/*! 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 76 Consider the following \((x, y)\... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider the following \((x, y)\) data. Calculate the correlation coefficient. Graph the data and comment on the relationship between \(x\) and \(y .\) Explain why the correlation coefficient does not detect the relationship between \(x\) and \(y\). $$\begin{array}{rrrr}\hline x & y & x & y \\\\\hline-4 & 0 & 0 & -4 \\\\-3 & -2.65 & 1 & 3.87 \\\\-3 & 2.65 & 1 & -3.87 \\\\-2 & -3.46 & 2 & 3.46 \\ -2 & 3.46 & 2 & -3.46 \\\\-1 & -3.87 & 3 & 2.65 \\\\-1 & 3.87 & 3 & -2.65 \\\0 & 4 & 4 & 0 \\\\\hline\end{array}$$

Short Answer

Expert verified
The correlation coefficient is 0. The data forms a circular pattern, which has no linear relationship, hence the coefficient doesn't detect it.

Step by step solution

01

Organize Data into Pairs

First, we will list the pairs of \((x, y)\) coordinates from the given data table:- \((-4, 0)\)- \((-3, -2.65)\)- \((-3, 2.65)\)- \((-2, -3.46)\)- \((-2, 3.46)\)- \((-1, -3.87)\)- \((-1, 3.87)\)- \((0, 4)\)- \((0, -4)\)- \((1, 3.87)\)- \((1, -3.87)\)- \((2, 3.46)\)- \((2, -3.46)\)- \((3, 2.65)\)- \((3, -2.65)\)- \((4, 0)\)
02

Calculate Mean of x and y

Calculate the mean of the \(x\) and \(y\) values separately:For \(x\), add the values and divide by 16: \[ \bar{x} = \frac{-4 - 3 - 3 - 2 - 2 - 1 - 1 + 0 + 0 + 1 + 1 + 2 + 2 + 3 + 3 + 4}{16} = 0 \]For \(y\), add the values and divide by 16: \[ \bar{y} = \frac{0 - 2.65 + 2.65 - 3.46 + 3.46 - 3.87 + 3.87 + 4 - 4 + 3.87 - 3.87 + 3.46 - 3.46 + 2.65 - 2.65 + 0}{16} = 0 \]
03

Compute Pearson Correlation Coefficient

The formula for the correlation coefficient \(r\) is: \[ r = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\sum (x_i - \bar{x})^2 \sum (y_i - \bar{y})^2}} \]Given that \(\bar{x} = 0\) and \(\bar{y} = 0\), this simplifies to:The numerator:\[ \sum x_i y_i = (-4)(0) + (-3)(-2.65) + (-3)(2.65) + (-2)(-3.46) + (-2)(3.46) + \cdots + (4)(0) = 0 \]The denominator:\[ \sqrt{\sum x_i^2 \sum y_i^2} = \sqrt{80 \cdot 85.0894} = \sqrt{0} \]Thus, since the numerator is 0 and both sums are non-zero, \[ r = 0 \].
04

Graph the Data

Plot the given \((x, y)\) pairs on a graph.While plotting, you might notice that the data takes a circular form, rather than a linear one. This setup shows that the pairs create a pattern that forms a circle, suggesting a non-linear relationship between \(x\) and \(y\).
05

Comment on the Correlation Coefficient

The Pearson correlation coefficient, \(r\), measures only the strength and direction of a linear relationship between two variables. Because the data is not linearly related (it forms a circular pattern), the correlation coefficient \(r = 0\) fails to detect the relationship between \(x\) and \(y\). The relationship is perfectly circular, implying zero correlation in a linear sense, though there is indeed a significant non-linear relationship.

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Key Concepts

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

Non-linear relationship
We often encounter various types of relationships between two variables, but one crucial distinction is whether the relationship is linear or non-linear. In a **non-linear relationship**, changes in one variable do not result in proportional changes in another. Unlike linear relationships that can be visualized as straight lines on a graph, non-linear relationships can take various shapes, such as curves, circles, or waves.

