/*! 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 13 The accompanying dataon \(x=\) c... [FREE SOLUTION] | 91Ó°ÊÓ

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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. $$ \begin{array}{l|rrrr} x & 20 & 40 & 60 & 80 \\ \hline y & .24 & 1.20 & 1.71 & 2.22 \end{array} $$

Short Answer

Expert verified
Yes, the strong correlation coefficient of approximately 0.994 supports the claim of a linear relationship.

Step by step solution

01

Plot the Data Points

First, let's plot the given data points on a graph to visually inspect the relationship. The data points are (20, 0.24), (40, 1.20), (60, 1.71), and (80, 2.22). Plot these points on a Cartesian coordinate system with \(x\) on the horizontal axis and \(y\) on the vertical axis.
02

Determine the Line of Best Fit

To investigate if a linear relationship exists, we'll need to see if we can fit a straight line through the data points. Calculate the slope (m) and the y-intercept (b) of the line using the least squares method. The formula for the slope is \(m = \frac{n(\sum{xy}) - (\sum{x})(\sum{y})}{n(\sum{x^2}) - (\sum{x})^2}\) and the y-intercept is \(b = \frac{(\sum{y}) - m(\sum{x})}{n}\), where \(n\) is the number of points.
03

Compute Slope and Intercept

Calculate the sums: \(\sum{x} = 200\), \(\sum{y} = 5.37\), \(\sum{xy} = 356.8\), and \(\sum{x^2} = 3600\). With these, compute the slope: \(m = \frac{4(356.8) - (200)(5.37)}{4(3600) - 200^2} = 0.0253\). Then compute the intercept: \(b = \frac{(5.37) - 0.0253(200)}{4} = -0.0095\). Thus, the line of best fit is \(y = 0.0253x - 0.0095\).
04

Evaluate Linear Fit

Assess how close the data points are to the calculated line of best fit. Calculate the correlation coefficient (r) to evaluate the strength of the linear relationship. This is given by \(r = \frac{n(\sum{xy}) - (\sum{x})(\sum{y})}{\sqrt{[n\sum{x^2} - (\sum{x})^2][n\sum{y^2} - (\sum{y})^2]}}\). A value of \(r\) close to 1 or -1 indicates a strong linear relationship.
05

Conclude Relationship

Calculate \(\sum{y^2} = 9.5067\). Using these, the correlation coefficient \(r = \frac{4(356.8) - (200)(5.37)}{\sqrt{[4(3600) - 200^2][4(9.5067) - 5.37^2]}} \approx 0.994\), showing a strong positive linear correlation. Hence, it is reasonable to agree with the author's claim that there is a linear relationship between current density and deposition rate.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Data Visualization
Data visualization is a powerful tool that helps us quickly understand and interpret data, especially when we're exploring the relationship between two variables. By plotting data points on a graph, we can easily see patterns and trends that might not be evident from a table of numbers alone.
In this exercise, we are examining data for current density and deposition rate. Each pair of data points (e.g., (20, 0.24)) is plotted on a Cartesian coordinate system, where current density is on the x-axis and deposition rate is on the y-axis. This visual representation allows us to observe the shape of the data distribution and get an initial impression of the possible relationship between the variables.
  • A straight line pattern suggests a potential linear relationship.
  • If points show a random scatter, the relationship might be non-linear or non-existent.
Visual inspection gives us a clue but isn't conclusive, so we proceed to statistical methods to confirm our observations.
Least Squares Method
The least squares method is a standard approach to find the line of best fit for a set of data points. It minimizes the sum of the squares of the differences between observed and predicted values, ensuring the most accurate linear model possible.
In finding the line of best fit, we calculate both the slope and y-intercept of the line. Here, we use the formulas: \[ m = \frac{n(\sum{xy}) - (\sum{x})(\sum{y})}{n(\sum{x^2}) - (\sum{x})^2} \] and \[ b = \frac{(\sum{y}) - m(\sum{x})}{n} \]These formulas help us derive the equation of the line that best represents the relationship between our variables. The data in this exercise was used to compute a slope \(m = 0.0253\) and a y-intercept \(b = -0.0095\), resulting in the line \(y = 0.0253x - 0.0095\).
  • The slope \(m\) indicates the rate of change in \(y\) for each unit of \(x\).
  • The y-intercept \(b\) is the value of \(y\) when \(x\) is zero.
The accuracy of this line determines how well we understand the relationship between current density and deposition rate.
Correlation Coefficient
The correlation coefficient, denoted as \(r\), measures the strength and direction of a linear relationship between two variables on a scale from -1 to 1. An \(r\) value close to 1 indicates a strong positive linear correlation, whereas an \(r\) value close to -1 signifies a strong negative correlation.
In our exercise, to compute \(r\), we use the formula: \[ r = \frac{n(\sum{xy}) - (\sum{x})(\sum{y})}{\sqrt{[n\sum{x^2} - (\sum{x})^2][n\sum{y^2} - (\sum{y})^2]}} \]Upon evaluating this for the given data, we found \(r \approx 0.994\). This value is very close to 1, which demonstrates a strong positive linear correlation between current density and deposition rate.
  • An \(r\) value around 0 suggests no linear relationship.
  • The larger the absolute value of \(r\), the stronger the linear relationship.
Calculating the correlation coefficient helps us to verify the findings from our visual and mathematical analyses.
Linear Relationship Analysis
Linear relationship analysis allows us to understand the nature of the relationship between two continuous variables. It involves analyzing data both visually and statistically to determine if the relationship between variables can be described as linear.
In this exercise, we analyzed the rate of deposition as a function of current density. We started by visually assessing the data points for a linear trend. Then, through the least squares method, we quantified this relationship with a line of best fit. Finally, by calculating the correlation coefficient, we confirmed the strength and direction of this relationship.
  • This analysis is confirmed by the line equation \(y = 0.0253x - 0.0095\) and a correlation coefficient \(r \approx 0.994\).
  • The close fit and high \(r\) value support the author's claim of a linear relationship.
Understanding how variables are related is crucial in predicting and making informed decisions using the data. This linear relationship analysis provides a comprehensive view and validation of the claim about the linear interaction between current density and deposition rate.

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Most popular questions from this chapter

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.

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An investigation was carried out to study the relationship between speed (ft/sec) and stride rate (number of steps taken/sec) among female marathon runners. Resulting summary quantities included \(n=11, \Sigma\) (speed) \(=205.4\), \(\Sigma(\text { speed })^{2}=3880.08, \Sigma\) (rate \()=35.16, \Sigma(\text { rate })^{2}=112.681\), and \(\Sigma(\) speed \()(\) rate \()=660.130 .\) a. Calculate the equation of the least squares line that you would use to predict stride rate from speed. b. Calculate the equation of the least squares line that you would use to predict speed from stride rate. c. Calculate the coefficient of determination for the regression of stride rate on speed of part (a) and for the regression of speed on stride rate of part (b). How are these related?

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