/*! 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 37 Magnetic resonance imaging (MRI)... [FREE SOLUTION] | 91影视

91影视

Magnetic resonance imaging (MRI) is well established as a tool for measuring blood velocities and volume flows. The article "Correlation Analysis of Stenotic Aortic Valve Flow Patterns Using Phase Contrast MRI,"鈥 referenced in Exercise 1.67, proposed using this methodology for determination of valve area in patients with aortic stenosis. The accompanying data on peak velocity \((\mathrm{m} / \mathrm{s})\) from scans of 23 patients in two different planes was read from a graph in the cited paper. $$ \begin{aligned} &\begin{array}{l|rrrrrrrr} \text { Level- } & .60 & .82 & .85 & .89 & .95 & 1.01 & 1.01 & 1.05 \\ \hline \text { Level- } & .50 & .68 & .76 & .64 & .68 & .86 & .79 & 1.03 \end{array}\\\ &\begin{array}{c|rrrrrrrr} \text { Level- } & 1.08 & 1.11 & 1.18 & 1.17 & 1.22 & 1.29 & 1.28 & 1.32 \\ \hline \text { Level-- } & .75 & .90 & .79 & .86 & .99 & .80 & 1.10 & 1.15 \end{array}\\\ &\begin{array}{l|lllllll} \text { Level- } & 1.37 & 1.53 & 1.55 & 1.85 & 1.93 & 1.93 & 2.14 \\ \hline \text { Level-- } & 1.04 & 1.16 & 1.28 & 1.39 & 1.57 & 1.39 & 1.32 \end{array} \end{aligned} $$ a. Does there appear to be a difference between true average velocity in the two different planes? Carry out an appropriate test of hypotheses (as did the authors of the article). b. The authors of the article also regressed level-velocity against level- velocity. The resulting estimated intercept and slope are .14701 and .65393, with corresponding estimated standard errors \(.07877\) and \(.05947\), coefficient of determination .852, and \(s=.110673\). The article included a comment that this regression showed evidence of a strong linear relationship but a regression slope well below 1. Do you agree?

Short Answer

Expert verified
A significant difference in average velocities likely exists. The regression indicates a strong linear relationship, but with a slope below 1.

Step by step solution

01

Organize the Data

First, label the data from the two different planes as Plane 1 and Plane 2 for clarity. Plane 1 is labeled 'Level- ' and Plane 2 labeled 'Level-- '. The data points are given in the exercise.
02

State the Hypotheses

We need to test if there is a difference in true average velocities between these two planes. The null hypothesis \(H_0\) states that the true average velocities for the different planes are equal, i.e., \( \mu_1 = \mu_2 \). The alternative hypothesis \(H_a\) states that the true averages are not equal, i.e., \( \mu_1 eq \mu_2 \).
03

Perform the t-test for Two Independent Samples

Calculate the sample means and standard deviations for both Plane 1 and Plane 2. Then, use these to compute the test statistic for a two-sample t-test. Use the formula:\[t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\]where \( \bar{x}_1\) and \( \bar{x}_2\) are the sample means, \( s_1\) and \( s_2\) are the sample standard deviations, and \( n_1\) and \( n_2\) are the number of observations per sample.
04

Evaluate and Conclude the Test

Determine the critical value from the t-distribution table at a chosen significance level (e.g., \( \alpha = 0.05 \)). Compare the calculated t-value to this critical value to decide whether to reject or not reject the null hypothesis. If \( |t| \) is greater than the critical value, reject \(H_0\), indicating a significant difference in velocities.
05

Interpret Regression Analysis

Analyze the given regression results: intercept = 0.14701, slope = 0.65393, standard errors, and coefficient of determination = 0.852. A coefficient of determination of 0.852 indicates a strong linear relationship, which supports the comment in the study. However, since the slope (0.65393) is significantly less than 1, it indicates that the increase in one level is not fully captured by a similar increase in the other, a point which aligns with the observation in the study.

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.

