/*! 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 7 The article "Some Field Experien... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 7

The article "Some Field Experience in the Use of an Accelerated Method in Estimating 28-Day Strength of Concrete" ( \(J\). of Amer. Concrete Institute, 1969: 895 ) considered regressing \(y=28\)-day standard-cured strength (psi) against \(x=\) accelerated strength (psi). Suppose the equation of the true regression line is \(y=1800+1.3 x\) a. What is the expected value of 28-day strength when accelerated strength \(=2500\) ? b. By how much can we expect 28-day strength to change when accelerated strength increases by 1 psi? c. Answer part (b) for an increase of 100 psi. d. Answer part (b) for a decrease of 100 psi.

Short Answer

Expert verified
a. 5050 psi; b. Increase by 1.3 psi; c. Increase by 130 psi; d. Decrease by 130 psi.

Step by step solution

01

Identify the Given Regression Equation

The problem provides the regression equation: \( y = 1800 + 1.3x \). This equation relates the 28-day standard-cured strength, \(y\), to the accelerated strength, \(x\).
02

Calculate the Expected Value for Accelerated Strength of 2500 psi

To find the expected 28-day strength when the accelerated strength \(x = 2500\), substitute \(2500\) into the regression equation: \ \( y = 1800 + 1.3 \times 2500 \). \ First, calculate \(1.3 \times 2500 = 3250\). Then, add \(1800\) to \(3250\) to get \(5050\). \ Thus, the expected value is \(5050\) psi.
03

Determine the Change in 28-day Strength for 1 psi Increase in Accelerated Strength

From the regression equation \(y = 1800 + 1.3x\), the coefficient of \(x\) is \(1.3\). This coefficient represents the change in \(y\) for a one-unit increase in \(x\). \ Therefore, the expected change in 28-day strength for an increase of 1 psi in accelerated strength is \(1.3\) psi.
04

Calculate the Change for 100 psi Increase in Accelerated Strength

Using the same coefficient \(1.3\) from the regression equation, for every 1 psi increase of \(x\), \(y\) increases by \(1.3\) psi. \ For a 100 psi increase, the change will be \(1.3 \times 100 = 130\) psi.
05

Calculate the Change for 100 psi Decrease in Accelerated Strength

Similarly, for a decrease of 100 psi, the change in \(y\) will still be \(130\) psi, but negative, reflecting a decrease. \ Thus, for a 100 psi decrease, the 28-day strength will decrease by \(130\) psi.

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.

Regression Equation
The regression equation is a mathematical expression that describes the relationship between two variables. In this case, the regression equation provided is \( y = 1800 + 1.3x \). Here, \( y \) represents the 28-day standard-cured strength of concrete, while \( x \) represents the accelerated strength.
  • The number "1800" is the y-intercept, which is the expected value of \( y \) when \( x \) is zero. It indicates where the line crosses the y-axis.
  • The coefficient "1.3" accompanying \( x \) signifies how much \( y \) changes for each unit increase in \( x \).
This form of the equation helps us predict the value of \( y \) based on any given \( x \), providing a foundation for analyzing relationships in data.
Expected Value Calculation
Calculating the expected value involves substituting a specific value of \( x \) into the regression equation to find the corresponding value of \( y \). This helps in predicting the dependent variable based on the independent one.In the exercise, we calculate the expected 28-day strength when the accelerated strength is 2500 psi:
  • Substitute 2500 for \( x \) in the equation: \( y = 1800 + 1.3 \times 2500 \).
  • First, compute \( 1.3 \times 2500 = 3250 \).
  • Add 3250 to 1800 to get 5050.
Thus, the expected 28-day strength for an accelerated strength of 2500 psi is 5050 psi.
Coefficient Interpretation
In regression equations, coefficients tell us how much the dependent variable changes when the independent variable increases by one unit. For the regression equation \( y = 1800 + 1.3x \), the coefficient is 1.3.
  • This means that for every 1 psi increase in accelerated strength, the expected 28-day strength increases by 1.3 psi.
  • It also works the other way around; for a decrease of 1 psi in accelerated strength, the expected 28-day strength would decrease by 1.3 psi.
Interpreting coefficients allows us to understand the sensitivity of \( y \) to changes in \( x \), which is crucial in predicting outcomes and making informed decisions.
Strength Estimation
Strength estimation involves using the regression equation to predict changes in strength with varying levels of accelerated strength. Let's consider different scenarios:
  • **For a 100 psi increase**: Since each unit increase in \( x \) leads to a 1.3 psi increase in \( y \), a 100 psi increase results in \( 1.3 \times 100 = 130 \) psi increase in the 28-day strength.
  • **For a 100 psi decrease**: Similarly, a 100 psi decrease in \( x \) leads to a \( 1.3 \times 100 = 130 \) psi decrease in the 28-day strength.
This understanding allows us to reliably predict how much the strength of concrete will change with different accelerated strengths.

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

For the past decade, rubber powder has been used in asphalt cement to improve performance. The article "Experimental Study of Recycled Rubber-Filled High- Strength Concrete" (Magazine of Concrete Res., 2009: 549-556) includes a regression of \(y=\) axial strength (MPa) on \(x=\) cube strength (MPa) based on the following sample data: $$ \begin{array}{r|rrrrrrrrrr} x & 112.3 & 97.0 & 92.7 & 86.0 & 102.0 & 99.2 & 95.8 & 103.5 & 89.0 & 86.7 \\ \hline y & 75.0 & 71.0 & 57.7 & 48.7 & 74.3 & 73.3 & 68.0 & 59.3 & 57.8 & 48.5 \end{array} $$ a. Obtain the equation of the least squares line, and interpret its slope. b. Calculate and interpret the coefficient of determination. c. Calculate and interpret an estimate of the error standard deviation \(\sigma\) in the simple linear regression model.

