\bar{X} - \lambda_0}{\sqrt{\frac{\lambda_0}{n}}}\]This particular Z-score formula lets us measure how far our sample mean (\(\bar{X}\)) is from the hypothetical population mean (\(\lambda_0\)) while accounting for sample size (\(n\)).
Once the test statistic is computed, we compare it with a critical value from the standard normal distribution to make a decision regarding the null hypothesis. This comparison helps us conclude if the observed effect is statistically significant, as was done in the open circuits test case.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The mean bond strength of a cement product must be at least 1000 psi. The process by which this material is manufactured must show equivalence to this standard. If the process can manufacture cement for which the mean bond strength is at least 9750 psi, it will be considered equivalent to the standard. A random sample of six observations is available, and the sample mean and standard deviation of bond strength are 9360 psi and 42.6 psi, respectively. (a) State the appropriate hypotheses that must be tested to demonstrate equivalence. (b) What are your conclusions using \(\alpha=0.05 ?\)

Consider the following frequency table of observations on the random variable \(X\) : $$\begin{array}{llrrrr}\text { Values } & 0 & 1 & 2 & 3 & 4 \\\\\text { Frequency } & 4 & 21 & 10 & 13 & 2\end{array}$$ (a) Based on these 50 observations, is a binomial distribution with \(n=6\) and \(p=0.25\) an appropriate model? Perform a goodness-of-fit procedure with \(\alpha=0.05 .\) (b) Calculate the \(P\) -value for this test.

An article in the ASCE Journal of Energy Engineering (1999, Vol. \(125,\) pp. \(59-75\) ) describes a study of the thermal inertia properties of autoclaved aerated concrete used as a building material. Five samples of the material were tested in a structure, and the average interior temperatures \(\left({ }^{\circ} \mathrm{C}\right)\) reported were as follows: \(23.01,22.22,22.04,22.62,\) and \(22.59 .\) (a) Test the hypotheses \(H_{0}: \mu=22.5\) versus \(H_{1}: \mu \neq 22.5,\) using \(\alpha=0.05 .\) Find the \(P\) -value. (b) Check the assumption that interior temperature is normally distributed. (c) Compute the power of the test if the true mean interior temperature is as high as 22.75 . (d) What sample size would be required to detect a true mean interior temperature as high as 22.75 if you wanted the power of the test to be at least \(0.9 ?\) (e) Explain how the question in part (a) could be answered by constructing a two-sided confidence interval on the mean interior temperature.

A manufacturer produces crankshafts for an automobile engine. The crankshafts wear after 100,000 miles (0.0001 inch) is of interest because it is likely to have an impact on warranty claims. A random sample of \(n=15\) shafts is tested and \(\bar{x}=2.78 .\) It is known that \(\sigma=0.9\) and that wear is normally distributed. (a) Test \(H_{0}: \mu=3\) versus \(H_{1}: \mu \neq 3\) using \(\alpha=0.05 .\) (b) What is the power of this test if \(\mu=3.25 ?\) (c) What sample size would be required to detect a true mean of 3.75 if we wanted the power to be at least \(0.9 ?\)

For the hypothesis test \(H_{0}: \mu=10\) against \(H_{1}: \mu>10\) and variance known, calculate the \(P\) -value for each of the following test statistics. (a) \(z_{0}=2.05\) (b) \(z_{0}=-1.84\) (c) \(z_{0}=0.4\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.