Hypothesis Testing
Hypothesis testing is a statistical method used to make inferences about populations based on sample data.
It begins with establishing a null hypothesis (H_0), which is a statement of no effect or no difference. In this exercise, the null hypothesis posits that the true average velocities in two different MRI planes are equal, or \( \mu_1 = \mu_2 \). The alternative hypothesis (H_a) suggests there is a difference, meaning \( \mu_1 eq \mu_2 \).
To determine if we can reject the null hypothesis, a t-test is performed for two independent samples.
  • Step 1: Calculate the sample means and standard deviations for each plane.
  • Step 2: Compute the test statistic using: \[t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\] Here, \( \bar{x}_1 \) and \( \bar{x}_2 \) represent the sample means, and \( s_1 \) and \( s_2 \) are the standard deviations. \( n_1 \) and \( n_2 \) are the sample sizes.
  • Step 3: Compare the calculated t-value with the critical value from the t-distribution table. A significant result (where the absolute t-value exceeds the critical value) leads to the rejection of H_0.
This t-test is pivotal in hypothesis testing as it determines whether observed data can refute the null hypothesis and supports conclusions about the population behavior.
Linear Regression
Linear regression is a statistical method used for modeling the relationship between two variables by fitting a linear equation to observed data.
Despite its simplicity, linear regression is a powerful tool to predict one variable based on the knowledge of another.
The key components of a linear regression model include:
  • Independent variable: The predictor, often denoted as x .
  • Dependent variable: The outcome, typically represented as y .
  • Intercept: The value of y when x is zero.
  • Slope: The change in the dependent variable for a one-unit change in the independent variable.
In this exercise, the regression was performed between velocities across two MRI planes. The intercept (0.14701) and slope (0.65393) were estimated using sample data, showing how changes in one plane predict changes in another.
The slope less than 1 reflects that the increase in level velocity in one plane does not fully translate to the same increase in another plane. The statistical significance of this relationship is evident from the results, which aligns with the article's conclusion.
Coefficient of Determination
The coefficient of determination, denoted as R^2 , is a key metric in assessing the quality of a linear regression model.
It describes the proportion of variance in the dependent variable that is predictable from the independent variable.
In this exercise, an R^2 value of 0.852 was obtained, signifying a strong linear relationship between the velocity measurements in the two MRI planes.
This means that 85.2% of the variance in one plane鈥檚 velocity can be explained by changes in the other plane's velocities.
An R^2 value closer to 1 indicates a strong predictive relationship, whereas a value closer to 0 suggests a weak one.
Understanding this metric is crucial because it reveals how well the regression model captures the actual dynamics in the observed data.
While a strong R^2 is promising, it鈥檚 essential to also consider the slope and other model diagnostics to get a comprehensive view of the regression efficacy.
This coefficient acts as a validation tool for linear models, confirming if the patterns observed can reliably predict future outcomes. So, knowing about and interpreting R^2 helps to assess the potential and limitations of regression insights into real-world phenomena.

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

In biofiltration of wastewater, air discharged from a treatment facility is passed through a damp porous membrane that causes contaminants to dissolve in water and be transformed into harmless products. The accompanying data on \(x=\) inlet temperature \(\left({ }^{\circ} \mathrm{C}\right)\) and \(y=\) removal efficiency (\%) was the basis for a scatterplot that appeared in the article "Treatment of Mixed Hydrogen Sulfide and Organic Vapors in a Rock Medium Biofilter" (Water Environment Research, 2001: 426-435). Calculated summary quantities are \(\Sigma x_{i}=384.26, \Sigma y_{i}=\) \(3149.04, \Sigma x_{i}^{2}=5099.2412, \Sigma x_{i} y_{i}=37,850.7762\), and \(\Sigma y_{i}^{2}=309,892.6548\). a. Does a scatterplot of the data suggest appropriateness of the simple linear regression model? b. Fit the simple linear regression model, obtain a point prediction of removal efficiency when temperature \(=10.50\), and calculate the value of the corresponding residual. c. Roughly what is the size of a typical deviation of points in the scatterplot from the least squares line? d. What proportion of observed variation in removal efficiency can be attributed to the model relationship? e. Estimate the slope coefficient in a way that conveys information about reliability and precision, and interpret your estimate. f. Personal communication with the authors of the article revealed that there was one additional observation that was not included in their scatterplot: (6.53, 96.55). What impact does this additional observation have on the equation of the least squares line and the values of \(s\) and \(r^{2} ?\)

