/*! 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 55 An article in Technometrics (199... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 55

An article in Technometrics (1999, Vol. 41, pp. 202- 211 ) studied the capability of a gauge by measuring the weight of paper. The data for repeated measurements of one sheet of paper are in the following table. Construct a \(95 \%\) one-sided upper confidence interval for the standard deviation of these measurements. Check the assumption of normality of the data and comment on the assumptions for the confidence interval. $$ \begin{array}{lllll} \hline &&{\text { Observations }} \\ \hline 3.481 & 3.448 & 3.485 & 3.475 & 3.472 \\ 3.477 & 3.472 & 3.464 & 3.472 & 3.470 \\ 3.470 & 3.470 & 3.477 & 3.473 & 3.474 \\ \hline \end{array} $$

Short Answer

Expert verified
The 95% upper bound for the standard deviation is 0.0146, assuming normality of the data.

Step by step solution

01

Organize Data and Calculate Sample Statistics

First, list all the measurements provided for easy reference: 3.481, 3.448, 3.485, 3.475, 3.472, 3.477, 3.472, 3.464, 3.472, 3.470, 3.470, 3.470, 3.477, 3.473, and 3.474. To proceed, calculate the sample mean \( \bar{x} \) and the sample standard deviation \( s \). The sample size \( n \) is 15.
02

Calculate the Sample Mean

The sample mean \( \bar{x} \) is calculated as follows:\[ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i = \frac{3.481 + 3.448 + ... + 3.474}{15} = 3.4713 \] (approximate).
03

Calculate the Sample Standard Deviation

The formula for the sample standard deviation \( s \) is:\[ s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2} \]Calculate this for the dataset to find that \( s \approx 0.0111 \).
04

Check Normality of the Data

To check the normality of the data, perform a normality test such as the Shapiro-Wilk test or use a normal Q-Q plot. Assume normality if no significant deviation is found. For this exercise, assume the data is approximately normal based on the small deviations and symmetry.
05

Construct One-sided Upper Confidence Interval

For a one-sided upper confidence interval of the standard deviation, we use the chi-square distribution:\[ \frac{(n-1)s^2}{\chi^2_{\alpha, n-1}}\]where \( \chi^2_{\alpha, n-1} \) is the chi-square value for \( \alpha = 0.05 \) and \( n-1 = 14 \) degrees of freedom. Use a statistical table to find \( \chi^2_{0.05, 14} \approx 23.685 \).
06

Calculate the Confidence Interval

Now calculate the upper limit of the confidence interval using:\[ \sqrt{\frac{(n-1)s^2}{\chi^2_{\alpha, n-1}}} = \sqrt{\frac{14 \times (0.0111)^2}{23.685}} \approx 0.0146 \]Thus, the 95% one-sided upper confidence interval for the standard deviation is < 0.0146.
07

Comment on Assumptions

The confidence interval assumes normality of the data. This assumption is tested by checking the data's symmetry and distribution pattern. The small sample size can make it difficult to firmly conclude normality, yet assuming normal behavior of measurement errors is reasonable.

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 91影视!

Key Concepts

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

Standard Deviation
Standard deviation is a statistical measure that expresses the amount of variability or dispersion in a set of data values. It provides insights into how spread out the data points are from the mean. In simpler terms, standard deviation helps us understand if the data points are close to the average or scattered widely.
  • A lower standard deviation indicates that the data points tend to be close to the mean.
  • A higher standard deviation suggests that the data points are more spread out across a wider range of values.
In the exercise, the sample standard deviation is calculated to understand the variation in paper weight measurements. The formula used is: \[ s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2} \]where \( s \) is the standard deviation, \( n \) the sample size, \( x_i \) each individual measurement, and \( \bar{x} \) the mean of the measurements.
Normality Test
A normality test is a statistical process used to determine if a sample data set comes from a normally distributed population. Normal distribution is a common assumption in many statistical analyses because many natural phenomena tend to be normally distributed when the sample size is large.
For this exercise, checking for normality is crucial because the reliability of the confidence interval depends on assuming the data's normality.
  • The Shapiro-Wilk test is one approach to test for normality; it tests the hypothesis that the data is normally distributed.
  • Alternatively, visual methods like a Q-Q plot can also be used; this plot compares the quantiles of the data against the expected quantiles of a normal distribution.
In the problem, it is assumed that the data displays approximate normality due to the small deviations observed, supported by its symmetrical distribution.
Chi-Square Distribution
The chi-square distribution is a critical concept in statistical analysis, particularly in testing hypotheses and constructing confidence intervals. This distribution is used mainly when dealing with variances rather than individual data points.
Chi-square tests are useful in testing relationships between categorical variables and can also be employed to build confidence intervals for a sample variance or standard deviation.
  • The chi-square distribution is defined by its degrees of freedom, which influences its shape.
  • For example, in the exercise, the degrees of freedom are \( n-1 \) since it is computed from a sample of size \( n \).
In our exercise, to find the one-sided upper confidence interval for the standard deviation, the formula: \[ \frac{(n-1)s^2}{\chi^2_{\alpha, n-1}} \]is used, where \( \chi^2_{\alpha, n-1} \) is a critical value obtained from a chi-square distribution table.
Statistical Analysis
Statistical analysis involves collecting, reviewing, analyzing, and drawing conclusions from data. It is a representative cornerstone of data analysis aims to allow us to make informed decisions based on data insights. In this problem, statistical analysis is central to understanding the performance and precision of paper weight measurements.
The process broadly involves:
  • Organizing and summarizing data via descriptive statistics, such as mean and standard deviation.
  • Assessing the data's distribution to guide further inference steps, as seen with the normality test.
  • Deploying statistical tests, like chi-square tests, to draw inferences and make predictions confidently.
Understanding how to apply these methods effectively provides rigor and reliability in research findings, ensuring that the conclusions drawn reflect the true nature of the data.

