{"id":5344,"date":"2021-10-13T18:33:18","date_gmt":"2021-10-13T18:33:18","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/chapter\/fitting-exponential-models-to-data\/"},"modified":"2022-04-25T19:46:42","modified_gmt":"2022-04-25T19:46:42","slug":"fitting-exponential-models-to-data","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/chapter\/fitting-exponential-models-to-data\/","title":{"raw":"Fitting Models to Data","rendered":"Fitting Models to Data"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use a graphing utility to create an exponential regression from a set of data.<\/li>\r\n \t<li>Use a graphing utility to create an logarithmic regression from a set of data.<\/li>\r\n \t<li>Use a graphing utility to create an logistic regression from a set of data.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1677675\">In previous sections of this chapter, we were either given a function explicitly to graph or evaluate, or we were given a set of points that were guaranteed to lie on the curve. Then we used algebra to find the equation that fit the points exactly. In this section, we use a modeling technique called <em>regression analysis<\/em> to find a curve that models data collected from real-world observations. With <strong>regression analysis<\/strong>, we don\u2019t expect all the points to lie perfectly on the curve. The idea is to find a model that best fits the data. Then we use the model to make predictions about future events.<\/p>\r\n<p id=\"fs-id1356390\">Do not be confused by the word <em>model<\/em>. In mathematics, we often use the terms <em>function<\/em>, <em>equation<\/em>, and <em>model<\/em> interchangeably, even though they each have their own formal definition. The term <em>model<\/em> is typically used to indicate that the equation or function approximates a real-world situation.<\/p>\r\n<p id=\"fs-id969059\">We will concentrate on three types of regression models in this section: exponential, logarithmic, and logistic. Having already worked with each of these functions gives us an advantage. Knowing their formal definitions, the behavior of their graphs, and some of their real-world applications gives us the opportunity to deepen our understanding. As each regression model is presented, key features and definitions of its associated function are included for review. Take a moment to rethink each of these functions, reflect on the work we\u2019ve done so far, and then explore the ways regression is used to model real-world phenomena.<\/p>\r\n\r\n<h2>Build an exponential model from data<\/h2>\r\n<p id=\"fs-id1637870\">As we\u2019ve learned, there are a multitude of situations that can be modeled by exponential functions, such as investment growth, radioactive decay, atmospheric pressure changes, and temperatures of a cooling object. What do these phenomena have in common? For one thing, all the models either increase or decrease as time moves forward. But that\u2019s not the whole story. It\u2019s the <em>way<\/em> data increase or decrease that helps us determine whether it is best modeled by an exponential equation. Knowing the behavior of exponential functions in general allows us to recognize when to use exponential regression, so let\u2019s review exponential growth and decay.<\/p>\r\n<p id=\"fs-id1365833\">Recall that exponential functions have the form [latex]y=a{b}^{x}[\/latex] or [latex]y={A}_{0}{e}^{kx}[\/latex]. When performing regression analysis, we use the form most commonly used on graphing utilities, [latex]y=a{b}^{x}[\/latex]. Take a moment to reflect on the characteristics we\u2019ve already learned about the exponential function [latex]y=a{b}^{x}[\/latex] (assume <em>a<\/em> &gt; 0):<\/p>\r\n\r\n<ul id=\"fs-id1694862\">\r\n \t<li><em>b<\/em>\u00a0must be greater than zero and not equal to one.<\/li>\r\n \t<li>The initial value of the model is <em>y\u00a0<\/em>= <em>a<\/em>.\r\n<ul id=\"fs-id1305009\">\r\n \t<li>If <em>b\u00a0<\/em>&gt; 1, the function models exponential growth. As <em>x<\/em>\u00a0increases, the outputs of the model increase slowly at first, but then increase more and more rapidly, without bound.<\/li>\r\n \t<li>If 0 &lt; <em>b\u00a0<\/em>&lt; 1, the function models <strong>exponential decay<\/strong>. As <em>x<\/em>\u00a0increases, the outputs for the model decrease rapidly at first and then level off to become asymptotic to the <em>x<\/em>-axis. In other words, the outputs never become equal to or less than zero.<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<p id=\"fs-id1373960\">As part of the results, your calculator will display a number known as the <em>correlation coefficient<\/em>, labeled by the variable <em>r<\/em>, or [latex]{r}^{2}[\/latex]. (You may have to change the calculator\u2019s settings for these to be shown.) The values are an indication of the \"goodness of fit\" of the regression equation to the data. We more commonly use the value of [latex]{r}^{2}[\/latex] instead of <em>r<\/em>, but the closer either value is to 1, the better the regression equation approximates the data.<\/p>\r\n\r\n<div id=\"fs-id1424609\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Exponential Regression<\/h3>\r\n<p id=\"fs-id1532328\"><em>Exponential regression<\/em> is used to model situations in which growth begins slowly and then accelerates rapidly without bound, or where decay begins rapidly and then slows down to get closer and closer to zero. We use the command \"ExpReg\" on a graphing utility to fit an exponential function to a set of data points. This returns an equation of the form, [latex]y=a{b}^{x}[\/latex]<\/p>\r\n<p id=\"fs-id1585350\">Note that:<\/p>\r\n\r\n<ul id=\"fs-id1534713\">\r\n \t<li><em>b<\/em>\u00a0must be non-negative.<\/li>\r\n \t<li>when <em>b\u00a0<\/em>&gt; 1, we have an exponential growth model.<\/li>\r\n \t<li>when 0 &lt; <em>b\u00a0<\/em>&lt; 1, we have an exponential decay model.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1375157\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1588505\">How To: Given a set of data, perform exponential regression using a graphing utility.<\/h3>\r\n<ol id=\"fs-id836076\">\r\n \t<li>Use the STAT then EDIT menu to enter given data.\r\n<ol id=\"fs-id1293089\">\r\n \t<li>Clear any existing data from the lists.<\/li>\r\n \t<li>List the input values in the L1 column.<\/li>\r\n \t<li>List the output values in the L2 column.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Graph and observe a scatter plot of the data using the STATPLOT feature.\r\n<ol id=\"fs-id900017\">\r\n \t<li>Use ZOOM [9] to adjust axes to fit the data.<\/li>\r\n \t<li>Verify the data follow an exponential pattern.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Find the equation that models the data.\r\n<ol id=\"fs-id1677252\">\r\n \t<li>Select \"ExpReg\" from the STAT then CALC menu.<\/li>\r\n \t<li>Use the values returned for <em>a<\/em> and <em>b<\/em> to record the model, [latex]y=a{b}^{x}[\/latex].<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Graph the model in the same window as the scatterplot to verify it is a good fit for the data.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<h2><\/h2>\r\nhttps:\/\/www.youtube.com\/watch?v=TkMQ5n6vWGg&amp;feature=youtu.be\r\n\r\n&nbsp;\r\n<div id=\"Example_04_08_01\" class=\"example\">\r\n<div id=\"fs-id1647375\" class=\"exercise\">\r\n<div id=\"fs-id1338736\" class=\"problem textbox shaded\">\r\n<h3>Example 1: Using Exponential Regression to Fit a Model to Data<\/h3>\r\n<p id=\"fs-id1424549\">In 2007, a university study was published investigating the crash risk of alcohol impaired driving. Data from 2,871 crashes were used to measure the association of a person\u2019s blood alcohol level (BAC) with the risk of being in an accident. The table below\u00a0shows results from the study.[footnote]Source: <em>Indiana University Center for Studies of Law in Action, 2007<\/em>[\/footnote] The <em>relative risk<\/em> is a measure of how many times more likely a person is to crash. So, for example, a person with a BAC of 0.09 is 3.54 times as likely to crash as a person who has not been drinking alcohol.<\/p>\r\n\r\n<table id=\"Table_04_08_01\" summary=\"Two rows and thirteen columns. The first row is labeled, \">\r\n<tbody>\r\n<tr>\r\n<td><strong>BAC<\/strong><\/td>\r\n<td>0<\/td>\r\n<td>0.01<\/td>\r\n<td>0.03<\/td>\r\n<td>0.05<\/td>\r\n<td>0.07<\/td>\r\n<td>0.09<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Relative Risk of Crashing<\/strong><\/td>\r\n<td>1<\/td>\r\n<td>1.03<\/td>\r\n<td>1.06<\/td>\r\n<td>1.38<\/td>\r\n<td>2.09<\/td>\r\n<td>3.54<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>BAC<\/strong><\/td>\r\n<td>0.11<\/td>\r\n<td>0.13<\/td>\r\n<td>0.15<\/td>\r\n<td>0.17<\/td>\r\n<td>0.19<\/td>\r\n<td>0.21<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Relative Risk of Crashing<\/strong><\/td>\r\n<td>6.41<\/td>\r\n<td>12.6<\/td>\r\n<td>22.1<\/td>\r\n<td>39.05<\/td>\r\n<td>65.32<\/td>\r\n<td>99.78<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<ol id=\"fs-id1326068\">\r\n \t<li>Let <em>x<\/em>\u00a0represent the BAC level, and let <em>y\u00a0<\/em>represent the corresponding relative risk. Use exponential regression to fit a model to these data.