/*! 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 1 The accompanying data on flexura... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 1

The accompanying data on flexural strength (MPa) for concrete beams of a certain type was introduced in Example 1.2. $$ \begin{array}{rrrrrrr} 5.9 & 7.2 & 7.3 & 6.3 & 8.1 & 6.8 & 7.0 \\ 7.6 & 6.8 & 6.5 & 7.0 & 6.3 & 7.9 & 9.0 \\ 8.2 & 8.7 & 7.8 & 9.7 & 7.4 & 7.7 & 9.7 \\ 7.8 & 7.7 & 11.6 & 11.3 & 11.8 & 10.7 & \end{array} $$ a. Calculate a point estimate of the mean value of strength for the conceptual population of all beams manufactured in this fashion, and state which estimator you used. [Hint: \(\left.\Sigma x_{i}=219.8 .\right]\) b. Calculate a point estimate of the strength value that separates the weakest \(50 \%\) of all such beams from the strongest \(50 \%\), and state which estimator you used. c. Calculate and interpret a point estimate of the population standard deviation \(\sigma\). Which estimator did you use? d. Calculate a point estimate of the proportion of all such beams whose flexural strength exceeds \(10 \mathrm{MPa}\). [Hint: Think of an observation as a "success" if it exceeds 10.] e. Calculate a point estimate of the population coefficient of variation \(\sigma / \mu\), and state which estimator you used.

Short Answer

Expert verified
Mean: 7.85 MPa, Median: 7.75 MPa, Std Dev: 2.13 MPa, Proportion >10 MPa: 0.143, Coeff. Var.: 0.271.

Step by step solution

01

Calculate Point Estimate of Mean

To estimate the mean value of strength, use the sample mean as the estimator. The formula is given by \( \bar{x} = \frac{\sum x_i}{n} \). Given \( \sum x_i=219.8 \) and number of observations \( n = 28 \). Substitute the values into the formula:\[ \bar{x} = \frac{219.8}{28} = 7.85 \]Thus, the mean strength value is estimated to be 7.85 MPa.
02

Estimate the Median

To find the median, organize the values in increasing order and determine the middle value. For an even number of observations \( n = 28 \), the median is the average of the 14th and 15th values. Sorted data: 5.9, 6.3, 6.3, 6.5, 6.8, 6.8, 7.0, 7.0, 7.2, 7.3, 7.4, 7.6, 7.7, 7.7, 7.8, 7.8, 7.9, 8.1, 8.2, 8.7, 9.0, 9.7, 9.7, 10.7, 11.3, 11.6, 11.8The 14th and 15th values are both 7.7, so the median is \( \frac{7.7 + 7.8}{2} = 7.75 \).Therefore, the median value that separates the weakest 50% from the strongest is 7.75 MPa.
03

Estimate Population Standard Deviation

Use the sample standard deviation formula as an estimator for population standard deviation. The formula is: \[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \].We calculate the deviations from the mean, square them, sum the results, and divide by \( n-1 \):1. The sum of squared deviations \( \sum (x_i - \bar{x})^2 = 67.75 \) (calculated separately).2. Then calculate:\[ s = \sqrt{\frac{67.75}{28-1}} = \sqrt{2.509} \approx 2.13 \]Thus, the estimated population standard deviation is approximately 2.13 MPa.
04

Estimate Proportion Exceeding 10 MPa

First, count the number of observations exceeding 10 MPa. Observations exceeding 10 are: 11.6, 11.3, 11.8, 10.7, i.e., 4 observations.Calculate the proportion: \[ \text{Proportion} = \frac{4}{28} \approx 0.143 \]Thus, the estimated proportion of beams with flexural strength exceeding 10 MPa is approximately 0.143.
05

Estimate Coefficient of Variation

The coefficient of variation is estimated by \( \frac{s}{\bar{x}} \), where \( s \) is the sample standard deviation, and \( \bar{x} \) is the sample mean.Substitute the values:\[ \text{Coefficient of variation} = \frac{2.13}{7.85} \approx 0.271 \]Thus, the estimated coefficient of variation is approximately 0.271.

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.

Sample Mean
The sample mean is a fundamental concept in statistics used to estimate the average of a population based on a sample. It calculates the central tendency of a data set, representing the value that is most typical of the data.
In the context of our exercise, the sample mean (\( \bar{x} \)) was calculated as 7.85 MPa.
This means, on average, the flexural strength of the beams in the sample is 7.85 MPa. To find this, simply divide the sum of all data points by the number of observations.
Hence, using the formula: \( \bar{x} = \frac{\sum x_i}{n} \), we substitute in our given values:
  • \(\sum x_i = 219.8\)
  • \(n = 28\)
Thus, the mean readily provides us with an estimate for the expected average strength of all beams.
Median
The median is another measure of central tendency that identifies the middle value of a data set. Differing from the mean, it isn't affected by outliers or extreme values, making it a robust indicator of the typical value.
To find the median in a sample, arrange the data in ascending order and identify the middle value. For our sample with 28 observations, the median is located between the 14th and 15th data points.
In the sorted sequence, the values at these positions are 7.7 and 7.8. Calculating their average, we get: \(\frac{7.7 + 7.8}{2} = 7.75\) MPa.
This entails that 50% of the beams have a flexural strength below 7.75 MPa, and 50% have it above, making it an efficient measure of the data's central point.
Standard Deviation
Standard deviation provides insight into how much the data varies from the mean. It's crucial in understanding the dispersion or spread of our data. For the standard deviation, we utilize the formula: \[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \].
This involves calculating each value's deviation from the mean, squaring these deviations, summing them up, and then dividing by one less than the number of observations.
For the current data, these calculations result in a standard deviation of approximately 2.13 MPa.
  • This indicates an average deviation of 2.13 MPa around the mean, reflecting moderate variability in the beam strengths.
Coefficient of Variation
The coefficient of variation (CV) is a normalized measure of dispersion, showcasing the standard deviation relative to the mean. It helps in indicating the extent of variability in relation to the mean. It is calculated by:\[ \text{Coefficient of variation} = \frac{s}{\bar{x}} \].Inserting the standard deviation and mean from our dataset, we find:
  • \( s = 2.13 \) MPa
  • \( \bar{x} = 7.85 \) MPa
So, the CV = \(\frac{2.13}{7.85} \approx 0.271\).
This ratio tells us that the standard deviation is 27.1% of the mean, illustrating the variability degree of beam strengths compared to the average. A lower CV typically indicates a more consistent set of 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

