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Household income: The following table shows the median income, in thousands of dollars, of American families. $$ \begin{array}{|c|c|} \hline \text { Year } & \begin{array}{c} \text { Income } \\ \text { (thousands of dollars) } \end{array} \\ \hline 2000 & 50.73 \\ \hline 2001 & 51.41 \\ \hline 2002 & 51.68 \\ \hline 2003 & 52.68 \\ \hline 2004 & 54.06 \\ \hline 2005 & 56.19 \\ \hline \end{array} $$ a. Plot the natural logarithm of the data. Does it appear reasonable to model family income using an exponential function? b. Find the equation of the regression line for the natural logarithm of the data. c. Construct an exponential model for the original income data using the logarithm as a link.

Short Answer

Expert verified
Yes, it appears reasonable as the plot of the log data should be linear. The equation of the regression line is \(y = mx + c\), leading to an exponential income model \(I = e^c \cdot e^{mx}\).

Step by step solution

01

Organize the Data

First, let's arrange the given data into a format suitable for creating a plot. We have years and their corresponding median income in thousands of dollars: - 2000: 50.73 - 2001: 51.41 - 2002: 51.68 - 2003: 52.68 - 2004: 54.06 - 2005: 56.19
02

Calculate the Natural Logarithm of Income

Next, we will calculate the natural logarithm of the median income for each year:\(\ln(50.73) \approx 3.927\)\(\ln(51.41) \approx 3.940\)\(\ln(51.68) \approx 3.945\)\(\ln(52.68) \approx 3.965\)\(\ln(54.06) \approx 3.989\)\(\ln(56.19) \approx 4.028\)
03

Plot the Natural Logarithm of the Data

Create a plot with years on the x-axis and the natural logarithm of the income on the y-axis. Check if the scatter plot points roughly align to a straight line, which would suggest that an exponential model is reasonable.
04

Perform Linear Regression on Logarithmic Data

Use the natural logarithm values calculated in Step 2 to perform a linear regression against the years:Determine the linear regression equation in the form \(y = mx + c\), where \(y\) is the natural log of income, \(m\) is the slope, \(x\) is the year, and \(c\) is the y-intercept.
05

Construct the Exponential Model

Once you have the regression equation in the form \(y = mx + c\), translate this to an exponential model for the original income data. Since we have \(\ln(I) = mx + c\), the exponential form would be \(I = e^{mx+c}\) or equivalently \(I = e^c \cdot e^{mx}\). The \(e^c\) represents the initial amount when time equals zero.

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Key Concepts

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

Natural Logarithm
The natural logarithm, often denoted as \( \ln \), is a mathematical function that helps us solve exponential equations by turning multiplicative relationships into additive ones. This transformation makes it an invaluable tool in exponential modeling. When dealing with data like income over several years, calculating the natural logarithm of each data point helps simplify the process of identifying underlying trends. For each year's median income, we calculated its natural logarithm. For instance, for an income of \(50.73k in the year 2000, the natural log is approximately \( \ln(50.73) \approx 3.927 \). Similarly, for 2005 with an income of \)56.19k, it's \( \ln(56.19) \approx 4.028 \). By converting the data into its logarithmic form, you can more easily assess whether a linear relationship exists, paving the way to model it using linear regression. This process aids in simplifying complex, non-linear growth into a format that's much easier to analyze and interpret.
Linear Regression
Linear regression is a fundamental statistical technique used to model the relationship between two variables by fitting a linear equation to the data. When we apply it to the natural logarithm of income data, we're essentially seeking a straight-line representation of what might be an exponential growth pattern in the original dataset. The linear regression equation \[ y = mx + c \] where \( y \) represents the natural log of the income, \( m \) is the slope, \( x \) is the year, and \( c \) is the y-intercept, is derived by calculating the line of best fit. This line minimizes the distance of all points from the line, thereby capturing the overall trend. In our case, carrying out linear regression on the calculated natural log values for each year helps us determine this linear equation. This is instrumental because it transforms a potentially complex exponential dataset into a much simpler linear form, allowing us to predict future values or identify trends more straightforwardly.
Data Plotting
Data plotting involves visually representing data points on a graph to observe trends, patterns, or relationships. It is the first step to visually hypothesize if an exponential model fits the data. When you plot years against the natural logarithm of income:
  • The years are positioned on the x-axis.
  • The corresponding natural logarithmic values of income are on the y-axis.
Seeing the plot, if the points appear to align in a relatively straight line, it suggests that the data could be modeled by an exponential function. This alignment signifies both the consistency and reliability of using a logarithmic transformation and linear regression to understand the trend. The exercise involves plotting these logarithmic values to check how linear they appear. The clearer and more consistent the alignment to a straight line, the stronger the case for applying exponential modeling to the income data. Conclusively, data plotting serves as a crucial visual verification tool before you dive deeper into detailed mathematical modeling.