<\/li>\r\n \t<li>After 6 drinks, a person weighing 160 pounds will have a BAC of about 0.16. How many times more likely is a person with this weight to crash if they drive after having a 6-pack of beer? Round to the nearest hundredth.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"482873\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"482873\"]\r\n<ol id=\"fs-id1588497\">\r\n \t<li>Using the STAT then EDIT menu on a graphing utility, list the BAC values in L1 and the relative risk values in L2. Then use the STATPLOT feature to verify that the scatterplot follows the exponential pattern shown in Figure 1:\r\n<figure id=\"CNX_Precalc_Figure_04_08_001\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010836\/CNX_Precalc_Figure_04_08_0012.jpg\" alt=\"Graph of a scattered plot.\" width=\"487\" height=\"475\" \/> <b>Figure 1<\/b>[\/caption]<\/figure>\r\n<p id=\"fs-id1300931\">Use the \"ExpReg\" command from the STAT then CALC menu to obtain the exponential model,<\/p>\r\n<p id=\"fs-id1300931\" style=\"text-align: center;\">[latex]y=0.58304829{\\left(2.20720213\\text{E}10\\right)}^{x}[\/latex]<\/p>\r\n<p id=\"fs-id1370824\">Converting from scientific notation, we have:<\/p>\r\n<p id=\"fs-id1370824\" style=\"text-align: center;\">[latex]y=0.58304829{\\left(\\text{22,072,021,300}\\right)}^{x}[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010836\/CNX_Precalc_Figure_04_08_0022.jpg\" alt=\"Graph of a scattered plot with an estimation line.\" width=\"487\" height=\"475\" \/> <b>Figure 2<\/b>[\/caption]\r\n<p id=\"fs-id1327158\">Notice that [latex]{r}^{2}\\approx 0.97[\/latex] which indicates the model is a good fit to the data. To see this, graph the model in the same window as the scatterplot to verify it is a good fit as shown in Figure 2:<span id=\"fs-id1646833\">\r\n<\/span><\/p>\r\n<\/li>\r\n \t<li>\r\n<p id=\"fs-id1598603\">Use the model to estimate the risk associated with a BAC of 0.16. Substitute 0.16 for <em>x<\/em>\u00a0in the model and solve for <em>y<\/em>.<\/p>\r\n<p id=\"fs-id1598603\" style=\"text-align: center;\">[latex]\\begin{align}y&amp; =0.58304829{\\left(\\text{22,072,021,300}\\right)}^{x}&amp;&amp; \\text{Use the regression model found in part (a).} \\\\ &amp; =0.58304829{\\left(\\text{22,072,021,300}\\right)}^{0.16}&amp;&amp; \\text{Substitute 0}\\text{.16 for }x\\text{.} \\\\ &amp; \\approx \\text{26}\\text{.35}&amp;&amp; \\text{Round to the nearest hundredth.} \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1644634\">If a 160-pound person drives after having 6 drinks, he or she is about 26.35 times more likely to crash than if driving while sober.<\/p>\r\n<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1345853\">The table below shows a recent graduate\u2019s credit card balance each month after graduation.<\/p>\r\n\r\n<table id=\"Table_04_08_02\" summary=\"Two rows and ten columns. The first row is labeled, \">\r\n<tbody>\r\n<tr>\r\n<td><strong>Month<\/strong><\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>3<\/td>\r\n<td>4<\/td>\r\n<td>5<\/td>\r\n<td>6<\/td>\r\n<td>7<\/td>\r\n<td>8<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Debt ($)<\/strong><\/td>\r\n<td>620.00<\/td>\r\n<td>761.88<\/td>\r\n<td>899.80<\/td>\r\n<td>1039.93<\/td>\r\n<td>1270.63<\/td>\r\n<td>1589.04<\/td>\r\n<td>1851.31<\/td>\r\n<td>2154.92<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p style=\"padding-left: 60px;\">a. Use exponential regression to fit a model to these data.<\/p>\r\n<p style=\"padding-left: 60px;\">b. If spending continues at this rate, what will the graduate\u2019s credit card debt be one year after graduating?<\/p>\r\n[reveal-answer q=\"153962\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"153962\"]\r\n\r\na. The exponential regression model that fits these data is [latex]y=522.88585984{\\left(1.19645256\\right)}^{x}[\/latex].\r\nb. If spending continues at this rate, the graduate\u2019s credit card debt will be $4,499.38 after one year.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1395488\" class=\"note precalculus qa textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<p id=\"eip-id1950428\"><strong>Is it reasonable to assume that an exponential regression model will represent a situation indefinitely?<\/strong><\/p>\r\n<p id=\"fs-id1693939\"><em>No. Remember that models are formed by real-world data gathered for regression. It is usually reasonable to make estimates within the interval of original observation (interpolation). However, when a model is used to make predictions, it is important to use reasoning skills to determine whether the model makes sense for inputs far beyond the original observation interval (extrapolation).<\/em><\/p>\r\n\r\n<\/div>\r\n<h2>\u00a0Build a logarithmic model from data<\/h2>\r\n<p id=\"fs-id1638611\">Just as with exponential functions, there are many real-world applications for logarithmic functions: intensity of sound, pH levels of solutions, yields of chemical reactions, production of goods, and growth of infants. As with exponential models, data modeled by logarithmic functions are either always increasing or always decreasing as time moves forward. Again, it is the <em>way<\/em> they increase or decrease that helps us determine whether a <strong>logarithmic model<\/strong> is best.<\/p>\r\n<p id=\"fs-id1294851\">Recall that logarithmic functions increase or decrease rapidly at first, but then steadily slow as time moves on. By reflecting on the characteristics we\u2019ve already learned about this function, we can better analyze real world situations that reflect this type of growth or decay. When performing logarithmic <strong>regression analysis<\/strong>, we use the form of the logarithmic function most commonly used on graphing utilities, [latex]y=a+b\\mathrm{ln}\\left(x\\right)[\/latex]. For this function<\/p>\r\n\r\n<ul id=\"fs-id1505796\">\r\n \t<li>All input values, <em>x<\/em>, must be greater than zero.<\/li>\r\n \t<li>The point (1, <em>a<\/em>) is on the graph of the model.<\/li>\r\n \t<li>If <em>b<\/em> &gt; 0, the model is increasing. Growth increases rapidly at first and then steadily slows over time.<\/li>\r\n \t<li>If <em>b\u00a0<\/em>&lt; 0, the model is decreasing. Decay occurs rapidly at first and then steadily slows over time.<\/li>\r\n<\/ul>\r\n<div id=\"fs-id1675578\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Logarithmic Regression<\/h3>\r\n<p id=\"fs-id882689\"><em>Logarithmic regression<\/em> is used to model situations where growth or decay accelerates rapidly at first and then slows over time. We use the command \"LnReg\" on a graphing utility to fit a logarithmic function to a set of data points. This returns an equation of the form,<\/p>\r\n\r\n<div id=\"eip-id1165132974342\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=a+b\\mathrm{ln}\\left(x\\right)[\/latex]<\/div>\r\n<p id=\"fs-id1638058\">Note that<\/p>\r\n\r\n<ul id=\"fs-id934921\">\r\n \t<li>all input values, <em>x<\/em>, must be non-negative.<\/li>\r\n \t<li>when <em>b\u00a0<\/em>&gt; 0, the model is increasing.<\/li>\r\n \t<li>when <em>b\u00a0<\/em>&lt; 0, the model is decreasing.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1530388\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1395706\">How To: Given a set of data, perform logarithmic regression using a graphing utility.<\/h3>\r\n<ol>\r\n \t<li style=\"list-style-type: none;\">\r\n<ol>\r\n \t<li>Use the STAT then EDIT menu to enter given data.\r\n<ol id=\"fs-id1528962\">\r\n \t<li>Clear any existing data from the lists.<\/li>\r\n \t<li>List the input values in the L1 column.<\/li>\r\n \t<li>List the output values in the L2 column.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Graph and observe a scatter plot of the data using the STATPLOT feature.\r\n<ol id=\"fs-id882304\">\r\n \t<li>Use ZOOM [9] to adjust axes to fit the data.<\/li>\r\n \t<li>Verify the data follow a logarithmic pattern.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Find the equation that models the data.\r\n<ol id=\"fs-id1107843\">\r\n \t<li>Select \"LnReg\" from the STAT then CALC menu.<\/li>\r\n \t<li>Use the values returned for <em>a<\/em> and <em>b<\/em> to record the model, [latex]y=a+b\\mathrm{ln}\\left(x\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Graph the model in the same window as the scatterplot to verify it is a good fit for the data.