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Chapter 5: Problem 12

The table on the following page shows the number, in millions, graduating from high school in the United States for selected years. \({ }^{48}\) $$ \begin{array}{|c|c|} \hline \text { Year } & \text { Number } \\ \hline 1986 & 2.79 \\ \hline 1987 & 2.65 \\ \hline 1989 & 2.45 \\ \hline 1990 & 2.36 \\ \hline 1991 & 2.28 \\ \hline 1992 & 2.40 \\ \hline 1994 & 2.52 \\ \hline 1996 & 2.66 \\ \hline 1998 & 2.80 \\ \hline \end{array} $$ a. Make a plot of the data and explain why a linear model is not appropriate. b. Use regression to find a linear model for the years 1986 through 1991 . (In this part and the next, round regression line parameters to three decimal places.) c. Use regression to find a linear model for the years 1992 through \(1998 .\) d. Write a formula for a model of the number, in millions, graduating as a piecewise-defined function using the linear models from part \(b\) and part c. e. Make a graph of the formula you found in part \(\mathrm{d}\). f. The number graduating in 1993 was \(2.34 \mathrm{mil}-\) lion. On the basis of your graph in part e, determine whether this is more or less than would be expected from your formula.

Short Answer

Expert verified
The 1993 graduating number (2.34 million) is less than the expected 2.43 million from our model.

Step by step solution

01

Plot the Data

Using the given table, plot the years on the x-axis and the number of graduates on the y-axis. This involves creating a scatter plot with points for each year, and you will observe that the data does not form a straight line, showing fluctuations and changes in trend over the years.
02

Analyze for Linear Model Appropriateness

Examine the scatter plot. A linear model is not appropriate if the data points do not follow a straight line trend. From our plot, it appears that a single linear line would not fit all the data well because the number initially decreases each year from 1986 to 1991 and then has an increasing trend from 1992 to 1998.
03

Perform Regression for 1986-1991

Using the data from 1986 to 1991, perform a linear regression to find the best linear fit. The equation will be of the form \( y = mx + b \). After calculations, the linear regression yields approximately: \( y = -0.102x + 207.06 \), where \( x \) is the year and \( y \) is the number of graduates in millions.
04

Perform Regression for 1992-1998

Now use the data from 1992 to 1998 to perform another linear regression. This will also yield an equation of the form \( y = mx + b \). The resulting equation for these years is approximately: \( y = 0.048x - 94.64 \).
05

Create Piecewise Function

Write a piecewise-defined function using the equations found: \[ f(x) = \begin{cases} -0.102x + 207.06, & \text{for } 1986 \leq x \leq 1991 \ 0.048x - 94.64, & \text{for } 1992 \leq x \leq 1998 \end{cases} \]
06

Graph the Piecewise Function

Plot both lines in the piecewise function from step 5 over their respective intervals. Ensure that the transitions at the year 1991 to 1992 are clear and distinct.
07

Compare 1993 Data to Model

Using the piecewise-defined function, estimate the expected value for 1993. Insert 1993 into the second linear equation since 1993 falls in the second interval. The expected number of graduates is \( 0.048 \times 1993 - 94.64 \approx 2.43 \) million, which is greater than the actual 2.34 million, indicating the actual was less than the expectation.

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Key Concepts

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

Linear Regression
Linear regression is a statistical method used to model the relationship between a dependent variable and one or more independent variables. In the context of the high school graduation data exercise, linear regression helps us find the best-fit line that represents the trend of the number of graduates over specific time intervals.

To construct the linear model, we use the equation of a line:
  • \( y = mx + b \)
  • where \( y \) is the dependent variable (number of graduates), \( x \) is the independent variable (year), \( m \) is the slope of the line, and \( b \) is the y-intercept.
The process involves using statistical software or manual calculations to determine the values of \( m \) and \( b \) that minimize the distance between the data points and the line.

In this exercise, linear regression was applied separately over two intervals: 1986-1991 and 1992-1998, due to changes in trends within those periods.
Scatter Plot
A scatter plot is an effective way to visually explore the relationship between two quantitative variables. In this exercise, the scatter plot displays the year on the x-axis and the number of high school graduates on the y-axis, with each point representing a specific year.

By plotting the data, we can visually assess patterns, trends, and the variability within the data set. With the scatter plot of high school graduates from 1986 to 1998, none of the points form a consistently straight line, indicating the appropriateness of a linear model is questionable.

This distinct visualization highlights that numbers initially decreased from 1986 to 1991 but then exhibited an upward trend from 1992 onward. This observation led to the decision to apply separate linear models for these time frames by splitting the data into different intervals for more accurate analysis.
High School Graduation Data
The high school graduation data from 1986 to 1998 captures the yearly number of graduates, measured in millions, across specified intervals. This data is crucial for understanding educational trends and making informed decisions about resource allocation and educational policies.

Analyzing this dataset involves recognizing the factors that might influence trends in graduation rates, such as changes in population demographics, educational policies, and economic conditions.

Studying the dataset helps highlight fluctuations in graduation numbers, with a noted decrease from 1986 to 1991 due to possible demographic shifts or educational reforms. An increase observed from 1992 onwards could reflect different external influences promoting education or societal shifts that encourage higher graduation rates.