The article "Characterization of Highway Runoff in Austin, Texas, Area" (J. of Envir. Engr., 1998: 131137) gave a scatterplot, along with the least squares line, of \(x=\) rainfull volume \(\left(\mathrm{m}^{3}\right)\) and \(y=\) runoff volume \(\left(\mathrm{m}^{3}\right)\) for a particular location. The accompanying values were read from the plot. $$ \begin{array}{l|rrrrrrrr} x & 5 & 12 & 14 & 17 & 23 & 30 & 40 & 47 \\ \hline y & 4 & 10 & 13 & 15 & 15 & 25 & 27 & 46 \\ x & 55 & 67 & 72 & 81 & 96 & 112 & 127 & \\ \hline y & 38 & 46 & 53 & 70 & 82 & 99 & 100 & \end{array} $$ a. Does a scatterplot of the data support the use of the simple linear regression model? b. Calculate point estimates of the slope and intercept of the population regression line. c. Calculate a point estimate of the true average runoff volume when rainfall volume is 50 . d. Calculate a point estimate of the standard deviation \(\sigma\). e. What proportion of the observed variation in runoff volume can be attributed to the simple linear regression relationship between runoff and rainfall?

The accompanying dataon \(x=\) current density \(\left(\mathrm{mA} / \mathrm{cm}^{2}\right)\) and \(y=\) rate of deposition \((\mu \mathrm{m} / \mathrm{min})\) appeared in the article "Plating of 60/40 Tin/Lead Solder for Head Termination Metallurgy" (Plating and Surface Finishing, Jan. 1997: 38-40). Do you agree with the claim by the article's author that "a linear relationship was obtained from the tin-lead rate of deposition as a function of current density"? Explain your reasoning. $$ \begin{array}{l|rrrr} x & 20 & 40 & 60 & 80 \\ \hline y & .24 & 1.20 & 1.71 & 2.22 \end{array} $$

Suppose that \(x\) and \(y\) are positive variables and that a sample of \(n\) pairs results in \(r \approx 1\). If the sample correlation coefficient is computed for the \(\left(x, y^{2}\right)\) pairs, will the resulting value also be approximately 1 ? Explain.

The accompanying data on \(x=\mathrm{UV}\) transparency index and \(y=\) maximum prevalence of infection was read from a graph in the article "Solar Radiation Decreases Parasitism in Daphnia" (Ecology Letters, 2012: 47-54): $$ \begin{array}{l|ccccccccc} x & 1.3 & 1.4 & 1.5 & 2.0 & 2.2 & 2.7 & 2.7 & 2.7 & 2.8 \\ \hline y & 16 & 3 & 32 & 1 & 13 & 0 & 8 & 16 & 2 \\ x & 2.9 & 3.0 & 3.6 & 3.8 & 3.8 & 4.6 & 5.1 & 5.7 \\ \hline y & 1 & 7 & 36 & 25 & 10 & 35 & 58 & 56 \end{array} $$ Summary quantities include \(S_{x x}=25.5224, S_{y y}=\) \(5593.0588\), and \(S_{x y}=264.4882 .\) a. Calculate and interpret the value of the sample correlation coefficient. b. If you decided to fit the simple linear regression model to this data, what proportion of observed variation in maximum prevalence could be explained by the model relationship? c. If you decided to regress UV transparency index on maximum prevalence (i.e., interchange the roles of \(x\) and \(y\) ), what proportion of observed variation could be attributed to the model relationship? d. Carry out a test of \(H_{0}: \rho=.5\) versus \(H_{\mathrm{a}}: \rho>.5\) using a significance level of .05.

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.