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<\/div>\r\n&nbsp;\r\n\r\nhttps:\/\/youtu.be\/YCRgsUSotEY\r\n\r\n&nbsp;\r\n<div id=\"Example_04_08_02\" class=\"example\">\r\n<div id=\"fs-id1616172\" class=\"exercise\">\r\n<div id=\"fs-id1586202\" class=\"problem textbox shaded\">\r\n<h3>Example 2: Using Logarithmic Regression to Fit a Model to Data<\/h3>\r\n<p id=\"fs-id1675089\">Due to advances in medicine and higher standards of living, life expectancy has been increasing in most developed countries since the beginning of the 20th century.<\/p>\r\n<p id=\"eip-id1165134068998\">The table below shows the average life expectancies, in years, of Americans from 1900\u20132010.[footnote]Source: <em>Center for Disease Control and Prevention, 2013<\/em>[\/footnote]<\/p>\r\n\r\n<table id=\"Table_04_08_03\" summary=\"Two rows and twelve columns. The first row is labeled, \">\r\n<tbody>\r\n<tr>\r\n<td><strong>Year<\/strong><\/td>\r\n<td>1900<\/td>\r\n<td>1910<\/td>\r\n<td>1920<\/td>\r\n<td>1930<\/td>\r\n<td>1940<\/td>\r\n<td>1950<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Life Expectancy(Years)<\/strong><\/td>\r\n<td>47.3<\/td>\r\n<td>50.0<\/td>\r\n<td>54.1<\/td>\r\n<td>59.7<\/td>\r\n<td>62.9<\/td>\r\n<td>68.2<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Year<\/strong><\/td>\r\n<td>1960<\/td>\r\n<td>1970<\/td>\r\n<td>1980<\/td>\r\n<td>1990<\/td>\r\n<td>2000<\/td>\r\n<td>2010<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Life Expectancy(Years)<\/strong><\/td>\r\n<td>69.7<\/td>\r\n<td>70.8<\/td>\r\n<td>73.7<\/td>\r\n<td>75.4<\/td>\r\n<td>76.8<\/td>\r\n<td>78.7<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<ol id=\"fs-id899809\">\r\n \t<li>Let <em>x<\/em>\u00a0represent time in decades starting with <em>x\u00a0<\/em>= 1 for the year 1900, <em>x\u00a0<\/em>= 2 for the year 1910, and so on. Let <em>y<\/em>\u00a0represent the corresponding life expectancy. Use logarithmic regression to fit a model to these data.<\/li>\r\n \t<li>Use the model to predict the average American life expectancy for the year 2030.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"181179\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"181179\"]\r\n<ol id=\"fs-id1601326\">\r\n \t<li>Using the STAT then EDIT menu on a graphing utility, list the years using values 1\u201312 in L1 and the corresponding life expectancy in L2. Then use the STATPLOT feature to verify that the scatterplot follows a logarithmic pattern.\r\n<figure id=\"CNX_Precalc_Figure_04_08_003\" class=\"medium\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010836\/CNX_Precalc_Figure_04_08_0032.jpg\" alt=\"Graph of a scattered plot.\" width=\"731\" height=\"437\" \/> <b>Figure 3<\/b>[\/caption]<\/figure>\r\n<p id=\"fs-id1381567\">Use the \"LnReg\" command from the STAT then CALC menu to obtain the logarithmic model,<\/p>\r\n<p id=\"fs-id1381567\" style=\"text-align: center;\">[latex]y=42.52722583+13.85752327\\mathrm{ln}\\left(x\\right)[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010837\/CNX_Precalc_Figure_04_08_0042.jpg\" alt=\"Graph of a scattered plot with an estimation line.\" width=\"731\" height=\"440\" \/> <b>Figure 4<\/b>[\/caption]\r\n<p id=\"fs-id1677824\">Next, graph the model in the same window as the scatterplot to verify it is a good fit.<span id=\"fs-id1157626\">\r\n<\/span><\/p>\r\n<\/li>\r\n \t<li>To predict the life expectancy of an American in the year 2030, substitute <em>x\u00a0<\/em>= 14 for the in the model and solve for <em>y<\/em>:<\/li>\r\n<\/ol>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}y&amp; =42.52722583+13.85752327\\mathrm{ln}\\left(x\\right)&amp;&amp; \\text{Use the regression model found in part (a).} \\\\ &amp; =42.52722583+13.85752327\\mathrm{ln}\\left(14\\right)&amp;&amp; \\text{Substitute 14 for }x\\text{.} \\\\ &amp; \\approx \\text{79}\\text{.1}&amp;&amp; \\text{Round to the nearest tenth.} \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1628396\">If life expectancy continues to increase at this pace, the average life expectancy of an American will be 79.1 by the year 2030.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id899893\">Sales of a video game released in the year 2000 took off at first, but then steadily slowed as time moved on. The table below\u00a0shows the number of games sold, in thousands, from the years 2000\u20132010.<\/p>\r\n\r\n<table id=\"Table_04_08_04\" summary=\"Two rows and twelve columns. The first row is labeled, \">\r\n<tbody>\r\n<tr>\r\n<td><strong>Year<\/strong><\/td>\r\n<td>2000<\/td>\r\n<td>2001<\/td>\r\n<td>2002<\/td>\r\n<td>2003<\/td>\r\n<td>2004<\/td>\r\n<td>2005<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Number Sold (thousands)<\/strong><\/td>\r\n<td>142<\/td>\r\n<td>149<\/td>\r\n<td>154<\/td>\r\n<td>155<\/td>\r\n<td>159<\/td>\r\n<td>161<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Year<\/strong><\/td>\r\n<td>2006<\/td>\r\n<td>2007<\/td>\r\n<td>2008<\/td>\r\n<td>2009<\/td>\r\n<td>2010<\/td>\r\n<td style=\"text-align: center;\">\u2014<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Number Sold (thousands)<\/strong><\/td>\r\n<td>163<\/td>\r\n<td>164<\/td>\r\n<td>164<\/td>\r\n<td>166<\/td>\r\n<td>167<\/td>\r\n<td style=\"text-align: center;\">\u2014<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p style=\"padding-left: 60px;\">a. Let <em>x<\/em> represent time in years starting with <em>x\u00a0<\/em>= 1 for the year 2000. Let <em>y<\/em>\u00a0represent the number of games sold in thousands. Use logarithmic regression to fit a model to these data.\r\nb. If games continue to sell at this rate, how many games will sell in 2015? Round to the nearest thousand.<\/p>\r\n[reveal-answer q=\"77368\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"77368\"]\r\n\r\na. The logarithmic regression model that fits these data is [latex]y=141.91242949+10.45366573\\mathrm{ln}\\left(x\\right)[\/latex]\r\nb. If sales continue at this rate, about 171,000 games will be sold in the year 2015.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>\u00a0Build a logistic model from data<\/h2>\r\n<p id=\"fs-id1316523\">Like exponential and logarithmic growth, logistic growth increases over time. One of the most notable differences with logistic growth models is that, at a certain point, growth steadily slows and the function approaches an upper bound, or <em>limiting value<\/em>. Because of this, logistic regression is best for modeling phenomena where there are limits in expansion, such as availability of living space or nutrients.<\/p>\r\n<p id=\"fs-id1677718\">It is worth pointing out that logistic functions actually model resource-limited exponential growth. There are many examples of this type of growth in real-world situations, including population growth and spread of disease, rumors, and even stains in fabric. When performing logistic <strong>regression analysis<\/strong>, we use the form most commonly used on graphing utilities:<\/p>\r\n\r\n<div id=\"eip-154\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=\\frac{c}{1+a{e}^{-bx}}[\/latex]<\/div>\r\n<p id=\"fs-id1310104\">Recall that:<\/p>\r\n\r\n<ul id=\"fs-id1294720\">\r\n \t<li>[latex]\\frac{c}{1+a}[\/latex] is the initial value of the model.<\/li>\r\n \t<li>when <em>b\u00a0<\/em>&gt; 0, the model increases rapidly at first until it reaches its point of maximum growth rate, [latex]\\left(\\frac{\\mathrm{ln}\\left(a\\right)}{b},\\frac{c}{2}\\right)[\/latex]. At that point, growth steadily slows and the function becomes asymptotic to the upper bound <em>y\u00a0<\/em>= <em>c<\/em>.<\/li>\r\n \t<li><em>c<\/em>\u00a0is the limiting value, sometimes called the <em>carrying capacity<\/em>, of the model.<\/li>\r\n<\/ul>\r\n<div id=\"fs-id1454974\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Logistic Regression<\/h3>\r\n<p id=\"fs-id1583226\"><em>Logistic regression<\/em> is used to model situations where growth accelerates rapidly at first and then steadily slows to an upper limit. We use the command \"Logistic\" on a graphing utility to fit a logistic function to a set of data points. This returns an equation of the form<\/p>\r\n\r\n<div id=\"eip-491\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=\\frac{c}{1+a{e}^{-bx}}[\/latex]<\/div>\r\n<p id=\"fs-id1701600\">Note that<\/p>\r\n\r\n<ul id=\"fs-id1358262\">\r\n \t<li>The initial value of the model is [latex]\\frac{c}{1+a}[\/latex].<\/li>\r\n \t<li>Output values for the model grow closer and closer to <em>y<\/em>\u00a0=\u00a0<em>c<\/em>\u00a0as time increases.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1361145\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1676582\">How To: Given a set of data, perform logistic regression using a graphing utility.<\/h3>\r\n<ol id=\"fs-id1690756\">\r\n \t<li>Use the STAT then EDIT menu to enter given data.\r\n<ol id=\"fs-id1562359\">\r\n \t<li>Clear any existing data from the lists.<\/li>\r\n \t<li>List the input values in the L1 column.