/*! 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 48 A rivet is to be inserted into a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A rivet is to be inserted into a hole. A random sample of \(n=15\) parts is selected, and the hole diameter is measured. The sample standard deviation of the hole diameter measurements is \(s=0.008\) millimeters. Construct a \(99 \%\) lower confidence bound for \(\sigma^{2}\).

Short Answer

Expert verified
The 99% lower confidence bound for \(\sigma^2\) is approximately 0.0000293 square millimeters.

Step by step solution

01

Understand the problem

We need to find the lower confidence bound for the population variance \(\sigma^2\) given a random sample. We are given the sample standard deviation \(s = 0.008\) millimeters and sample size \(n = 15\). The confidence level is \(99\%\).
02

Define the Chi-Square Distribution

The statistic used here is the sample variance \(s^2\), which follows a chi-square distribution \(\chi^2_{n-1}\) when scaled by \((n-1)\sigma^2\). The degrees of freedom \(u\) is \(n - 1 = 14\).
03

Calculate the Sample Variance

The sample variance \(s^2\) is the square of the sample standard deviation. Thus, \(s^2 = (0.008)^2 = 0.000064\) square millimeters.
04

Determine the Chi-Square Critical Value

For a \(99\%\) lower confidence bound, we need the critical value \(\chi^2_{\alpha/2, u}\). Here \(\alpha = 0.01\), so \(\alpha/2 = 0.005\). We use a chi-square distribution table, finding \(\chi^2_{0.005, 14} = 30.58\).
05

Compute the Lower Confidence Bound

The formula for the lower confidence bound of \(\sigma^2\) is \(\frac{(n-1)s^2}{\chi^2_{\alpha/2, u}}\). Substituting, we have \(\frac{14 \times 0.000064}{30.58} \approx 0.0000293\) square millimeters.

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.

Chi-Square Distribution
The Chi-Square distribution is a fundamental concept in statistics, especially when dealing with variance. It is primarily used to assess how a sample variance might represent the population variance. This statistical distribution is denoted as \[ \chi^2_k \]where \(k\) is the degrees of freedom, which we will discuss further in another section.

The Chi-Square distribution is important because it takes sample variances and allows us to make inferences about the population variance. For example, in our original problem, the Chi-Square distribution was used to help construct a confidence bound for a population variance. This is crucial in determining the precision of a machinery component like a rivet or hole's diameter, ensuring they fit precisely.
  • The shape of the Chi-Square distribution depends on its degrees of freedom.
  • It is not symmetrical, skewing right for lower degrees of freedom.
  • As the degrees of freedom increase, the distribution approaches a normal distribution.
Understanding this distribution allows us to evaluate how likely it is that our sample variance approximates the true population variance.
Sample Variance
Sample variance, denoted as \(s^2\), is a measure of the variability or spread within a sample dataset. It quantifies the degree by which individual data points differ from the sample mean. In statistical terms, it serves as an estimate for the population variance. To compute the sample variance, you take the sum of the squared differences between each data point and the sample mean, and then divide by the sample size minus one (n - 1).

In the context of our problem, the sample variance was calculated from a measured standard deviation of the hole diameters. Knowing the sample variance allows us to explore how consistent the hole sizes are and if they can acceptably fit the rivet. The formula for sample variance is:
  • Calculate mean of your data.
  • Find the difference between each data point and the mean.
  • Square these differences.
  • Sum up all these squared differences.
  • Divide by \(n-1\), where \(n\) is the sample size.
This calculated sample variance helps us to derive inferential statistics about the population, especially when accessing population variance.
Population Variance
Population variance, denoted as \(\sigma^2\), is a measure of the spread of a set of data points in a population. It signifies how data points differ from the population mean, providing a comprehensive understanding of variability across the entire population. Unlike sample variance, which is calculated from a subset of data, population variance accounts for every individual in a dataset.