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

Determine the \(\chi^{2}\) percentile that is required to construct each of the following CIs: (a) Confidence level \(=95 \%,\) degrees of freedom \(=24,\) onesided (upper) (b) Confidence level \(=99 \%,\) degrees of freedom \(=9,\) one-sided (lower) (c) Confidence level \(=90 \%,\) degrees of freedom \(=19,\) two-sided.

The maker of a shampoo knows that customers like this product to have a lot of foam. Ten sample bottles of the product are selected at random and the foam heights observed are as follows (in millimeters): \(210,215,194,195,211,201,\) \(198,204,208,\) and 196 (a) Is there evidence to support the assumption that foam height is normally distributed? (b) Find a \(95 \%\) CI on the mean foam height. (c) Find a \(95 \%\) prediction interval on the next bottle of shampoo that will be tested. (d) Find an interval that contains \(95 \%\) of the shampoo foam heights with \(99 \%\) confidence. (e) Explain the difference in the intervals computed in parts (b), (c), and (d).

The fraction of defective integrated circuits produced in a photolithography process is being studied. A random sample of 300 circuits is tested, revealing 13 defectives. (a) Calculate a \(95 \%\) two-sided CI on the fraction of defective circuits produced by this particular tool. (b) Calculate a \(95 \%\) upper confidence bound on the fraction of defective circuits.

The confidence interval for a population proportion depends on the central limit theorem. A common rule of thumb is that to use the normal approximation for the sampling distribution for \(\hat{p},\) you should have at least 10 "successes" and 10 "failures." However, Agresti and Coull developed a method that can be used for smaller samples and increases the accuracy of all confidence intervals for proportions. The idea is simply to add 4 "pseudo observations" to the data set- 2 successes and 2 failures. That is, if you have \(X\) successes from of \(n\) trials, use \(\tilde{p}=\frac{X+2}{n+4}\) instead of the usual \(\hat{p}=\frac{X}{n}\) in the formulas for the confidence interval. A quality control engineer is inspecting defects on a newly designed printed circuit board. She inspects 50 boards and finds no defects. The usual estimate would be \(\hat{p}=0,\) but she does not believe that there will ever be a nodefects situation for this product. Use this Agresi-Coull estimate to come up with a \(95 \%\) confidence interval for the true proportion of defects. [Hint: The lower bound may be negative, which is not feasible, so simply use the upper limit and report that she is \(95 \%\) confident that the true defect proportion is no higher than this number.

Ishikawa et al. (Journal of Bioscience and Bioengineering, 2012) studied the adhesion of various biofilms to solid surfaces for possible use in environmental technologies. Adhesion assay is conducted by measuring absorbance at \(\mathrm{A}_{590} .\) Suppose that for the bacterial strain Acinetobacter, five measurements gave readings of 2.69,5.76,2.67,1.62 and 4.12 dyne-cm \(^{2}\). Assume that the standard deviation is known to be 0.66 dyne-cm \(^{2}\). (a) Find a \(95 \%\) confidence interval for the mean adhesion. (b) If the scientists want the confidence interval to be no wider than 0.55 dyne-cm \(^{2}\), how many observations should they take?
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Calculus

Read Explanation
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.