<\/li>\r\n \t<li>List the output values in the L2 column.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Graph and observe a scatter plot of the data using the STATPLOT feature.\r\n<ol id=\"fs-id1431005\">\r\n \t<li>Use ZOOM [9] to adjust axes to fit the data.<\/li>\r\n \t<li>Verify the data follow a logistic pattern.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Find the equation that models the data.\r\n<ol id=\"fs-id1585412\">\r\n \t<li>Select \"Logistic\" from the STAT then CALC menu.<\/li>\r\n \t<li>Use the values returned for <em>a<\/em>, <em>b<\/em>, and <em>c<\/em>\u00a0to record the model, [latex]y=\\frac{c}{1+a{e}^{-bx}}[\/latex].<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Graph the model in the same window as the scatterplot to verify it is a good fit for the data.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<h2><\/h2>\r\nhttps:\/\/youtu.be\/fRiTk2ywiio\r\n\r\n&nbsp;\r\n<div id=\"Example_04_08_03\" class=\"example\">\r\n<div id=\"fs-id1646727\" class=\"exercise\">\r\n<div id=\"fs-id1523395\" class=\"problem textbox shaded\">\r\n<h3>Example 3: Using Logistic Regression to Fit a Model to Data<\/h3>\r\n<p id=\"fs-id1422087\">Mobile telephone service has increased rapidly in America since the mid 1990s. Today, almost all residents have cellular service. The table below\u00a0shows the percentage of Americans with cellular service between the years 1995 and 2012.[footnote]Source: <em>The World Bank, 2013<\/em>[\/footnote]<\/p>\r\n\r\n<table id=\"Table_04_08_05\" summary=\"Nineteen rows and two columns. The first column is labeled, \">\r\n<thead>\r\n<tr>\r\n<th>Year<\/th>\r\n<th>Americans with Cellular Service (%)<\/th>\r\n<th>Year<\/th>\r\n<th>Americans with Cellular Service (%)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>1995<\/td>\r\n<td>12.69<\/td>\r\n<td>2004<\/td>\r\n<td>62.852<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>1996<\/td>\r\n<td>16.35<\/td>\r\n<td>2005<\/td>\r\n<td>68.63<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>1997<\/td>\r\n<td>20.29<\/td>\r\n<td>2006<\/td>\r\n<td>76.64<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>1998<\/td>\r\n<td>25.08<\/td>\r\n<td>2007<\/td>\r\n<td>82.47<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>1999<\/td>\r\n<td>30.81<\/td>\r\n<td>2008<\/td>\r\n<td>85.68<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>2000<\/td>\r\n<td>38.75<\/td>\r\n<td>2009<\/td>\r\n<td>89.14<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>2001<\/td>\r\n<td>45.00<\/td>\r\n<td>2010<\/td>\r\n<td>91.86<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>2002<\/td>\r\n<td>49.16<\/td>\r\n<td>2011<\/td>\r\n<td>95.28<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>2003<\/td>\r\n<td>55.15<\/td>\r\n<td>2012<\/td>\r\n<td>98.17<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<ol id=\"fs-id1624806\">\r\n \t<li>Let <em>x<\/em>\u00a0represent time in years starting with <em>x\u00a0<\/em>= 0 for the year 1995. Let <em>y<\/em>\u00a0represent the corresponding percentage of residents with cellular service. Use logistic regression to fit a model to these data.<\/li>\r\n \t<li>Use the model to calculate the percentage of Americans with cell service in the year 2013. Round to the nearest tenth of a percent.<\/li>\r\n \t<li>Discuss the value returned for the upper limit, <em>c<\/em>. What does this tell you about the model? What would the limiting value be if the model were exact?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"581544\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"581544\"]\r\n<ol id=\"fs-id1455761\">\r\n \t<li>Using the STAT then EDIT menu on a graphing utility, list the years using values 0\u201315 in L1 and the corresponding percentage in L2. Then use the STATPLOT feature to verify that the scatterplot follows a logistic pattern as shown in Figure 5:\r\n<figure id=\"CNX_Precalc_Figure_04_08_005\">\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"805\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010837\/CNX_Precalc_Figure_04_08_0052.jpg\" alt=\"Graph of a scattered plot.\" width=\"805\" height=\"396\" \/> <b>Figure 5<\/b>[\/caption]<\/figure>\r\n<p id=\"fs-id1366003\">Use the \"Logistic\" command from the STAT then CALC menu to obtain the logistic model,<\/p>\r\n<p id=\"fs-id1366003\" style=\"text-align: center;\">[latex]y=\\frac{105.7379526}{1+6.88328979{e}^{-0.2595440013x}}[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"806\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010837\/CNX_Precalc_Figure_04_08_0062.jpg\" alt=\"Graph of a scattered plot with an estimation line.\" width=\"806\" height=\"396\" \/> <b>Figure 6<\/b>[\/caption]\r\n<p id=\"fs-id1650001\">Next, graph the model in the same window as shown in Figure 6\u00a0to verify it is a good fit:<span id=\"fs-id1246113\">\r\n<\/span><\/p>\r\n<\/li>\r\n \t<li>To approximate the percentage of Americans with cellular service in the year 2013, substitute <em>x\u00a0<\/em>= 18 for the in the model and solve for\u00a0<em>y<\/em>:\r\n<p id=\"fs-id1410550\" style=\"text-align: center;\">[latex]\\begin{align}y&amp; =\\frac{105.7379526}{1+6.88328979{e}^{-0.2595440013x}}&amp;&amp; \\text{Use the regression model found in part (a)}.\\\\ &amp; =\\frac{105.7379526}{1+6.88328979{e}^{-0.2595440013\\left(18\\right)}}&amp;&amp; \\text{Substitute 18 for }x. \\\\ &amp; \\approx \\text{99}\\text{.3 }&amp;&amp; \\text{Round to the nearest tenth} \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1569617\">According to the model, about 98.8% of Americans had cellular service in 2013.<\/p>\r\n<\/li>\r\n \t<li>\r\n<p id=\"fs-id1428422\">The model gives a limiting value of about 105. This means that the maximum possible percentage of Americans with cellular service would be 105%, which is impossible. (How could over 100% of a population have cellular service?) If the model were exact, the limiting value would be <em>c\u00a0<\/em>= 100 and the model\u2019s outputs would get very close to, but never actually reach 100%. After all, there will always be someone out there without cellular service!<\/p>\r\n<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1638072\">The table below shows the population, in thousands, of harbor seals in the Wadden Sea over the years 1997 to 2012.<\/p>\r\n\r\n<table id=\"Table_04_08_06\" summary=\"Seventeen rows and two columns. The first column is labeled, \">\r\n<thead>\r\n<tr>\r\n<th>Year<\/th>\r\n<th>Seal Population (Thousands)<\/th>\r\n<th>Year<\/th>\r\n<th>Seal Population (Thousands)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>1997<\/td>\r\n<td>3.493<\/td>\r\n<td>2005<\/td>\r\n<td>19.590<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>1998<\/td>\r\n<td>5.282<\/td>\r\n<td>2006<\/td>\r\n<td>21.955<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>1999<\/td>\r\n<td>6.357<\/td>\r\n<td>2007<\/td>\r\n<td>22.862<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>2000<\/td>\r\n<td>9.201<\/td>\r\n<td>2008<\/td>\r\n<td>23.869<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>2001<\/td>\r\n<td>11.224<\/td>\r\n<td>2009<\/td>\r\n<td>24.243<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>2002<\/td>\r\n<td>12.964<\/td>\r\n<td>2010<\/td>\r\n<td>24.344<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>2003<\/td>\r\n<td>16.226<\/td>\r\n<td>2011<\/td>\r\n<td>24.919<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>2004<\/td>\r\n<td>18.137<\/td>\r\n<td>2012<\/td>\r\n<td>25.108<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p style=\"padding-left: 60px;\">a. Let <em>x<\/em>\u00a0represent time in years starting with <em>x<\/em> = 0\u00a0for the year 1997. Let <em>y<\/em>\u00a0represent the number of seals in thousands. Use logistic regression to fit a model to these data.<\/p>\r\n<p style=\"padding-left: 60px;\">b. Use the model to predict the seal population for the year 2020.<\/p>\r\n<p style=\"padding-left: 60px;\">c. To the nearest whole number, what is the limiting value of this model?<\/p>\r\n[reveal-answer q=\"152985\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"152985\"]\r\n\r\na. The logistic regression model that fits these data is [latex]y=\\frac{25.65665979}{1+6.113686306{e}^{-0.3852149008x}}[\/latex].\r\nb. If the population continues to grow at this rate, there will be about 25,634 seals in 2020.\r\nc. To the nearest whole number, the carrying capacity is 25,657\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2><\/h2>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use a graphing utility to create an exponential regression from a set of data.<\/li>\n<li>Use a graphing utility to create an logarithmic regression from a set of data.<\/li>\n<li>Use a graphing utility to create an logistic regression from a set of data.