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Chapter 17: Problem 668

Within certain limits, slime mold is hypothesized to grow at a rate proportional to its size. That is, if \(\mathrm{y}(\mathrm{t})=\) size of slime mold at time t. \([(\mathrm{dy}) /(\mathrm{dt})]=\mathrm{k} \mathrm{y}(\mathrm{t})\), where \([(\mathrm{dy}) /(\mathrm{dt})]\) is the rate of growth. If the size of the slime mold is measured at time intervals of 1 day for 300 days, we have 300 pairs of data. How could linear regression be used to check the significance of this hypothesized growth model?

Short Answer

Expert verified
In order to use linear regression to check the significance of the hypothesized slime mold growth model, first integrate the given differential equation and obtain the expression \(y(t) = Ae^{kt}\). Apply the natural logarithm to obtain a linear expression: \(v(t) = \ln(y(t)) = B + kt\). Plot \(v(t)\) against time t using the 300 pairs of data and apply linear regression to find the best-fitting line. Assess the validity of the growth model by examining the correlation coefficient (R-squared), slope (k), intercept (B), and their statistical significance. A strong positive correlation (close to 1) would support the hypothesis that the slime mold's growth rate is proportional to its size.

Step by step solution

01

Integrate the given differential equation

First, we have to integrate the differential equation to obtain an expression for the size of the slime mold. We can rewrite the equation as \[\frac{dy}{y} = k \, dt.\] Integrating both sides with respect to their respective variables gives us: \[\ln(y) = kt + C,\] where C is the constant of integration, representing the initial size of the slime mold.
02

Invert the natural logarithm function

To solve for y(t), we can raise both sides of the equation to the power of \(e\): \[y(t) = e^{kt + C}.\] To simplify further, let \(A = e^C\). Then our expression becomes: \[y(t) = Ae^{kt}.\]
03

Apply the natural logarithm to obtain a linear expression

Now, let's apply the natural logarithm to both sides of the equation so we can obtain a linear expression that can be used in a linear regression setting: \[\ln(y(t)) = \ln(Ae^{kt}).\] By applying the logarithmic rules, we get: \[\ln(y(t)) = \ln(A) + kt.\] Now, let \(v(t) = \ln(y(t))\) and \(B = \ln(A)\). Our linear expression becomes: \[v(t) = B + kt.\]
04

Apply linear regression to check the significance

Using the 300 pairs of data, plot \(v(t) = \ln(y(t))\) against time t. Then, apply linear regression to find the best-fitting line through these data points. Linear regression will yield an estimated slope (k) and intercept (B), along with their respective standard errors and correlation coefficient (R-squared). The validity of the hypothesized growth model can be assessed by looking at the correlation coefficient (R-squared), slope (k), and intercept (B) obtained from the linear regression. A strong positive correlation (close to 1) would mean that the slime mold's growth rate is indeed proportional to its size, as initially assumed. The slope (k), intercept (B), and standard errors obtained from regression can also be used to perform further statistical tests, such as t-tests or confidence intervals, to check the significance of the model. In summary, linear regression can be applied to check the significance of the hypothesized growth model by analyzing the correlation between the natural logarithm of the size of the slime mold and the time in days, and then examining the resulting regression coefficients and their significance.

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Key Concepts

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

Differential Equations
A differential equation is an equation that relates a function with its derivatives. It's like a recipe that describes how a quantity changes over time. Consider it as a way to model the behavior of dynamic systems. In this exercise, the differential equation is given by:
a \( \frac{dy}{dt} = ky(t) \).
This indicates that the rate of change of the slime mold's size (\(\frac{dy}{dt}\)) is directly proportional to its current size (\(y(t)\)).

Solutions to Differential Equations

To solve such equations, integration is applied to find the function describing how the system evolves over time.
For our slime mold, integrating the equation results in:
\[ \ln(y) = kt + C \]where \(C\) is a constant representing initial conditions.
This solution tells us how the slime mold grows, following an exponential pattern until it reaches a limiting factor, as modeled in the next concept.
Exponential Growth
Exponential growth occurs when the growth rate of a system is proportional to its current size. In mathematical terms, it forms an exponential function.