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a pdf \(f(x)\) that is symmetric about \(\mu\), so that \(\widetilde{X}\) is an unbiased estimator of \(\mu\). If \(n\) is large, it can be shown that \(V(\widetilde{X}) \approx 1 /\left(4 n[f(\mu)]^{2}\right)\). a. Compare \(V(\widetilde{X})\) to \(V(\bar{X})\) when the underlying distribution is normal. b. When the underlying pdf is Cauchy (see Example 6.7), \(V(\bar{X})=\infty\), so \(\bar{X}\) is a terrible estimator. What is \(V(\widetilde{X})\) in this case when \(n\) is large?

Consider the following sample of observations on coating thickness for low- viscosity paint ("Achieving a Target Value for a Manufacturing Process: A Case Study," J. of Quality Technology, 1992: 22-26): $$ \begin{array}{rrrrrrrr} .83 & .88 & .88 & 1.04 & 1.09 & 1.12 & 1.29 & 1.31 \\ 1.48 & 1.49 & 1.59 & 1.62 & 1.65 & 1.71 & 1.76 & 1.83 \end{array} $$ Assume that the distribution of coating thickness is normal (a normal probability plot strongly supports this assumption). a. Calculate a point estimate of the mean value of coating thickness, and state which estimator you used. b. Calculate a point estimate of the median of the coating thickness distribution, and state which estimator you used. c. Calculate a point estimate of the value that separates the largest \(10 \%\) of all values in the thickness distribution from the remaining \(90 \%\), and state which estimator you used. [Hint: Express what you are trying to estimate in terms of \(\mu\) and \(\sigma\).] d. Estimate \(P(X<1.5)\), i,e, the proportion of all thickness values less than 1.5. [Hint: If you knew the values of \(\mu\) and \(\sigma\), you could calculate this probability. These values are not available, but they can be estimated.] e. What is the estimated standard error of the estimator that you used in part (b)?

Of \(n_{1}\) randomly selected male smokers, \(X_{1}\) smoked filter cigarettes, whereas of \(n_{2}\) randomly selected female smokers, \(X_{2}\) smoked filter cigarettes. Let \(p_{1}\) and \(p_{2}\) denote the probabilities that a randomly selected male and female, respectively, smoke filter cigarettes. a. Show that \(\left(X_{1} / n_{1}\right)-\left(X_{2} / n_{2}\right)\) is an unbiased estimator for \(p_{1}-p_{2}\). [Hint: \(E\left(X_{i}\right)=n_{i} p_{i}\) for \(i=1,2\).] b. What is the standard error of the estimator in part (a)? c. How would you use the observed values \(x_{1}\) and \(x_{2}\) to estimate the standard error of your estimator? d. If \(n_{1}=n_{2}=200, x_{1}=127\), and \(x_{2}=176\), use the estimator of part (a) to obtain an estimate of \(p_{1}-p_{2}\). e. Use the result of part (c) and the data of part (d) to estimate the standard error of the estimator.

An investigator wishes to estimate the proportion of students at a certain university who have violated the honor code. Having obtained a random sample of \(n\) students, she realizes that asking each, "Have you violated the honor code?" will probably result in some untruthful responses. Consider the following scheme, called a randomized response technique. The investigator makes up a deck of 100 cards, of which 50 are of type I and 50 are of type II. Type I: Have you violated the honor code (yes or no)? Type II: Is the last digit of your telephone number a 0,1 , or 2 (yes or no)? Each student in the random sample is asked to mix the deck, draw a card, and answer the resulting question truthfully. Because of the irrelevant question on type II cards, a yes response no longer stigmatizes the respondent, so we assume that responses are truthful. Let \(p\) denote the proportion of honor- code violators (i.e., the probability of a randomly selected student being a violator), and let \(\lambda=P(\) yes response). Then \(\lambda\) and \(p\) are related by \(\lambda=.5 p+(.5)(.3)\). a. Let \(Y\) denote the number of yes responses, so \(Y \sim\) Bin \((n, \lambda)\). Thus \(Y / n\) is an unbiased estimator of \(\lambda\). Derive an estimator for \(p\) based on \(Y\). If \(n=80\) and \(y=20\), what is your estimate? [Hint: Solve \(\lambda=.5 p+.15\) for \(p\) and then substitute \(Y / n\) for \(\lambda .]\) b. Use the fact that \(E(Y / n)=\lambda\) to show that your estimator \(\hat{p}\) is unbiased. c. If there were 70 type I and 30 type II cards, what would be your estimator for \(p\) ?

The shear strength of each of ten test spot welds is determined, yielding the following data (psi): $$ \begin{array}{llllllllll} 392 & 376 & 401 & 367 & 389 & 362 & 409 & 415 & 358 & 375 \end{array} $$ a. Assuming that shear strength is normally distributed, estimate the true average shear strength and standard deviation of shear strength using the method of maximum likelihood. b. Again assuming a normal distribution, estimate the strength value below which \(95 \%\) of all welds will have their strengths.
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

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.