<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1677675\">In previous sections of this chapter, we were either given a function explicitly to graph or evaluate, or we were given a set of points that were guaranteed to lie on the curve. Then we used algebra to find the equation that fit the points exactly. In this section, we use a modeling technique called <em>regression analysis<\/em> to find a curve that models data collected from real-world observations. With <strong>regression analysis<\/strong>, we don\u2019t expect all the points to lie perfectly on the curve. The idea is to find a model that best fits the data. Then we use the model to make predictions about future events.<\/p>\n<p id=\"fs-id1356390\">Do not be confused by the word <em>model<\/em>. In mathematics, we often use the terms <em>function<\/em>, <em>equation<\/em>, and <em>model<\/em> interchangeably, even though they each have their own formal definition. The term <em>model<\/em> is typically used to indicate that the equation or function approximates a real-world situation.<\/p>\n<p id=\"fs-id969059\">We will concentrate on three types of regression models in this section: exponential, logarithmic, and logistic. Having already worked with each of these functions gives us an advantage. Knowing their formal definitions, the behavior of their graphs, and some of their real-world applications gives us the opportunity to deepen our understanding. As each regression model is presented, key features and definitions of its associated function are included for review. Take a moment to rethink each of these functions, reflect on the work we\u2019ve done so far, and then explore the ways regression is used to model real-world phenomena.<\/p>\n<h2>Build an exponential model from data<\/h2>\n<p id=\"fs-id1637870\">As we\u2019ve learned, there are a multitude of situations that can be modeled by exponential functions, such as investment growth, radioactive decay, atmospheric pressure changes, and temperatures of a cooling object. What do these phenomena have in common? For one thing, all the models either increase or decrease as time moves forward. But that\u2019s not the whole story. It\u2019s the <em>way<\/em> data increase or decrease that helps us determine whether it is best modeled by an exponential equation. Knowing the behavior of exponential functions in general allows us to recognize when to use exponential regression, so let\u2019s review exponential growth and decay.<\/p>\n<p id=\"fs-id1365833\">Recall that exponential functions have the form [latex]y=a{b}^{x}[\/latex] or [latex]y={A}_{0}{e}^{kx}[\/latex]. When performing regression analysis, we use the form most commonly used on graphing utilities, [latex]y=a{b}^{x}[\/latex]. Take a moment to reflect on the characteristics we\u2019ve already learned about the exponential function [latex]y=a{b}^{x}[\/latex] (assume <em>a<\/em> &gt; 0):<\/p>\n<ul id=\"fs-id1694862\">\n<li><em>b<\/em>\u00a0must be greater than zero and not equal to one.<\/li>\n<li>The initial value of the model is <em>y\u00a0<\/em>= <em>a<\/em>.\n<ul id=\"fs-id1305009\">\n<li>If <em>b\u00a0<\/em>&gt; 1, the function models exponential growth. As <em>x<\/em>\u00a0increases, the outputs of the model increase slowly at first, but then increase more and more rapidly, without bound.<\/li>\n<li>If 0 &lt; <em>b\u00a0<\/em>&lt; 1, the function models <strong>exponential decay<\/strong>. As <em>x<\/em>\u00a0increases, the outputs for the model decrease rapidly at first and then level off to become asymptotic to the <em>x<\/em>-axis. In other words, the outputs never become equal to or less than zero.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p id=\"fs-id1373960\">As part of the results, your calculator will display a number known as the <em>correlation coefficient<\/em>, labeled by the variable <em>r<\/em>, or [latex]{r}^{2}[\/latex]. (You may have to change the calculator\u2019s settings for these to be shown.) The values are an indication of the &#8220;goodness of fit&#8221; of the regression equation to the data. We more commonly use the value of [latex]{r}^{2}[\/latex] instead of <em>r<\/em>, but the closer either value is to 1, the better the regression equation approximates the data.<\/p>\n<div id=\"fs-id1424609\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Exponential Regression<\/h3>\n<p id=\"fs-id1532328\"><em>Exponential regression<\/em> is used to model situations in which growth begins slowly and then accelerates rapidly without bound, or where decay begins rapidly and then slows down to get closer and closer to zero. We use the command &#8220;ExpReg&#8221; on a graphing utility to fit an exponential function to a set of data points. This returns an equation of the form, [latex]y=a{b}^{x}[\/latex]<\/p>\n<p id=\"fs-id1585350\">Note that:<\/p>\n<ul id=\"fs-id1534713\">\n<li><em>b<\/em>\u00a0must be non-negative.<\/li>\n<li>when <em>b\u00a0<\/em>&gt; 1, we have an exponential growth model.<\/li>\n<li>when 0 &lt; <em>b\u00a0<\/em>&lt; 1, we have an exponential decay model.<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1375157\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1588505\">How To: Given a set of data, perform exponential regression using a graphing utility.<\/h3>\n<ol id=\"fs-id836076\">\n<li>Use the STAT then EDIT menu to enter given data.\n<ol id=\"fs-id1293089\">\n<li>Clear any existing data from the lists.<\/li>\n<li>List the input values in the L1 column.<\/li>\n<li>List the output values in the L2 column.<\/li>\n<\/ol>\n<\/li>\n<li>Graph and observe a scatter plot of the data using the STATPLOT feature.\n<ol id=\"fs-id900017\">\n<li>Use ZOOM [9] to adjust axes to fit the data.<\/li>\n<li>Verify the data follow an exponential pattern.<\/li>\n<\/ol>\n<\/li>\n<li>Find the equation that models the data.\n<ol id=\"fs-id1677252\">\n<li>Select &#8220;ExpReg&#8221; from the STAT then CALC menu.<\/li>\n<li>Use the values returned for <em>a<\/em> and <em>b<\/em> to record the model, [latex]y=a{b}^{x}[\/latex].<\/li>\n<\/ol>\n<\/li>\n<li>Graph the model in the same window as the scatterplot to verify it is a good fit for the data.<\/li>\n<\/ol>\n<\/div>\n<h2><\/h2>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Exponential Regression on the TI84 - Example 1\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/TkMQ5n6vWGg?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>&nbsp;<\/p>\n<div id=\"Example_04_08_01\" class=\"example\">\n<div id=\"fs-id1647375\" class=\"exercise\">\n<div id=\"fs-id1338736\" class=\"problem textbox shaded\">\n<h3>Example 1: Using Exponential Regression to Fit a Model to Data<\/h3>\n<p id=\"fs-id1424549\">In 2007, a university study was published investigating the crash risk of alcohol impaired driving. Data from 2,871 crashes were used to measure the association of a person\u2019s blood alcohol level (BAC) with the risk of being in an accident. The table below\u00a0shows results from the study.<a class=\"footnote\" title=\"Source: Indiana University Center for Studies of Law in Action, 2007\" id=\"return-footnote-5344-1\" href=\"#footnote-5344-1\" aria-label=\"Footnote 1\"><sup class=\"footnote\">[1]<\/sup><\/a> The <em>relative risk<\/em> is a measure of how many times more likely a person is to crash. So, for example, a person with a BAC of 0.09 is 3.54 times as likely to crash as a person who has not been drinking alcohol.<\/p>\n<table id=\"Table_04_08_01\" summary=\"Two rows and thirteen columns. The first row is labeled,\">\n<tbody>\n<tr>\n<td><strong>BAC<\/strong><\/td>\n<td>0<\/td>\n<td>0.01<\/td>\n<td>0.03<\/td>\n<td>0.05<\/td>\n<td>0.07<\/td>\n<td>0.09<\/td>\n<\/tr>\n<tr>\n<td><strong>Relative Risk of Crashing<\/strong><\/td>\n<td>1<\/td>\n<td>1.03<\/td>\n<td>1.06<\/td>\n<td>1.38<\/td>\n<td>2.09<\/td>\n<td>3.54<\/td>\n<\/tr>\n<tr>\n<td><strong>BAC<\/strong><\/td>\n<td>0.11<\/td>\n<td>0.13<\/td>\n<td>0.15<\/td>\n<td>0.17<\/td>\n<td>0.19<\/td>\n<td>0.21<\/td>\n<\/tr>\n<tr>\n<td><strong>Relative Risk of Crashing<\/strong><\/td>\n<td>6.41<\/td>\n<td>12.6<\/td>\n<td>22.1<\/td>\n<td>39.05<\/td>\n<td>65.32<\/td>\n<td>99.78<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<ol id=\"fs-id1326068\">\n<li>Let <em>x<\/em>\u00a0represent the BAC level, and let <em>y\u00a0<\/em>represent the corresponding relative risk. Use exponential regression to fit a model to these data.<\/li>\n<li>After 6 drinks, a person weighing 160 pounds will have a BAC of about 0.16. How many times more likely is a person with this weight to crash if they drive after having a 6-pack of beer? Round to the nearest hundredth.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q482873\">Show Solution<\/span><\/p>\n<div id=\"q482873\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1588497\">\n<li>Using the STAT then EDIT menu on a graphing utility, list the BAC values in L1 and the relative risk values in L2. Then use the STATPLOT feature to verify that the scatterplot follows the exponential pattern shown in Figure 1:<br \/>\n<figure id=\"CNX_Precalc_Figure_04_08_001\" class=\"small\">\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010836\/CNX_Precalc_Figure_04_08_0012.jpg\" alt=\"Graph of a scattered plot.