Suppose that in a certain chemical process the reaction time \(y(\mathrm{hr})\) is related to the temperature \(\left({ }^{\circ} \mathrm{F}\right)\) in the chamber in which the reaction takes place according to the simple linear regression model with equation \(y=\) \(5.00-.01 x\) and \(\sigma=.075\). a. What is the expected change in reaction time for a \(1^{\circ} \mathrm{F}\) increase in temperature? For a \(10^{\circ} \mathrm{F}\) increase in temperature? b. What is the expected reaction time when temperature is \(200^{\circ} \mathrm{F}\) ? When temperature is \(250^{\circ} \mathrm{F}\) ? c. Suppose five observations are made independently on reaction time, each one for a temperature of \(250^{\circ} \mathrm{F}\). What is the probability that all five times are between \(2.4\) and \(2.6 \mathrm{hr}\) ? d. What is the probability that two independently observed reaction times for temperatures \(1^{\circ}\) apart are such that the time at the higher temperature exceeds the time at the lower temperature?

No-fines concrete, made from a uniformly graded coarse aggregate and a cement- water paste, is beneficial in areas prone to excessive rainfall because of its excellent drainage properties. The article "Pavement Thickness Design for No- Fines Concrete Parking Lots," J. of Trans. Engr., 1995: 476-484) employed a least squares analysis in studying how \(y=\) porosity (\%) is related to \(x=\) unit weight (pcf) in concrete specimens. Consider the following representative data: $$ \begin{array}{l|rrrrrrrr} x & 99.0 & 101.1 & 102.7 & 103.0 & 105.4 & 107.0 & 108.7 & 110.8 \\ \hline y & 28.8 & 27.9 & 27.0 & 25.2 & 22.8 & 21.5 & 20.9 & 19.6 \\ x & 112.1 & 112.4 & 113.6 & 113.8 & 115.1 & 115.4 & 120.0 \\ \hline y & 17.1 & 18.9 & 16.0 & 16.7 & 13.0 & 13.6 & 10.8 \\ \text { Relevant } & \text { summary } & \text { quantities } & \text { are } & \Sigma x_{i}=1640.1, \\ \Sigma y_{i}=299.8, \quad \Sigma x_{i}^{2}=179,849.73, & \Sigma x_{i} y_{i}=32,308.59, \\ \Sigma y_{i}^{2}=6430.06 . \end{array} $$ a. Obtain the equation of the estimated regression line. Then create a scatterplot of the data and graph the estimated line. Does it appear that the model relationship will explain a great deal of the observed variation in \(y\) ? b. Interpret the slope of the least squares line. c. What happens if the estimated line is used to predict porosity when unit weight is 135 ? Why is this not a good idea? d. Calculate the residuals corresponding to the first two observations. e. Calculate and interpret a point estimate of \(\sigma\). f. What proportion of observed variation in porosity can be attributed to the approximate linear relationship between unit weight and porosity?

The article "Quantitative Estimation of Clay Mineralogy in Fine-Grained Soils" (J. of Geotechnical and Geoenvironmental Engr., 2011: 997-1008) reported on various chemical properties of natural and artificial soils. Here are observations on \(x=\) cation exchange capacity (CEC, in meq/100 g) and \(y=\) specific surface area (SSA, in \(\mathrm{m}^{2} / \mathrm{g}\) ) of 20 natural soils. $$ \begin{array}{c|cccccccccc} x & 66 & 121 & 134 & 101 & 77 & 89 & 63 & 57 & 117 & 118 \\ \hline y & 175 & 324 & 460 & 288 & 205 & 210 & 295 & 161 & 314 & 265 \\ x & 76 & 125 & 75 & 71 & 133 & 104 & 76 & 96 & 58 & 109 \\ \hline y & 236 & 355 & 240 & 133 & 431 & 306 & 132 & 269 & 158 & 303 \end{array} $$ Minitab gave the following output in response to a request for \(r\) : Normal probability plots of \(x\) and \(y\) are quite straight. a. Carry out a test of hypotheses to see if there is a positive linear association in the population from which the sample data was selected. b. With \(n=20\), how small would the value of \(r\) have to be in order for the null hypothesis in the test of (a) to not be rejected at significance level .01? c. Calculate a confidence interval for \(\rho\) using a \(95 \%\) confidence level.

Torsion during hip external rotation and extension may explain why acetabular labral tears occur in professional athletes. The article "Hip Rotational Velocities During the Full Golf Swing" (J. of Sports Science and Med., 2009: 296-299) reported on an investigation in which lead hip internal peak rotational velocity \((x)\) and trailing hip peak external rotational velocity \((y)\) were determined for a sample of 15 golfers. Data provided by the article's authors was used to calculate the following summary quantities: $$ \begin{aligned} \sum\left(x_{i}-\bar{x}\right)^{2}=& 64,732.83, \sum\left(y_{i}-\bar{y}\right)^{2}=130,566.96, \\ & \sum\left(x_{i}-\bar{x}\right)\left(y_{i}-\bar{y}\right)=44,185.87 \end{aligned} $$ Separate normal probability plots showed very substantial linear patterns. a. Calculate a point estimate for the population correlation coefficient. b. Carry out a test at significance level .01 to decide whether there is a linear relationship between the two velocities in the sampled population. c. Would the conclusion of (b) have changed if you had tested appropriate hypotheses to decide whether there is a positive linear association in the population? What if a significance level of .05 rather than \(.01\) had been used?
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

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.