/*! 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 15 A civil engineer is analyzing th... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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).

Short Answer

Expert verified
The 95% CI is [3231.11, 3268.89] psi, and the 99% CI is [3226.47, 3273.53] psi; 99% CI is wider.

Step by step solution

01

Understanding the Problem

We need to construct two confidence intervals for the mean compressive strength using the given sample mean (\(\bar{x} = 3250\) psi), sample size (\(n = 12\)), and population variance (\(\sigma^2 = 1000\,\text{psi}^2\)). The confidence levels required are 95% and 99%.
02

Calculate Standard Deviation

The standard deviation \(\sigma\) is the square root of the variance. Given \(\sigma^2 = 1000\), we find \(\sigma = \sqrt{1000} \approx 31.62\,\text{psi}\).
03

Calculate the Standard Error

The standard error (SE) is calculated as \(\text{SE} = \frac{\sigma}{\sqrt{n}}\), where \(n = 12\) is the sample size. Plugging in the numbers, we have \(\text{SE} = \frac{31.62}{\sqrt{12}} \approx 9.13\).
04

Find the Z-Score for 95% Confidence Interval

For a 95% confidence interval, the Z-score corresponding to \(\alpha/2 = 0.025\) is approximately 1.96. This can be found using standard normal distribution tables.
05

Calculate the 95% Confidence Interval

Using the formula for a confidence interval, \(\bar{x} \pm Z\cdot\text{SE}\), we substitute \(\bar{x} = 3250\), \(Z = 1.96\), and \(\text{SE} = 9.13\). The interval is calculated as \(3250 \pm 1.96 \times 9.13\), resulting in \([3231.11, 3268.89]\).
06

Find the Z-Score for 99% Confidence Interval

For a 99% confidence interval, the Z-score corresponding to \(\alpha/2 = 0.005\) is approximately 2.576.
07

Calculate the 99% Confidence Interval

Using the same formula, \(\bar{x} \pm Z\cdot\text{SE}\), with \(Z = 2.576\), we calculate \(3250 \pm 2.576 \times 9.13\), resulting in \([3226.47, 3273.53]\).
08

Compare the Widths of Confidences Intervals

The width of the 95% CI is \(3268.89 - 3231.11 = 37.78\) psi, while the width of the 99% CI is \(3273.53 - 3226.47 = 47.06\) psi. The 99% confidence interval is wider than the 95% interval, which is typical as increasing the confidence level increases the interval width.

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.

Normal Distribution
When working with statistics, the normal distribution is a vital concept. It's a probability distribution that is symmetrical and bell-shaped, which means that data tends to cluster around a central mean. In many natural phenomena, data follows this type of distribution, making it crucial for various statistical analyses.

For instance, in the problem of analyzing compressive strength of concrete, it is assumed that the data follows a normal distribution. This assumption allows engineers to make predictions about population parameters using sample data. One key property of a normal distribution is that it is defined by two parameters: the mean and the standard deviation. These parameters help to describe the center of the data and how spread out it is around this center.

Understanding normal distribution is essential because it allows us to apply tools like confidence intervals and hypothesis tests, which are based on this distribution.
Z-score
In statistics, the Z-score is a very useful tool. It tells us how many standard deviations an element is from the mean. When constructing confidence intervals, Z-scores help us determine the extent to which a sample mean deviates from the population mean.

The importance of Z-scores arises when calculating confidence intervals. For example, when a sample mean is needed to estimate a population mean, we use Z-scores from the standard normal distribution. These scores correspond to the desired confidence level, like 95% or 99%, and help adjust the boundaries of our estimates accordingly.

To illustrate with this exercise, for a 95% confidence interval, the Z-score is approximately 1.96. This value indicates that we expect the sample mean to fall within 1.96 standard deviations from the population mean 95% of the time. Similarly, for a 99% confidence interval, the Z-score is about 2.576, suggesting a wider interval to be more confident in capturing the true mean.
Standard Error
The standard error (SE) is an indication of the variability or precision of the sample mean as an estimate of the population mean. It is calculated as the standard deviation divided by the square root of the sample size.

For the given exercise, the formula for SE is: \( SE = \frac{\sigma}{\sqrt{n}} \), where \(\sigma\) is the standard deviation of the population, and \(n\) is the sample size. In this context, it helps understand how much the sample mean is likely to vary from the population mean. The smaller the standard error, the more precise the sample mean is as an estimate.

In practical terms, when constructing confidence intervals, the standard error plays a crucial role. A smaller SE will result in a narrower confidence interval, reflecting greater precision, whereas a larger SE will lead to a wider interval, indicating less precision.
Population Variance
Population variance (\(\sigma^2\)) measures how data points in a population deviate from the population mean. It is the average of the squared differences from the mean, providing a sense of the dispersion within the entire population.

In the context of the exercise related to compressive strength, the population variance is given as 1000 (\( ext{psi}^2\)). This number alone provides a wealth of information about the variability of compressive strengths across concrete specimens.

Why is understanding population variance important? It's because it gives insight into how much individual data points are expected to fluctuate. This level of variability directly impacts the standard deviation, which subsequently affects other statistical calculations, like the standard error.
  • A high variance suggests data points are more spread out, indicating a wider range of values.
  • A low variance indicates data points are closer to the mean, suggesting less variability.

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

An article in Nuclear Engineering International (February \(1988,\) p. 33 ) describes several characteristics of fuel rods used in a reactor owned by an electric utility in Norway. Measurements on the percentage of enrichment of 12 rods were reported as follows: \(\begin{array}{llllll}2.94 & 3.00 & 2.90 & 2.75 & 3.00 & 2.95 \\ 2.90 & 2.75 & 2.95 & 2.82 & 2.81 & 3.05\end{array}\) (a) Use a normal probability plot to check the normality assumption. (b) Find a \(99 \%\) two-sided confidence interval on the mean percentage of enrichment. Are you comfortable with the statement that the mean percentage of enrichment is \(2.95 \%\) ? Why?

The percentage of titanium in an alloy used in aerospace castings is measured in 51 randomly selected parts. The sample standard deviation is \(s=0.37\). Construct a \(95 \%\) two sided confidence interval for \(\sigma .\)

Suppose that \(n=100\) random samples of water from a freshwater lake were taken and the calcium concentration (milligrams per liter) measured. A \(95 \%\) CI on the mean calcium concentration is \(0.49 \leq \mu \leq 0.82\). (a) Would a \(99 \%\) CI calculated from the same sample data be longer or shorter? (b) Consider the following statement: There is a \(95 \%\) chance that \(\mu\) is between 0.49 and \(0.82 .\) Is this statement correct? Explain your answer. (c) Consider the following statement: If \(n=100\) random samples of water from the lake were taken and the \(95 \% \mathrm{CI}\) on \(\mu\) computed, and this process were repeated 1000 times, 950 of the CIs would contain the true value of \(\mu\). Is this statement correct? Explain your answer.

Determine the \(t\) -percentile that is required to construct each of the following two-sided confidence intervals: (a) Confidence level \(=95 \%,\) degrees of freedom \(=12\) (b) Confidence level \(=95 \%,\) degrees of freedom \(=24\) (c) Confidence level \(=99 \%,\) degrees of freedom \(=13\) (d) Confidence level \(=99.9 \%,\) degrees of freedom \(=15\)

An Izod impact test was performed on 20 specimens of PVC pipe. The sample mean is \(\bar{x}=1.25\) and the sample standard deviation is \(s=0.25 .\) Find a \(99 \%\) lower confidence bound on Izod impact strength.

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.