\" width=\"487\" height=\"475\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 1<\/b><\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1300931\">Use the &#8220;ExpReg&#8221; command from the STAT then CALC menu to obtain the exponential model,<\/p>\n<p id=\"fs-id1300931\" style=\"text-align: center;\">[latex]y=0.58304829{\\left(2.20720213\\text{E}10\\right)}^{x}[\/latex]<\/p>\n<p id=\"fs-id1370824\">Converting from scientific notation, we have:<\/p>\n<p id=\"fs-id1370824\" style=\"text-align: center;\">[latex]y=0.58304829{\\left(\\text{22,072,021,300}\\right)}^{x}[\/latex]<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010836\/CNX_Precalc_Figure_04_08_0022.jpg\" alt=\"Graph of a scattered plot with an estimation line.\" width=\"487\" height=\"475\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 2<\/b><\/p>\n<\/div>\n<p id=\"fs-id1327158\">Notice that [latex]{r}^{2}\\approx 0.97[\/latex] which indicates the model is a good fit to the data. To see this, graph the model in the same window as the scatterplot to verify it is a good fit as shown in Figure 2:<span id=\"fs-id1646833\"><br \/>\n<\/span><\/p>\n<\/li>\n<li>\n<p id=\"fs-id1598603\">Use the model to estimate the risk associated with a BAC of 0.16. Substitute 0.16 for <em>x<\/em>\u00a0in the model and solve for <em>y<\/em>.<\/p>\n<p id=\"fs-id1598603\" style=\"text-align: center;\">[latex]\\begin{align}y& =0.58304829{\\left(\\text{22,072,021,300}\\right)}^{x}&& \\text{Use the regression model found in part (a).} \\\\ & =0.58304829{\\left(\\text{22,072,021,300}\\right)}^{0.16}&& \\text{Substitute 0}\\text{.16 for }x\\text{.} \\\\ & \\approx \\text{26}\\text{.35}&& \\text{Round to the nearest hundredth.} \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1644634\">If a 160-pound person drives after having 6 drinks, he or she is about 26.35 times more likely to crash than if driving while sober.<\/p>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1345853\">The table below shows a recent graduate\u2019s credit card balance each month after graduation.<\/p>\n<table id=\"Table_04_08_02\" summary=\"Two rows and ten columns. The first row is labeled,\">\n<tbody>\n<tr>\n<td><strong>Month<\/strong><\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<td>4<\/td>\n<td>5<\/td>\n<td>6<\/td>\n<td>7<\/td>\n<td>8<\/td>\n<\/tr>\n<tr>\n<td><strong>Debt ($)<\/strong><\/td>\n<td>620.00<\/td>\n<td>761.88<\/td>\n<td>899.80<\/td>\n<td>1039.93<\/td>\n<td>1270.63<\/td>\n<td>1589.04<\/td>\n<td>1851.31<\/td>\n<td>2154.92<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"padding-left: 60px;\">a. Use exponential regression to fit a model to these data.<\/p>\n<p style=\"padding-left: 60px;\">b. If spending continues at this rate, what will the graduate\u2019s credit card debt be one year after graduating?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q153962\">Show Solution<\/span><\/p>\n<div id=\"q153962\" class=\"hidden-answer\" style=\"display: none\">\n<p>a. The exponential regression model that fits these data is [latex]y=522.88585984{\\left(1.19645256\\right)}^{x}[\/latex].<br \/>\nb. If spending continues at this rate, the graduate\u2019s credit card debt will be $4,499.38 after one year.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1395488\" class=\"note precalculus qa textbox\">\n<h3>Q &amp; A<\/h3>\n<p id=\"eip-id1950428\"><strong>Is it reasonable to assume that an exponential regression model will represent a situation indefinitely?<\/strong><\/p>\n<p id=\"fs-id1693939\"><em>No. Remember that models are formed by real-world data gathered for regression. It is usually reasonable to make estimates within the interval of original observation (interpolation). However, when a model is used to make predictions, it is important to use reasoning skills to determine whether the model makes sense for inputs far beyond the original observation interval (extrapolation).<\/em><\/p>\n<\/div>\n<h2>\u00a0Build a logarithmic model from data<\/h2>\n<p id=\"fs-id1638611\">Just as with exponential functions, there are many real-world applications for logarithmic functions: intensity of sound, pH levels of solutions, yields of chemical reactions, production of goods, and growth of infants. As with exponential models, data modeled by logarithmic functions are either always increasing or always decreasing as time moves forward. Again, it is the <em>way<\/em> they increase or decrease that helps us determine whether a <strong>logarithmic model<\/strong> is best.<\/p>\n<p id=\"fs-id1294851\">Recall that logarithmic functions increase or decrease rapidly at first, but then steadily slow as time moves on. By reflecting on the characteristics we\u2019ve already learned about this function, we can better analyze real world situations that reflect this type of growth or decay. When performing logarithmic <strong>regression analysis<\/strong>, we use the form of the logarithmic function most commonly used on graphing utilities, [latex]y=a+b\\mathrm{ln}\\left(x\\right)[\/latex]. For this function<\/p>\n<ul id=\"fs-id1505796\">\n<li>All input values, <em>x<\/em>, must be greater than zero.<\/li>\n<li>The point (1, <em>a<\/em>) is on the graph of the model.<\/li>\n<li>If <em>b<\/em> &gt; 0, the model is increasing. Growth increases rapidly at first and then steadily slows over time.<\/li>\n<li>If <em>b\u00a0<\/em>&lt; 0, the model is decreasing. Decay occurs rapidly at first and then steadily slows over time.<\/li>\n<\/ul>\n<div id=\"fs-id1675578\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Logarithmic Regression<\/h3>\n<p id=\"fs-id882689\"><em>Logarithmic regression<\/em> is used to model situations where growth or decay accelerates rapidly at first and then slows over time. We use the command &#8220;LnReg&#8221; on a graphing utility to fit a logarithmic function to a set of data points. This returns an equation of the form,<\/p>\n<div id=\"eip-id1165132974342\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=a+b\\mathrm{ln}\\left(x\\right)[\/latex]<\/div>\n<p id=\"fs-id1638058\">Note that<\/p>\n<ul id=\"fs-id934921\">\n<li>all input values, <em>x<\/em>, must be non-negative.<\/li>\n<li>when <em>b\u00a0<\/em>&gt; 0, the model is increasing.<\/li>\n<li>when <em>b\u00a0<\/em>&lt; 0, the model is decreasing.<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1530388\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1395706\">How To: Given a set of data, perform logarithmic regression using a graphing utility.<\/h3>\n<ol>\n<li style=\"list-style-type: none;\">\n<ol>\n<li>Use the STAT then EDIT menu to enter given data.\n<ol id=\"fs-id1528962\">\n<li>Clear any existing data from the lists.<\/li>\n<li>List the input values in the L1 column.<\/li>\n<li>List the output values in the L2 column.<\/li>\n<\/ol>\n<\/li>\n<li>Graph and observe a scatter plot of the data using the STATPLOT feature.\n<ol id=\"fs-id882304\">\n<li>Use ZOOM [9] to adjust axes to fit the data.<\/li>\n<li>Verify the data follow a logarithmic pattern.<\/li>\n<\/ol>\n<\/li>\n<li>Find the equation that models the data.\n<ol id=\"fs-id1107843\">\n<li>Select &#8220;LnReg&#8221; from the STAT then CALC menu.<\/li>\n<li>Use the values returned for <em>a<\/em> and <em>b<\/em> to record the model, [latex]y=a+b\\mathrm{ln}\\left(x\\right)[\/latex].<\/li>\n<\/ol>\n<\/li>\n<li>Graph the model in the same window as the scatterplot to verify it is a good fit for the data.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/div>\n<p>&nbsp;<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Logarithmic Regression on the TI84\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/YCRgsUSotEY?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>&nbsp;<\/p>\n<div id=\"Example_04_08_02\" class=\"example\">\n<div id=\"fs-id1616172\" class=\"exercise\">\n<div id=\"fs-id1586202\" class=\"problem textbox shaded\">\n<h3>Example 2: Using Logarithmic Regression to Fit a Model to Data<\/h3>\n<p id=\"fs-id1675089\">Due to advances in medicine and higher standards of living, life expectancy has been increasing in most developed countries since the beginning of the 20th century.<\/p>\n<p id=\"eip-id1165134068998\">The table below shows the average life expectancies, in years, of Americans from 1900\u20132010.