Famously, a non-linear relationship is often more complex and cannot be easily summarized by a simple equation like a linear one. Here are some key characteristics of non-linear relationships:
  • They can involve variables that change at rates that are not constant.
  • A graph representing them will not be a straight line.
  • They can demonstrate varied patterns, such as exponential growth or decay, and even cyclic patterns.
Understanding non-linear relationships is vital in fields such as physics, biology, and economics, where systems often interact in complex, non-linear ways, leading to diverse and interesting behavior between variables.
Pearson correlation
The **Pearson correlation coefficient** is a statistical measure that describes the linear relationship between two variables. It quantifies how well two variables are related through a straight line. Ranging from -1 to 1, a Pearson correlation of 1 or -1 represents perfect linear relationships where:
  • **1** indicates a perfect positive linear relationship.
  • **-1** indicates a perfect negative linear relationship.
  • **0** indicates no linear relationship.
Mathematically, it is calculated using the formula: \[ r = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\sum (x_i - \bar{x})^2 \sum (y_i - \bar{y})^2}} \] This formula helps in standardizing data, allowing for results that can be easily interpreted. However, the limitation is that it only measures linear dependencies. In situations where the relationship is non-linear, like a circular pattern, Pearson might misleadingly suggest that there's no relationship (as seen with a result of 0), when in fact, a non-linear dependency is present.
Graphing data
**Graphing data** is a powerful tool for visualizing relationships and patterns that might not be immediately obvious from raw numbers. Various types of graphs help depict different kinds of data relationships, including linear, non-linear, or even more complex ones. For our example:
  • Plot the data points on a two-dimensional plane, with one axis representing each variable.
  • By joining these points or observing their spread, we can visually discern the type of relationship.
When we plot our example data set, we notice that the pattern forms a shape that looks circular, not aligned or following a straight path. This deviation from a line suggests a non-linear relationship. Graphing helps in quickly identifying such patterns by drawing our attention to clusters, trends, or shapes among data points that tabular data cannot convey instantly. This visualization gives context that numerical calculations like the Pearson correlation might overlook.
Circular data pattern
A **circular data pattern** is a fascinating visual format where plotted data points form a circle. In our exercise data, this kind of pattern hints at a unique, non-linear relationship between two variables. Unlike straight or curved lines resulting from linear or simple quadratic data, circular patterns break away from these norms, indicating that one variable's effect on another completes a loop or cycle.

Such patterns may emerge in phenomena like orbit paths in physics or periodic trends in time series data. Due to their non-linear nature, these circular patterns cannot be adequately described using linear regression or correlation tools like Pearson.
  • They are better analyzed with qualitative visual methods or specialized non-linear correlation metrics.
  • Understanding and identifying a circular pattern requires us to look beyond standard metrics, focusing instead on visual diagnostic tools or other sophisticated models designed for complex relationships.
By recognizing the nature of such circular arrangements, we gain insight into potential cyclic or periodic behaviors in our data.

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Most popular questions from this chapter

Show that, for the simple linear regression model, the following statements are true: (a) \(\sum_{i=1}^{n}\left(y_{i}-\hat{y}_{i}\right)=0\) (b) \(\sum_{i=1}^{n}\left(y_{i}-\hat{y}_{i}\right) x_{i}=0\) (c) \(\frac{1}{n} \sum_{i=1}^{n} \hat{y}_{i}=\bar{y}\)

An article in the Journal of the Environmental Engineering Division ["Least Squares Estimates of BOD Parameters" (1980, Vol. 106, pp. \(1197-1202\) ) ] took a sample from the Holston River below Kingport, Tennessee, during August 1977 . The biochemical oxygen demand (BOD) test is conducted over a period of time in days. The resulting data are shown below: Time (days): \(\begin{array}{lllllllll}1 & 2 & 4 & 6 & 8 & 10 & 12 & 14 & 16\end{array}\) \(18 \quad 20\) BOD (mg/liter): \(\begin{array}{llll}0.6 & 0.7 & 1.5 & 1.9\end{array}\) \(\begin{array}{ll}2.1 & 2.6\end{array}\) \(\begin{array}{lll}2.9 & 3.7 & 3.5\end{array}\) \(\begin{array}{ll}3.7 & 3.8\end{array}\) (a) Assuming that a simple linear regression model is appropriate, fit the regression model relating \(\mathrm{BOD}(y)\) to the time \((x) .\) What is the estimate of \(\sigma^{2} ?\) (b) What is the estimate of expected BOD level when the time is 15 days? (c) What change in mean \(\mathrm{BOD}\) is expected when the time changes by three days? (d) Suppose the time used is six days. Calculate the fitted value of \(y\) and the corresponding residual. (e) Calculate the fitted \(\hat{y}_{i}\) for each value of \(x_{i}\) used to fit the model. Then construct a graph of \(\hat{y}_{i}\) versus the corresponding observed values \(y_{i}\) and comment on what this plot would look like if the relationship between \(y\) and \(x\) was a deterministic (no random error) straight line. Does the plot actually obtained indicate that time is an effective regressor variable in predicting BOD?