Knowing population variance helps in making accurate predictions and inferences about the entire population, which is crucial in quality control and other analytical fields. This concept is particularly relevant in designing and manufacturing, where consistency and precision are critical. In our exercise, we aimed to find a confidence bound for the population variance based on a sample's data. This helps in determining how well the equipment or product will perform in the real world.
  • A lower population variance implies that data points are close to the mean, showcasing less variability.
  • A higher population variance signifies widespread data dispersion.
  • In many practical situations, especially when the entire population data is unavailable, we use sample data as an estimate for the population variance.
Understanding population variance allows practitioners to make informed decisions about production standards and process improvements.
Degrees of Freedom
Degrees of freedom, often denoted as \(df\) or \(u\), are an integral concept when performing statistical analyses, especially in variance estimation. Degrees of freedom refer to the number of independent values or quantities which can vary in an analysis without breaking any constraints.

In the sample variance calculation, the degrees of freedom are typically \(n-1\), where \(n\) represents the sample size. This adjustment is necessary because estimating the population variance involves estimating the mean from the same data set, forcing a loss of one degree of freedom.
  • Degrees of freedom are crucial when making statistical inferences.
  • They help adjust the variance estimate, ensuring it is unbiased.
  • In hypothesis testing and constructing confidence intervals, it determines certain distribution characteristics, like the shape of the Chi-Square distribution.
In our problem, the degrees of freedom were 14, given the sample size of 15. This influenced how we calculated the confidence bounds for the population variance, underscoring the importance of correctly identifying the degrees of freedom in statistical tests and interval estimations.

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

Information on a packet of seeds claims that \(93 \%\) of them will germinate. Of the 200 seeds that I planted, only 180 germinated. (a) Find a \(95 \%\) confidence interval for the true proportion of seeds that germinate based on this sample. (b) Does this seem to provide evidence that the claim is wrong?

A normal population has known mean \(\mu=50\) and variance \(\sigma^{2}=5\). What is the approximate probability that the sample variance is greater than or equal to \(7.44 ?\) less than or equal to \(2.56 ?\) For a random sample of size (a) \(n=16\) (b) \(n=30\) (c) \(n=71\) (d) Compare your answers to parts (a)-(c) for the approximate probability that the sample variance is greater than or equal to \(7.44 .\) Explain why this tail probability is increasing or decreasing with increased sample size. (e) Compare your answers to parts (a)-(c) for the approximate probability that the sample variance is less than or equal to \(2.56 .\) Explain why this tail probability is increasing or decreasing with increased sample size.

A particular brand of diet margarine was analyzed to determine the level of polyunsaturated fatty acid (in percentages). A sample of six packages resulted in the following data: 16.8,17.2,17.4,16.9,16.5,17.1 (a) Check the assumption that the level of polyunsaturated fatty acid is normally distributed. (b) Calculate a \(99 \%\) confidence interval on the mean \(\mu\). Provide a practical interpretation of this interval. (c) Calculate a \(99 \%\) lower confidence bound on the mean. Compare this bound with the lower bound of the two-sided confidence interval and discuss why they are different.

A civil engineer is analyzing the compressive strength of concrete. Compressive strength is normally distributed with \(\sigma^{2}=1000(\mathrm{psi})^{2} .\) A random sample of 12 specimens has a mean compressive strength of \(\bar{x}=3250\) psi. (a) Construct a \(95 \%\) two-sided confidence interval on mean compressive strength. (b) Construct a \(99 \%\) two-sided confidence interval on mean compressive strength. Compare the width of this confidence interval with the width of the one found in part (a).

A random sample of 50 suspension helmets used by motorcycle riders and automobile race-car drivers was subjected to an impact test, and some damage was observed on 18 of these helmets. (a) Find a \(95 \%\) two-sided confidence interval on the true proportion of helmets that would show damage from this test. (b) Using the point estimate of \(p\) from the 50 helmets, how many helmets must be tested to be \(95 \%\) confident that the error in estimating \(p\) is less than \(0.02 ?\) (c) How large must the sample be if we wish to be at least \(95 \%\) confident that the error in estimating \(p\) is less than 0.02 regardless of the true value of \(p ?\)

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.