<a class=\"footnote\" title=\"Source: Center for Disease Control and Prevention, 2013\" id=\"return-footnote-5344-2\" href=\"#footnote-5344-2\" aria-label=\"Footnote 2\"><sup class=\"footnote\">[2]<\/sup><\/a><\/p>\n<table id=\"Table_04_08_03\" summary=\"Two rows and twelve columns. The first row is labeled,\">\n<tbody>\n<tr>\n<td><strong>Year<\/strong><\/td>\n<td>1900<\/td>\n<td>1910<\/td>\n<td>1920<\/td>\n<td>1930<\/td>\n<td>1940<\/td>\n<td>1950<\/td>\n<\/tr>\n<tr>\n<td><strong>Life Expectancy(Years)<\/strong><\/td>\n<td>47.3<\/td>\n<td>50.0<\/td>\n<td>54.1<\/td>\n<td>59.7<\/td>\n<td>62.9<\/td>\n<td>68.2<\/td>\n<\/tr>\n<tr>\n<td><strong>Year<\/strong><\/td>\n<td>1960<\/td>\n<td>1970<\/td>\n<td>1980<\/td>\n<td>1990<\/td>\n<td>2000<\/td>\n<td>2010<\/td>\n<\/tr>\n<tr>\n<td><strong>Life Expectancy(Years)<\/strong><\/td>\n<td>69.7<\/td>\n<td>70.8<\/td>\n<td>73.7<\/td>\n<td>75.4<\/td>\n<td>76.8<\/td>\n<td>78.7<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<ol id=\"fs-id899809\">\n<li>Let <em>x<\/em>\u00a0represent time in decades starting with <em>x\u00a0<\/em>= 1 for the year 1900, <em>x\u00a0<\/em>= 2 for the year 1910, and so on. Let <em>y<\/em>\u00a0represent the corresponding life expectancy. Use logarithmic regression to fit a model to these data.<\/li>\n<li>Use the model to predict the average American life expectancy for the year 2030.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q181179\">Show Solution<\/span><\/p>\n<div id=\"q181179\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1601326\">\n<li>Using the STAT then EDIT menu on a graphing utility, list the years using values 1\u201312 in L1 and the corresponding life expectancy in L2. Then use the STATPLOT feature to verify that the scatterplot follows a logarithmic pattern.<br \/>\n<figure id=\"CNX_Precalc_Figure_04_08_003\" class=\"medium\">\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010836\/CNX_Precalc_Figure_04_08_0032.jpg\" alt=\"Graph of a scattered plot.\" width=\"731\" height=\"437\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 3<\/b><\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1381567\">Use the &#8220;LnReg&#8221; command from the STAT then CALC menu to obtain the logarithmic model,<\/p>\n<p id=\"fs-id1381567\" style=\"text-align: center;\">[latex]y=42.52722583+13.85752327\\mathrm{ln}\\left(x\\right)[\/latex]<\/p>\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010837\/CNX_Precalc_Figure_04_08_0042.jpg\" alt=\"Graph of a scattered plot with an estimation line.\" width=\"731\" height=\"440\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 4<\/b><\/p>\n<\/div>\n<p id=\"fs-id1677824\">Next, graph the model in the same window as the scatterplot to verify it is a good fit.<span id=\"fs-id1157626\"><br \/>\n<\/span><\/p>\n<\/li>\n<li>To predict the life expectancy of an American in the year 2030, substitute <em>x\u00a0<\/em>= 14 for the in the model and solve for <em>y<\/em>:<\/li>\n<\/ol>\n<p style=\"text-align: center;\">[latex]\\begin{align}y& =42.52722583+13.85752327\\mathrm{ln}\\left(x\\right)&& \\text{Use the regression model found in part (a).} \\\\ & =42.52722583+13.85752327\\mathrm{ln}\\left(14\\right)&& \\text{Substitute 14 for }x\\text{.} \\\\ & \\approx \\text{79}\\text{.1}&& \\text{Round to the nearest tenth.} \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1628396\">If life expectancy continues to increase at this pace, the average life expectancy of an American will be 79.1 by the year 2030.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id899893\">Sales of a video game released in the year 2000 took off at first, but then steadily slowed as time moved on. The table below\u00a0shows the number of games sold, in thousands, from the years 2000\u20132010.<\/p>\n<table id=\"Table_04_08_04\" summary=\"Two rows and twelve columns. The first row is labeled,\">\n<tbody>\n<tr>\n<td><strong>Year<\/strong><\/td>\n<td>2000<\/td>\n<td>2001<\/td>\n<td>2002<\/td>\n<td>2003<\/td>\n<td>2004<\/td>\n<td>2005<\/td>\n<\/tr>\n<tr>\n<td><strong>Number Sold (thousands)<\/strong><\/td>\n<td>142<\/td>\n<td>149<\/td>\n<td>154<\/td>\n<td>155<\/td>\n<td>159<\/td>\n<td>161<\/td>\n<\/tr>\n<tr>\n<td><strong>Year<\/strong><\/td>\n<td>2006<\/td>\n<td>2007<\/td>\n<td>2008<\/td>\n<td>2009<\/td>\n<td>2010<\/td>\n<td style=\"text-align: center;\">\u2014<\/td>\n<\/tr>\n<tr>\n<td><strong>Number Sold (thousands)<\/strong><\/td>\n<td>163<\/td>\n<td>164<\/td>\n<td>164<\/td>\n<td>166<\/td>\n<td>167<\/td>\n<td style=\"text-align: center;\">\u2014<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"padding-left: 60px;\">a. Let <em>x<\/em> represent time in years starting with <em>x\u00a0<\/em>= 1 for the year 2000. Let <em>y<\/em>\u00a0represent the number of games sold in thousands. Use logarithmic regression to fit a model to these data.<br \/>\nb. If games continue to sell at this rate, how many games will sell in 2015? Round to the nearest thousand.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q77368\">Show Solution<\/span><\/p>\n<div id=\"q77368\" class=\"hidden-answer\" style=\"display: none\">\n<p>a. The logarithmic regression model that fits these data is [latex]y=141.91242949+10.45366573\\mathrm{ln}\\left(x\\right)[\/latex]<br \/>\nb. If sales continue at this rate, about 171,000 games will be sold in the year 2015.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>\u00a0Build a logistic model from data<\/h2>\n<p id=\"fs-id1316523\">Like exponential and logarithmic growth, logistic growth increases over time. One of the most notable differences with logistic growth models is that, at a certain point, growth steadily slows and the function approaches an upper bound, or <em>limiting value<\/em>. Because of this, logistic regression is best for modeling phenomena where there are limits in expansion, such as availability of living space or nutrients.<\/p>\n<p id=\"fs-id1677718\">It is worth pointing out that logistic functions actually model resource-limited exponential growth. There are many examples of this type of growth in real-world situations, including population growth and spread of disease, rumors, and even stains in fabric. When performing logistic <strong>regression analysis<\/strong>, we use the form most commonly used on graphing utilities:<\/p>\n<div id=\"eip-154\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=\\frac{c}{1+a{e}^{-bx}}[\/latex]<\/div>\n<p id=\"fs-id1310104\">Recall that:<\/p>\n<ul id=\"fs-id1294720\">\n<li>[latex]\\frac{c}{1+a}[\/latex] is the initial value of the model.<\/li>\n<li>when <em>b\u00a0<\/em>&gt; 0, the model increases rapidly at first until it reaches its point of maximum growth rate, [latex]\\left(\\frac{\\mathrm{ln}\\left(a\\right)}{b},\\frac{c}{2}\\right)[\/latex]. At that point, growth steadily slows and the function becomes asymptotic to the upper bound <em>y\u00a0<\/em>= <em>c<\/em>.<\/li>\n<li><em>c<\/em>\u00a0is the limiting value, sometimes called the <em>carrying capacity<\/em>, of the model.<\/li>\n<\/ul>\n<div id=\"fs-id1454974\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Logistic Regression<\/h3>\n<p id=\"fs-id1583226\"><em>Logistic regression<\/em> is used to model situations where growth accelerates rapidly at first and then steadily slows to an upper limit. We use the command &#8220;Logistic&#8221; on a graphing utility to fit a logistic function to a set of data points. This returns an equation of the form<\/p>\n<div id=\"eip-491\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=\\frac{c}{1+a{e}^{-bx}}[\/latex]<\/div>\n<p id=\"fs-id1701600\">Note that<\/p>\n<ul id=\"fs-id1358262\">\n<li>The initial value of the model is [latex]\\frac{c}{1+a}[\/latex].<\/li>\n<li>Output values for the model grow closer and closer to <em>y<\/em>\u00a0=\u00a0<em>c<\/em>\u00a0as time increases.<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1361145\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1676582\">How To: Given a set of data, perform logistic regression using a graphing utility.<\/h3>\n<ol id=\"fs-id1690756\">\n<li>Use the STAT then EDIT menu to enter given data.\n<ol id=\"fs-id1562359\">\n<li>Clear any existing data from the lists.<\/li>\n<li>List the input values in the L1 column.<\/li>\n<li>List the output values in the L2 column.<\/li>\n<\/ol>\n<\/li>\n<li>Graph and observe a scatter plot of the data using the STATPLOT feature.\n<ol id=\"fs-id1431005\">\n<li>Use ZOOM [9] to adjust axes to fit the data.<\/li>\n<li>Verify the data follow a logistic pattern.<\/li>\n<\/ol>\n<\/li>\n<li>Find the equation that models the data.\n<ol id=\"fs-id1585412\">\n<li>Select &#8220;Logistic&#8221; from the STAT then CALC menu.<\/li>\n<li>Use the values returned for <em>a<\/em>, <em>b<\/em>, and <em>c<\/em>\u00a0to record the model, [latex]y=\\frac{c}{1+a{e}^{-bx}}[\/latex].<\/li>\n<\/ol>\n<\/li>\n<li>Graph the model in the same window as the scatterplot to verify it is a good fit for the data.<\/li>\n<\/ol>\n<\/div>\n<h2><\/h2>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Logistic Regression on the TI84\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/fRiTk2ywiio?