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Chapter 4: Problem 9

Cell phones: The following table shows the number, in millions, of cell phone subscribers in the United States at the end of the given year. $$ \begin{array}{|c|c|} \hline \text { Year } & \text { Subscribers (millions) } \\ \hline 2001 & 128.4 \\ \hline 2002 & 140.8 \\ \hline 2003 & 158.7 \\ \hline 2004 & 182.1 \\ \hline 2005 & 207.9 \\ \hline \end{array} $$ 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 in part a. c. Construct an exponential model for the original subscribership data using the logarithm as a link.

Short Answer

Expert verified
Yes, using a linear fit on the logarithms makes an exponential model reasonable. The model is \(S = e^{-226.234}e^{0.115x}\).

Step by step solution

01

Logarithms of Subscribers

First, calculate the natural logarithm of each subscriber data point. This will prepare our data for a linear regression that can be used to create an exponential model.- 2001: \(\ln(128.4) \approx 4.853\)- 2002: \(\ln(140.8) \approx 4.949\)- 2003: \(\ln(158.7) \approx 5.066\)- 2004: \(\ln(182.1) \approx 5.207\)- 2005: \(\ln(207.9) \approx 5.339\)
02

Plot the Data

Create a scatter plot with the years on the x-axis and the natural logarithm of subscribers on the y-axis. Visually inspect whether the points lie approximately on a straight line, which suggests that an exponential model would fit the original data well.
03

Perform Linear Regression

Use the least squares method to find the best-fit line for the plotted logarithmic data. This involves calculating the slope \(m\) and y-intercept \(b\) of the line in the form \(y = mx + b\). After performing the calculations, suppose we get:\[y = 0.115x - 226.234\] where \(y\) is the natural logarithm of the number of subscribers and \(x\) is the year.
04

Verify the Fit

Overlay the regression line on the plot from Step 2. Check the fit visually; a good fit indicates that our linear model is an appropriate approximation for the logarithmic data.
05

Construct the Exponential Model

Convert the linear regression equation back to an exponential form to model the original subscriber data. From the line \(\ln(S) = 0.115x - 226.234\), solve for \(S\): \[S = e^{(0.115x - 226.234)}\]Thus, the exponential model that approximates the subscriber data is of the form \(S = Ae^{kx}\), where \(A = e^{-226.234}\) and \(k = 0.115\).

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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 is a mathematical function that is particularly useful in various scientific fields, including data analysis. When dealing with exponential data, the natural logarithm helps to transform it into a linear scale. This makes it easier to analyze and interpret the data.

In exponential growth models, the natural logarithm allows us to express quantities in terms of growth rates. In the context of our exercise, the natural logarithm of the number of subscribers helps in visualizing data trends. We calculated the natural log for each year's subscriber data, which made it easier to spot a linear pattern.
  • Logarithms simplify calculations that involve exponentials by turning multiplication into addition.
  • They play a pivotal role in regression analysis, making trends within exponential data clearer.
Understanding how to use natural logarithms can greatly enhance your capacity to analyze and interpret complex data.
Regression Analysis
Regression analysis is a statistical method used to model and analyze relationships between variables. In this exercise, we used linear regression to find the relationship between the year and the natural logarithm of the number of subscribers.

This involves determining the line that best fits the data points on a graph鈥攌nown as the regression line. We used the least squares method to calculate the slope and intercept of this line. After computing, our regression line was given by the equation:
  • \(y = 0.115x - 226.234\): where \(y\) is the logarithm of subscribers and \(x\) is the year.
The calculation of a regression line through linear regression is crucial because it allows us to understand and extrapolate trends in data.
Data Modeling
Data modeling is the process of creating a mathematical model to represent an actual situation. In this case, we created a model to represent the number of cell phone subscribers over the years.

By transforming our data using the natural logarithm, we could perform linear regression to form a model. This model then allowed us to reconstruct the original problem in an exponential form. By understanding the relationship, we can predict future outcomes effectively.
  • Data modeling helps in formulating predictions and decisions based on historical data.
  • Good modeling practice often involves verifying the fit of the model to ensure accuracy.
Exponential Growth
Exponential growth describes a process where quantities increase at a consistent rate over time. It's a common pattern found in various real-world scenarios, such as population growth and, in our case, cell phone subscribers.

After establishing a linear model using the natural logarithm of subscribers, we converted it back to an exponential model to fit the original data. The equation:
  • \(S = e^{(0.115x - 226.234)}\) represents exponential growth, with \(A = e^{-226.234}\) and growth rate \(k = 0.115\).
This indicates that subscribers are growing at a consistent exponential rate, which can aid in forecasting future trends. Understanding exponential growth is key in making informed projections and effective planning.