For the slime mold, starting from the differential equation, when you solve it, you get:\[ y(t) = Ae^{kt} \]This equation means that the slime mold's size at time \(t\), \(y(t)\), grows exponentially, with \(A\) and \(k\) being constants.

Characteristics of Exponential Growth

  • **Rapid Growth:** Initially slow but accelerates quickly.
  • **Doubling Time:** The time it takes for a quantity to double in size is constant.
  • **Real-Life Examples:** Population growth, compound interest.
Understanding exponential growth is crucial in predicting how quickly and to what extent the slime mold can expand, assuming unlimited resources.
Correlation Coefficient
The correlation coefficient, often denoted by \(R^2\), helps us understand the relationship between two variables, in this case, time and the logarithm of the slime mold size.

In the linear regression applied in our solution, the correlation coefficient quantifies how well the observed data align with the model, which is the linear relationship \(v(t) = B + kt\).

Analyzing Correlation Coefficient:

  • **Interpreting \(R^2\):** A value of 1 implies a perfect fit, meaning the data perfectly follow the predicted trend.
  • **Significance:** A high \(R^2\) validates the exponential growth model, indicating strong proportional growth as assumed in the differential equation.
  • **Practical Use:** It allows scientists to predict future trends in data and assess the reliability of the growth model.
In summary, the correlation coefficient is a critical measure to verify the relationship modeled by the linear regression, informing us about the accuracy and efficiency of predicting slime mold size over time.

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Most popular questions from this chapter

Economists have noticed a relationship between an individual's income and consumption pattern. Describe the functional form such a relationship might take and how linear regression might be used to estimate the consumption function.

A data set relates proportional limit and tensile strength in certain alloys of gold collected for presentation at a Dentistry Convention. (Proportional limit is the load in psi at which the elongation of a sample no longer obeys Hooke's Law.) Let \(\left(\mathrm{X}_{i}, \mathrm{Y}_{\mathrm{i}}\right)\) be an observed ordered pair consisting of \(\mathrm{X}_{\mathrm{i}}\), an observed tensile strength, and \(\mathrm{Y}_{\mathrm{i}}\) an observed proportional limit, each measured in pounds per square inch (psi). After 25 observations of this sort the following summary statistics are: $$ \begin{array}{ll} \dot{\sum} \mathrm{X}_{\mathrm{i}}=2,991,300, & \underline{\mathrm{X}}=119,652 \\\ \sum \mathrm{X}_{\mathrm{i}}^{2}=372,419,750,000 . & \\ \sum \mathrm{Y}_{\mathrm{i}}=2,131,200, & \underline{\mathrm{Y}}=85,248 \\ \sum \mathrm{Y}_{\mathrm{i}}^{2}=196,195,960,000 & \\ \sum \mathrm{X}_{\mathrm{i}} \mathrm{Y}_{\mathrm{i}}=269,069,420,000 & \end{array} $$ Compute the regression coefficients relating proportional limit and tensile strength.

Discover whether the maximum likelihood estimates found in the previous problem are unbiased. That is, does \(E\left(a^{\wedge}\right)=a\), \(E\left(\beta^{\wedge}\right)=\beta\) ? Is \(E\left(\sigma^{\wedge}\right)=\sigma^{2} ?\) If an estimate is biased, how can it be adjusted so that it is unbiased?

Historically, factor analysis arose from the idea that given a covariance matrix that showed how a set of observed variables covaried, the underlying common factors could be deduced. a) Show by example how this program is carried out. b) What are some conceptual complications that prevent the idea from being fully realized?

A data set with 37 observations has the following statistics: \(\mathrm{r}_{\mathrm{xy}}=.80, \underline{\mathrm{X}}=200, \mathrm{~S}_{\mathrm{x}}=20, \underline{\mathrm{Y}}=150, \mathrm{~S}_{\mathrm{y}}=15\) (a) Compute the regression equation. If \(\mathrm{X}=180\), compute \(\mathrm{Y}\). (b) Suppose the Y-scores are normally distributed. Describe the distribution of \(\mathrm{Y}\) scores for subjects with \(\mathrm{X}=180 .\) (c) If a subject's X-score is 180, what is the probability that his Y-score is 150 or more?
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