The use of mathematical modeling provides a deeper insight into these period-specific variations.
Trend Analysis
Trend analysis involves examining data to detect growth patterns, fluctuations, or sequences over a given time period. In the case of high school graduation data, trend analysis helps in understanding the underlying changes in graduation numbers.

The analysis begins with a visual inspection through scatter plots, helping to identify periods where data trends diverge. Following this, mathematical methods like linear regression are implemented to quantify trends. Here, it was determined that different linear trends existed before and after 1991.

Recognizing these periods of decline and growth allows for a more nuanced understanding. For example, a declining trend might suggest challenges in the educational environment during that time, whereas periods of growth could indicate successful policies or increased educational aspirations.

Using these insights, policymakers and educators can strategize better to enhance the educational environment and address periods of decline more effectively.

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Most popular questions from this chapter

Roughly \(90 \%\) of all stars are main-sequence stars. Exceptions include supergiants, giants, and dwarfs. For main-sequence stars (including the sun) there is an important relationship called the mass-luminosity relation between the relative luminosity \({ }^{25} L\) and the mass \(M\) in terms of solar masses. Relative masses and luminosities of several main-sequence stars are reported in the accompanying table. a. Find a power model for the data in this table. (Round the power and the coefficient to one decimal place.) The function you find is known to astronomers as the mass-luminosity relation. b. Kruger 60 is a main-sequence star that is about \(0.11\) solar mass. Use functional notation to express the relative luminosity of Kruger 60 , and then calculate that value. $$ \begin{array}{|l|c|c|} \hline \text { Star } & \begin{array}{c} \text { Solar mass } \\ M \end{array} & \begin{array}{c} \text { Luminosity } \\ L \end{array} \\ \hline \text { Spica } & 7.3 & 1050 \\ \hline \text { Vega } & 3.1 & 55 \\ \hline \text { Altair } & 1 & 1.1 \\ \hline \text { The Sun } & 1 & 1 \\ \hline 61 \text { Cygni A } & 0.17 & 0.002 \\ \hline \end{array} $$ c. Wolf 359 has a relative luminosity of about \(0.0001\). How massive is Wolf 359 ? d. If one star is 3 times as massive as another, how do their luminosities compare?

Enzymes are proteins that act as catalysts converting one type of substance, the substrate, into another type. An example of an enzyme is invertase, an enzyme in your body, which converts sucrose into fructose and glucose. Enzymes can act very rapidly; under the right circumstances, a single molecule of an enzyme can convert millions of molecules of the substrate per minute. The Michaelis-Menten relation expresses the initial speed of the reaction as a rational function of the initial concentration of the substrate: $$ v=\frac{V s}{s+K_{m}}, $$ where \(v\) is the initial speed of the reaction (in moles per liter per second), \(s\) is the initial concentration of the substrate (in moles per liter), and \(V\) and \(K_{m}\) are constants that are important measures of the kinetic properties of the enzyme. \({ }^{74}\) For this exercise, graph the Michaelis-Menten relation giving \(v\) as a function of \(s\) for two different values of \(V\) and of \(K_{m}\). a. On the basis of your graphs, what is the horizontal asymptote of \(v\) ? b. On the basis of your graphs, what value of \(s\) makes \(v(s)=V / 2\) ? How is that value related to \(K_{m}\) ? c. In practice you don't know the values of \(V\) or \(K_{m}\). Instead, you take measurements and find the graph of \(v\) as a function \(s\). Then you use the graph to determine \(V\) and \(K_{m}\). If you have the graph, how will that enable you to determine \(V\) ? How will that enable you to determine \(K_{m}\) ?

Studies to fit a logistic model to the Eastern Pacific yellowfin tuna population have yielded $$ N=\frac{148}{1+3.6 e^{-2.61 t}}, $$ where \(t\) is measured in years and \(N\) is measured in thousands of tons of fish. \({ }^{7}\) a. What is the \(r\) value for the Eastern Pacific yellowfin tuna? b. What is the carrying capacity \(K\) for the Eastern Pacific yellowfin tuna? c. What is the optimum yield level? d. Use your calculator to graph \(N\) against \(t\). e. At what time was the population growing the most rapidly?

The speed at which certain animals run is a power function of their stride length, and the power is \(k=1.7\). (See Figure 5.39.) If one animal has a stride length three times as long as another, how much faster does it run?

In a study by R. I. Van Hook \({ }^{41}\) of a grassland ecosystem in Tennessee, the rate \(O\) of energy loss to respiration of consumers and predators was initially modeled using $$ O=a W^{B}, $$ where \(W\) is weight and \(a\) and \(B\) are constants. The model was then corrected for temperature by multiplying \(O\) by \(C=1.07^{T-20}\), where \(T\) is temperature in degrees Celsius. a. What effect does the correction factor have on energy loss due to respiration if the temperature is larger than 20 degrees Celsius? b. What effect does the correction factor have on energy loss due to respiration if the temperature is exactly 20 degrees Celsius? c. What effect does the correction factor have on energy loss due to respiration if the temperature is less than 20 degrees Celsius?
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