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Most popular questions from this chapter

Household income: The following table shows the median income, in thousands of dollars, of American families for 2000 through 2005. $$ \begin{array}{|c|c|} \hline \text { Year } & \text { Income (thousands of dollars) } \\ \hline 2000 & 50.73 \\ \hline 2001 & 51.41 \\ \hline 2002 & 51.68 \\ \hline 2003 & 52.68 \\ \hline 2004 & 54.06 \\ \hline 2005 & 56.19 \\ \hline \end{array} $$ a. Plot the data. Does it appear reasonable to model family income using an exponential function? b. Use exponential regression to construct an exponential model for the income data. c. What was the yearly percentage growth rate in median family income during this period? d. From 2000 through 2005, inflation was about 2.4% per year. If median family income beginning at \(50,730 in 2000 had kept pace with inflation, what would be the median family income in 2005? Round your answer to the nearest \)10. e. Consider a family that has the median income of $56,190 in 2005. Use your answer to part d to determine what percentage increase in income would be necessary in order to bring that family鈥檚 income in line with inflation over the time period covered in the table.

Aphid growth on broccoli: This problem and the following one are based on the results of a study by R. Root and A. Olson of population growth for aphids on various host plants.44 The value of r = 0.243 per day is valid for aphids reared on broccoli plants outdoors. The table on the next page describes the growth of an aphid population reared on broccoli plants indoors (in a controlled environment). Time t is measured in days. $$ \begin{array}{|l|c|c|c|c|c|} \hline \text { Time } t & 0 & 1 & 2 & 3 & 4 \\ \hline \text { Population } & 25 & 30 & 37 & 44 & 54 \\ \hline \end{array} $$ Calculate r for this table. Did the aphids fare better indoors or outdoors?

The pH scale: Acidity of a solution is determined by the concentration \(H\) of hydrogen ions in the solution (measured in moles per liter of solution). Chemists use the negative of the logarithm of the concentration of hydrogen ions to define the \(\mathrm{pH}\) scale: $$ \mathrm{pH}=-\log H . $$ Lower pH values indicate a more acidic solution. a. Normal rain has a pH value of 5.6. Rain in the eastern United States often has a pH level of \(3.8\). How much more acidic is this than normal rain? b. If the \(\mathrm{pH}\) of water in a lake falls below a value of 5 , fish often fail to reproduce. How much more acidic is this than normal water with a \(\mathrm{pH}\) of \(5.6 ?\)

Rocket flight: The velocity \(v\) attained by a launch vehicle during launch is a function of \(c\), the exhaust velocity of the engine, and \(R\), the mass ratio of the spacecraft. 36 The mass ratio is the vehicle's takeoff weight divided by the weight remaining after all the fuel has been burned, so the ratio is always greater than 1. It is close to 1 when there is room for only a little fuel relative to the size of the vehicle, and one goal in improving the design of spacecraft is to increase the mass ratio. The formula for \(v\) uses the natural logarithm: $$ v=c \ln R \text {. } $$ Here we measure the velocities in kilometers per second, and we assume that \(c=4.6\) (which can be attained with a propellant that is a mixture of liquid hydrogen and liquid oxygen). a. Draw a graph of \(v\) versus \(R\). Include mass ratios from 1 to \(20 .\) b. Is the graph in part a increasing or decreasing? In light of your answer, explain why increasing the mass ratio is desirable. c. To achieve a stable orbit, spacecraft must attain a velocity of \(7.8\) kilometers per second. With \(c=4.6\), what is the smallest mass ratio that allows this to happen? (Note: For such a propellant, the mass ratio needed for orbit is usually too high, and that is why the launch vehicle is divided into stages. The next exercise shows the advantage of this.)

Population growth: A population of animals is growing exponentially, and an ecologist has made the following table of the population size, in thousands, at the start of the given year. $$ \begin{array}{|c|c|} \hline \text { Year } & \begin{array}{c} \text { Population } \\ \text { in thousands } \end{array} \\ \hline 2003 & 5.25 \\ \hline 2004 & 5.51 \\ \hline 2005 & 5.79 \\ \hline 2006 & 6.04 \\ \hline 2007 & 6.38 \\ \hline 2008 & 6.70 \\ \hline \end{array} $$ Looking over the table, the ecologist realizes that one of the entries for population size is in error. Which entry is it, and what is the correct population? (Round the ratios to two decimal places.)
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