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Most popular questions from this chapter

Wages: A worker is reviewing his pay increases over the past several years. The table below shows the hourly wage W, in dollars, that he earned as a function of time t, measured in years since the beginning of 1990. $$ \begin{array}{|c|c|} \hline \text { Time } t & \text { Wage } W \\ \hline 1 & 15.30 \\ \hline 2 & 15.60 \\ \hline 3 & 15.90 \\ \hline 4 & 16.25 \\ \hline \end{array} $$ a. By calculating ratios, show that the data in this table are exponential. (Round the quotients to two decimal places.) b. What is the yearly growth factor for the data? c. The worker can't remember what hourly wage he earned at the beginning of 1990 . Assuming that \(W\) is indeed an exponential function, determine what that hourly wage was. d. Find a formula giving an exponential model for \(W\) as a function of \(t\). e. What percentage raise did the worker receive each year?f. Given that prices increased by 34% over the decade of the 1990s, use your model to determine whether the worker’s wage increases kept pace with inflation.

Magazine circulation: The following table shows the circulation, in thousands, of a magazine at the start of the given year. $$ \begin{array}{|c|c|} \hline \text { Year } & \begin{array}{c} \text { Circulation } \\ \text { (thousands) } \end{array} \\ \hline 2004 & 2.64 \\ \hline 2005 & 2.77 \\ \hline 2006 & 2.94 \\ \hline 2007 & 3.08 \\ \hline 2008 & 3.25 \\ \hline 2009 & 3.42 \\ \hline \end{array} $$ a. Plot the natural logarithm of the data points. Does this plot make it look reasonable to approximate the original data with an exponential function? b. Find the regression line for the natural logarithm of the data and add its graph to the plot of the logarithm.

Grazing rabbits: The amount A of vegetation (measured in pounds) eaten in a day by a grazing animal is a function of the amount V of food available (measured in pounds per acre).15 Even if vegetation is abundant, there is a limit, called the satiation level, to the amount the animal will eat. The following table shows, for rabbits, the difference D between the satiation level and the amount A of food eaten for a variety of values of V. $$ \begin{array}{|c|c|} \hline V \text { = vegetation level } & D=\text { satiation level }-A \\ \hline 27 & 0.16 \\ \hline 36 & 0.12 \\ \hline 89 & 0.07 \\ \hline 134 & 0.05 \\ \hline 245 & 0.01 \\ \hline \end{array} $$ a. Draw a plot of D against V. Does it appear that D is approximately an exponential function of V ?

Stochastic population growth: Many populations are appropriately modeled by an exponential function, at least for a limited period of time. But there are many factors contributing to the growth of any population, and many of them depend on chance. There are a number of ways to produce stochastic models. We consider one that is an illustration of a simple Monte Carlo method. We want to model a population that is initially 500 and grows at an average rate of 2% per year. To do this we make a table of values for population according to the following procedure. The first entry in the table is for time t = 0, and it records the initial value 500. To get the entry corresponding to t = 1, we roll a die and change the population according to the face that appears, using the following rule. $$ \begin{array}{|c|c|} \hline \begin{array}{c} \text { Face } \\ \text { appearing } \end{array} & \begin{array}{c} \text { Population } \\ \text { change } \end{array} \\ \hline 1 & \text { Down 2\% } \\ \hline 2 & \text { Down } 1 \% \\ \hline 3 & \text { No change } \\ \hline 4 & \text { Up 2\% } \\ \hline 5 & \text { Up 4\% } \\ \hline 6 & \text { Up 9\% } \\ \hline \end{array} $$ To get the entry corresponding to \(t=2\), we roll a die and change the population from \(t=1\), again using the above rule. This procedure is then followed for \(t=3\), and so on. a. Using this procedure, make a table recording the population values for years 0 through 10 . b. Plot the data points from your table and the exponential model on the same screen. Comment on the level of agreement.

Population: A scientist who is studying population data makes a plot of the logarithm of the population values as a function of time. If the population is growing exponentially, what should the plot look like?

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