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>&nbsp;<\/p>\n<div id=\"Example_04_08_03\" class=\"example\">\n<div id=\"fs-id1646727\" class=\"exercise\">\n<div id=\"fs-id1523395\" class=\"problem textbox shaded\">\n<h3>Example 3: Using Logistic Regression to Fit a Model to Data<\/h3>\n<p id=\"fs-id1422087\">Mobile telephone service has increased rapidly in America since the mid 1990s. Today, almost all residents have cellular service. The table below\u00a0shows the percentage of Americans with cellular service between the years 1995 and 2012.<a class=\"footnote\" title=\"Source: The World Bank, 2013\" id=\"return-footnote-5344-3\" href=\"#footnote-5344-3\" aria-label=\"Footnote 3\"><sup class=\"footnote\">[3]<\/sup><\/a><\/p>\n<table id=\"Table_04_08_05\" summary=\"Nineteen rows and two columns. The first column is labeled,\">\n<thead>\n<tr>\n<th>Year<\/th>\n<th>Americans with Cellular Service (%)<\/th>\n<th>Year<\/th>\n<th>Americans with Cellular Service (%)<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>1995<\/td>\n<td>12.69<\/td>\n<td>2004<\/td>\n<td>62.852<\/td>\n<\/tr>\n<tr>\n<td>1996<\/td>\n<td>16.35<\/td>\n<td>2005<\/td>\n<td>68.63<\/td>\n<\/tr>\n<tr>\n<td>1997<\/td>\n<td>20.29<\/td>\n<td>2006<\/td>\n<td>76.64<\/td>\n<\/tr>\n<tr>\n<td>1998<\/td>\n<td>25.08<\/td>\n<td>2007<\/td>\n<td>82.47<\/td>\n<\/tr>\n<tr>\n<td>1999<\/td>\n<td>30.81<\/td>\n<td>2008<\/td>\n<td>85.68<\/td>\n<\/tr>\n<tr>\n<td>2000<\/td>\n<td>38.75<\/td>\n<td>2009<\/td>\n<td>89.14<\/td>\n<\/tr>\n<tr>\n<td>2001<\/td>\n<td>45.00<\/td>\n<td>2010<\/td>\n<td>91.86<\/td>\n<\/tr>\n<tr>\n<td>2002<\/td>\n<td>49.16<\/td>\n<td>2011<\/td>\n<td>95.28<\/td>\n<\/tr>\n<tr>\n<td>2003<\/td>\n<td>55.15<\/td>\n<td>2012<\/td>\n<td>98.17<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<ol id=\"fs-id1624806\">\n<li>Let <em>x<\/em>\u00a0represent time in years starting with <em>x\u00a0<\/em>= 0 for the year 1995. Let <em>y<\/em>\u00a0represent the corresponding percentage of residents with cellular service. Use logistic regression to fit a model to these data.<\/li>\n<li>Use the model to calculate the percentage of Americans with cell service in the year 2013. Round to the nearest tenth of a percent.<\/li>\n<li>Discuss the value returned for the upper limit, <em>c<\/em>. What does this tell you about the model? What would the limiting value be if the model were exact?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q581544\">Show Solution<\/span><\/p>\n<div id=\"q581544\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1455761\">\n<li>Using the STAT then EDIT menu on a graphing utility, list the years using values 0\u201315 in L1 and the corresponding percentage in L2. Then use the STATPLOT feature to verify that the scatterplot follows a logistic pattern as shown in Figure 5:<br \/>\n<figure id=\"CNX_Precalc_Figure_04_08_005\">\n<div style=\"width: 815px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010837\/CNX_Precalc_Figure_04_08_0052.jpg\" alt=\"Graph of a scattered plot.\" width=\"805\" height=\"396\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 5<\/b><\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1366003\">Use the &#8220;Logistic&#8221; command from the STAT then CALC menu to obtain the logistic model,<\/p>\n<p id=\"fs-id1366003\" style=\"text-align: center;\">[latex]y=\\frac{105.7379526}{1+6.88328979{e}^{-0.2595440013x}}[\/latex]<\/p>\n<div style=\"width: 816px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010837\/CNX_Precalc_Figure_04_08_0062.jpg\" alt=\"Graph of a scattered plot with an estimation line.\" width=\"806\" height=\"396\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 6<\/b><\/p>\n<\/div>\n<p id=\"fs-id1650001\">Next, graph the model in the same window as shown in Figure 6\u00a0to verify it is a good fit:<span id=\"fs-id1246113\"><br \/>\n<\/span><\/p>\n<\/li>\n<li>To approximate the percentage of Americans with cellular service in the year 2013, substitute <em>x\u00a0<\/em>= 18 for the in the model and solve for\u00a0<em>y<\/em>:\n<p id=\"fs-id1410550\" style=\"text-align: center;\">[latex]\\begin{align}y& =\\frac{105.7379526}{1+6.88328979{e}^{-0.2595440013x}}&& \\text{Use the regression model found in part (a)}.\\\\ & =\\frac{105.7379526}{1+6.88328979{e}^{-0.2595440013\\left(18\\right)}}&& \\text{Substitute 18 for }x. \\\\ & \\approx \\text{99}\\text{.3 }&& \\text{Round to the nearest tenth} \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1569617\">According to the model, about 98.8% of Americans had cellular service in 2013.<\/p>\n<\/li>\n<li>\n<p id=\"fs-id1428422\">The model gives a limiting value of about 105. This means that the maximum possible percentage of Americans with cellular service would be 105%, which is impossible. (How could over 100% of a population have cellular service?) If the model were exact, the limiting value would be <em>c\u00a0<\/em>= 100 and the model\u2019s outputs would get very close to, but never actually reach 100%. After all, there will always be someone out there without cellular service!<\/p>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1638072\">The table below shows the population, in thousands, of harbor seals in the Wadden Sea over the years 1997 to 2012.<\/p>\n<table id=\"Table_04_08_06\" summary=\"Seventeen rows and two columns. The first column is labeled,\">\n<thead>\n<tr>\n<th>Year<\/th>\n<th>Seal Population (Thousands)<\/th>\n<th>Year<\/th>\n<th>Seal Population (Thousands)<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>1997<\/td>\n<td>3.493<\/td>\n<td>2005<\/td>\n<td>19.590<\/td>\n<\/tr>\n<tr>\n<td>1998<\/td>\n<td>5.282<\/td>\n<td>2006<\/td>\n<td>21.955<\/td>\n<\/tr>\n<tr>\n<td>1999<\/td>\n<td>6.357<\/td>\n<td>2007<\/td>\n<td>22.862<\/td>\n<\/tr>\n<tr>\n<td>2000<\/td>\n<td>9.201<\/td>\n<td>2008<\/td>\n<td>23.869<\/td>\n<\/tr>\n<tr>\n<td>2001<\/td>\n<td>11.224<\/td>\n<td>2009<\/td>\n<td>24.243<\/td>\n<\/tr>\n<tr>\n<td>2002<\/td>\n<td>12.964<\/td>\n<td>2010<\/td>\n<td>24.344<\/td>\n<\/tr>\n<tr>\n<td>2003<\/td>\n<td>16.226<\/td>\n<td>2011<\/td>\n<td>24.919<\/td>\n<\/tr>\n<tr>\n<td>2004<\/td>\n<td>18.137<\/td>\n<td>2012<\/td>\n<td>25.108<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"padding-left: 60px;\">a. Let <em>x<\/em>\u00a0represent time in years starting with <em>x<\/em> = 0\u00a0for the year 1997. Let <em>y<\/em>\u00a0represent the number of seals in thousands. Use logistic regression to fit a model to these data.<\/p>\n<p style=\"padding-left: 60px;\">b. Use the model to predict the seal population for the year 2020.<\/p>\n<p style=\"padding-left: 60px;\">c. To the nearest whole number, what is the limiting value of this model?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q152985\">Show Solution<\/span><\/p>\n<div id=\"q152985\" class=\"hidden-answer\" style=\"display: none\">\n<p>a. The logistic regression model that fits these data is [latex]y=\\frac{25.65665979}{1+6.113686306{e}^{-0.3852149008x}}[\/latex].<br \/>\nb. If the population continues to grow at this rate, there will be about 25,634 seals in 2020.<br \/>\nc. To the nearest whole number, the carrying capacity is 25,657<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2><\/h2>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-5344\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section><hr class=\"before-footnotes clear\" \/><div class=\"footnotes\"><ol><li id=\"footnote-5344-1\">Source: <em>Indiana University Center for Studies of Law in Action, 2007<\/em> <a href=\"#return-footnote-5344-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><li id=\"footnote-5344-2\">Source: <em>Center for Disease Control and Prevention, 2013<\/em> <a href=\"#return-footnote-5344-2\" class=\"return-footnote\" aria-label=\"Return to footnote 2\">&crarr;<\/a><\/li><li id=\"footnote-5344-3\">Source: <em>The World Bank, 2013<\/em> <a href=\"#return-footnote-5344-3\" class=\"return-footnote\" aria-label=\"Return to footnote 3\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":167848,"menu_order":13,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-5344","chapter","type-chapter","status-publish","hentry"],"part":5352,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/5344","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/wp\/v2\/users\/167848"}],"version-history":[{"count":9,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/5344\/revisions"}],"predecessor-version":[{"id":5776,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/5344\/revisions\/5776"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/5352"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/5344\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/wp\/v2\/media?parent=5344"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=5344"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/wp\/v2\/contributor?post=5344"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/wp\/v2\/license?post=5344"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}