An article in the Journal of Sound and Vibration (Vol. \(151,1991,\) pp. \(383-394\) ) described a study investigating the relationship between noise exposure and hypertension. The following data are representative of those reported in the article. $$\begin{aligned}&\begin{array}{c|c|c|c|c|c|c|c|c|c|c}y & 1 & 0 & 1 & 2 & 5 & 1 & 4 & 6 & 2 & 3 \\\\\hline x & 60 & 63 & 65 & 70 & 70 & 70 & 80 & 90 & 80 & 80\end{array}\\\&\begin{array}{c|c|c|c|c|c|c|c|c|c|c}y & 5 & 4 & 6 & 8 & 4 & 5 & 7 & 9 & 7 & 6 \\\\\hline x & 85 & 89 & 90 & 90 & 90 & 90 & 94 & 100 & 100 & 100\end{array}\end{aligned}$$ (a) Draw a scatter diagram of \(y\) (blood pressure rise in millimeters of mercury) versus \(x\) (sound pressure level in decibels). Does a simple linear regression model seem reasonable in this situation? (b) Fit the simple linear regression model using least squares. Find an estimate of \(\sigma^{2}\) (c) Find the predicted mean rise in blood pressure level associated with a sound pressure level of 85 decibels.

A study was performed to investigate new automobile purchases. A sample of 20 families was selected. Each family was surveyed to determine the age of their oldest vehicle and their total family income. A follow-up survey was conducted six months later to determine if they had actually purchased a new vehicle during that time period \((y=1\) indicates yes and \(y=0\) indicates no). The data from this study are shown in the following table. $$\begin{array}{ccc|ccc}\text { Income, } x_{1} & \text { Age, } x_{2} & y & \text { Income, } x_{1} & \text { Age, } x_{2} & y \\\\\hline 45,000 & 2 & 0 & 37,000 & 5 & 1 \\\40,000 & 4 & 0 & 31,000 & 7 & 1 \\\60,000 & 3 & 1 & 40,000 & 4 & 1 \\\50,000 & 2 & 1 & 75,000 & 2 & 0 \\\55,000 & 2 & 0 & 43,000 & 9 & 1 \\\50,000 & 5 & 1 & 49,000 & 2 & 0 \\\35,000 & 7 & 1 & 37,500 & 4 & 1 \\\65,000 & 2 & 1 & 71,000 & 1 & 0 \\ 53,000 & 2 & 0 & 34,000 & 5 & 0 \\\48,000 & 1 & 0 & 27,000 & 6 & 0 \\\\\hline\end{array}$$ (a) Fit a logistic regression model to the data. (b) Is the logistic regression model in part (a) adequate? (c) Interpret the model coefficients \(\beta_{1}\) and \(\beta_{2}\). (d) What is the estimated probability that a family with an income of \(\$ 45,000\) and a car that is five years old will purchase a new vehicle in the next six months? (e) Expand the linear predictor to include an interaction term. Is there any evidence that this term is required in the model?

Suppose that we are fitting a line and we wish to make the variance of the regression coefficient \(\hat{\beta}_{1}\) as small as possible. Where should the observations \(x_{i}, i=1,2, \ldots, n,\) be taken so as to minimize \(V\left(\hat{\beta}_{1}\right) ?\) Discuss the practical implications of this allocation of the \(x_{i}\).

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