(*********************************************************************** Mathematica-Compatible Notebook This notebook can be used on any computer system with Mathematica 3.0, MathReader 3.0, or any compatible application. The data for the notebook starts with the line of stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a file with a name ending in .nb, then open the file inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). 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For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. ***********************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 142996, 4169]*) (*NotebookOutlinePosition[ 144101, 4203]*) (* CellTagsIndexPosition[ 144057, 4199]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell["L4 Finding a Model for Given Data", "Subtitle", FontSize->16, FontWeight->"Bold"], Cell["\<\ To illustrate the procedure we will use the data from the example at the \ beginning of Chapter 3 about the median age of first marriage of women in the \ United States.\ \>", "Text"], Cell[CellGroupData[{ Cell[TextData[StyleBox["Getting Started"]], "Subsection"], Cell[TextData[{ "First you need to enter the data into ", StyleBox["Mathematica", FontSlant->"Italic"], ". Either copy the median age data from the previous ", StyleBox["Mathematica", FontSlant->"Italic"], " lesson, or reenter it using ", StyleBox["Input", FontWeight->"Bold"], " \[Rule] ", StyleBox["Create Table/Matrix/Palette ", FontWeight->"Bold"], "and name it A. Evaluate the cell using the action key.\n" }], "Text"], Cell[BoxData[ RowBox[{ RowBox[{"A", "=", GridBox[{ {\("\"\), \("\"\)}, {"1960", "20.3"}, {"1965", "20.6"}, {"1970", "20.8"}, {"1975", "21.1"}, {"1980", "22"}, {"1985", "23.3"}, {"1990", "23.9"}, {"1995", "24.5"} }]}], ";"}]], "Input"], Cell[TextData[{ "The next step is to plot the data in order to determine possible function \ types based on the shape of the graph and/or numerical properties. (You may \ want to refer to the function library in Section 3.7.) Open the ", StyleBox["DataFitP", FontWeight->"Bold"], " palette (", StyleBox["File \[Rule] Palettes \[Rule] DataFitP", FontWeight->"Bold"], ") and move the palette over to the side so that it will not overlap with \ your notebook. Click on the topmost button to paste the command for loading \ the DataFit package, then evaluate the cell containing the command. Make \ sure you see the In[#] tag!" }], "Text"], Cell[BoxData[ \(<< DataFit`\)], "Input"], Cell[TextData[{ "Now you can plot the data using the function ", StyleBox["PlotData", FontWeight->"Bold"], "." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(PlotData[A]\)], "Input"], Cell[GraphicsData["PostScript", "\<\ %! %%Creator: Mathematica %%AspectRatio: .61803 MathPictureStart /Mabs { Mgmatrix idtransform Mtmatrix dtransform } bind def /Mabsadd { Mabs 3 -1 roll add 3 1 roll add exch } bind def %% Graphics /Courier findfont 10 scalefont setfont % Scaling calculations -48.3745 0.0247372 -2.51806 0.127403 [ [.11039 .00222 -12 -9 ] [.11039 .00222 12 0 ] [.35776 .00222 -12 -9 ] [.35776 .00222 12 0 ] [.60513 .00222 -12 -9 ] [.60513 .00222 12 0 ] [.8525 .00222 -12 -9 ] [.8525 .00222 12 0 ] [1.025 .01472 0 -7.5 ] [1.025 .01472 28 7.5 ] [.01131 .03 -12 -4.5 ] [.01131 .03 0 4.5 ] [.01131 .15741 -12 -4.5 ] [.01131 .15741 0 4.5 ] [.01131 .28481 -12 -4.5 ] [.01131 .28481 0 4.5 ] [.01131 .41221 -12 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Point[{1965, 20.6}], Point[{1970, 20.8}], Point[{1975, 21.1}], Point[{1980, 22}], Point[{1985, 23.3}], Point[{1990, 23.89999999999999}], Point[{1995, 24.5}]}}, {Line[{{1956.5, 19.88}, {1960, 19.88}}], Line[{{1956.5, 19.88}, {1956.5, 20.3}}]}}, {DisplayFunction -> (Display[$Display, #1] & ), PlotRange -> Automatic, AspectRatio -> GoldenRatio^(-1), DisplayFunction :> Identity, ColorOutput -> Automatic, Axes -> Automatic, AxesOrigin -> {1956.5, \ 19.88}, PlotLabel -> None, AxesLabel -> {\"year\", \"age\"}, Ticks -> Automatic, GridLines -> None, Prolog -> {}, Epilog -> {}, AxesStyle -> Automatic, Background -> Automatic, DefaultColor -> Automatic, DefaultFont :> \ $DefaultFont, RotateLabel -> True, Frame -> False, FrameStyle -> Automatic, FrameTicks -> Automatic, FrameLabel -> None, PlotRegion -> Automatic, ImageSize -> Automatic, TextStyle :> $TextStyle, FormatType :> \ $FormatType}]\ \>", "\<\ -Graphics-\ \>"], "Output"] }, Open ]], Cell["\<\ Of the possible types of functions, we can immediately exclude the sine and \ cubic functions as they seem unlikely from the context. In the case of the \ sine function, the median age of marriage would increase and decrease in a \ cycle. For the cubic function, the median age of marriage would have to \ decrease eventally, leading to age zero as median age of marriage at some \ point in the future. We will now check the other possibilities: linear, \ quadratic, exponential, and logistic one by one. \ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Selecting the Most Likely Type(s) of Functions", "Subsection"], Cell[CellGroupData[{ Cell["Linear Function", "Subsubsection"], Cell[TextData[{ "Even though the graph is only somewhat convincing for a linear function, \ we will perform the numerical test, namely computing the first unit \ differences. This can be accomplished by using the palette function ", StyleBox["FirstUnitDiff", FontWeight->"Bold"], " whose input is the list of data. " }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(FirstUnitDiff[A]\)], "Input"], Cell[OutputFormData["\<\ {0.06000000000000014, 0.03999999999999985, 0.06000000000000014, \ 0.1799999999999997, 0.2600000000000001, 0.1199999999999995, 0.1200000000000002}\ \>", "\<\ {0.06, 0.04, 0.06, 0.18, 0.26, 0.12, 0.12}\ \>"], "Output"] }, Open ]], Cell[TextData[StyleBox[ "Looking at these values, we cannot declare them almost constant - the \ largest value 0.26 is more than six times the size of the smallest value \ 0.04. This makes a linear function highly unlikely."]], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Quadratic Function", "Subsubsection"], Cell[TextData[{ "Again, the graph deviates somewhat from a quadratic function, but let's \ check the second unit differences using the palette function ", StyleBox["SecondUnitDiff", FontWeight->"Bold"], "." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(SecondUnitDiff[A]\)], "Input"], Cell[OutputFormData["\<\ {-0.002000000000000028, 0.002000000000000028, 0.01199999999999995, 0.00800000000000004, -0.01400000000000005, 7.077671781985372*^-17}\ \>", "\<\ -17 {-0.002, 0.002, 0.012, 0.008, -0.014, 7.07767 10 }\ \>"], "Output"] }, Open ]], Cell[TextData[ "All these numbers are very close to zero. In fact, if we round to two \ decimal places (twice as many as the data), the differences are either 0 or \ \[PlusMinus] 0.01, which is not too bad. A quadratic function is definitely a \ possibility."], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Exponential Function", "Subsubsection"], Cell[TextData[{ "Since we might be seeing only the increasing half of a quadratic function, \ an exponential function may also be likely. In this case we check whether the \ unit ratios are almost constant. The corresponding palette function is ", StyleBox["UnitRatios", FontWeight->"Bold"], ", with the data list as its input. However, ", StyleBox["UnitRatio", FontWeight->"Bold"], " works only when we can assume that the data has horizontal asymptote at \ level zero. From the context it is unlikely that the median age levels off \ at zero. The graph suggests a leveling off more around level 20. To use the \ numerical test for constant unit ratios, we need to adjust the output values. \ The function ", StyleBox["ShiftOutput", FontWeight->"Bold"], " takes as its arguments the data list and the value by which the output \ values should be increased or decreased. In this case, we want to subtract \ 20, so the adjustment is -20." }], "Text"], Cell[BoxData[ \(\(A2 = ShiftOutput[A, \(-20\)]; \)\)], "Input"], Cell["Now we can test the modified data for constant unit ratios.", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(UnitRatios[A2]\)], "Input"], Cell[OutputFormData["\<\ {1.148698354997035, 1.059223841048811, 1.065762756635474, 1.127009202097924, 1.105342296492869, 1.033975226531949, 1.029033661071188}\ \>", "\<\ {1.1487, 1.05922, 1.06576, 1.12701, 1.10534, 1.03398, 1.02903}\ \>"], "Output"] }, Open ]], Cell["\<\ Rounded to two decimal places, these values vary between 1.03 and 1.15. \ Again, not an optimal choice; the quadratic function looks more likely.\ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Logistic Function", "Subsubsection"], Cell["\<\ For this type of function we don't have a numerical test. However, if we look \ at the context it seems likely that the median age of first marriage would \ eventually level off (due to the biological clock for women), even though \ life expectancy may continue to increase.\ \>", "Text"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Fitting a Function", "Subsection"], Cell[TextData[{ "Now that we have narrowed the possible function types to just quadratic or \ logistic, we can compute the best fit for each type. For each type of \ function, there is a correponding palette function whose name ends in ", StyleBox["FitGraph", FontWeight->"Bold"], ". Using this function will display a graph of the data together with the \ fitted function. To get the functional expression of the fitted function, use \ the respective palette function whose name ends with ", StyleBox["FitFunc", FontWeight->"Bold"], ". " }], "Text"], Cell[CellGroupData[{ Cell["Quadratic Fit", "Subsubsection"], Cell[TextData[{ "A quadratic function is a polynomial of degree 2, thus we use ", StyleBox["PolyFitGraph", FontWeight->"Bold"], " to compute the least squares fit and to see the graph of the fitted \ function displayed together with the data. The input to the function ", StyleBox["PolyFitGraph", FontWeight->"Bold"], " consists of the data, followed by the degree of the polynomial to be \ fitted. To get the formula of the fitted quadratic function, we use ", StyleBox["PolyFitFunc", FontWeight->"Bold"], ". 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One of the \ constants is quite large, whereas the factor for the squared term is quite \ small. This is not too surprising once you think about it - our input values \ (= years) are very big! And if we square them, they become even bigger. One \ remedy to this problem is to redefine the meaning of the independent \ variable. So far, the input value denotes the ", StyleBox["actual year", FontVariations->{"Underline"->True}], ". We can change the meaning of the independent variable to stand for ", StyleBox["years after 1960", FontVariations->{"Underline"->True}], ". Then we have the following correspondence:\n\t\t\t1960 \ \[LongLeftRightArrow] 0\n\t\t\t1965 \[LongLeftRightArrow] 5\n\t\t\t1970 \ \[LongLeftRightArrow] 10\n\t\t\t\t\[VerticalEllipsis]\n\t\t\t1995 \ \[LongLeftRightArrow] 35\nThis makes the values of the independent variable \ much smaller, but we have to adjust our data correspondingly before we \ compute the best fit. In order to not have to retype all the data, we can use \ the function ", StyleBox["ShiftInput", FontWeight->"Bold"], ". This function shifts the input values by a given amount and has as \ entries the data to be modified and the amount to be added or subtracted. We \ just have to figure out how the modified values can be computed from the \ original ones. In the above example, we get the new values by ", StyleBox["subtracting 1960", FontWeight->"Bold"], " from the original input values. 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Cell[BoxData[ \(LogisticFitFunc[A2]\)], "Input"], Cell[BoxData[ \(6.41889634693810684` \/\(1 + 2.58312248271390298`*^97\ E\^\(\(-0.112903835761244408`\)\ x\)\)\)], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Comparison of Fitted Functions", "Subsection"], Cell[TextData[{ "Just looking at the graphs of the fitted functions, both the quadratic and \ the logistic functions look reasonable and it is hard to decide just from the \ graphs which one fits better. We can compute the square error for comparisons \ using the function ", StyleBox["FitComp", FontWeight->"Bold"], ". This function uses as input the data together with a list of fitted \ functions that are to be compared. The function ", StyleBox["FitComp", FontWeight->"Bold"], " will display the total squared error for each of the fitted functions, as \ well as a table displaying the data, the predicted values, and the \ differences between the data and the prediction. Note, however, that we can \ compare only functions that were fitted using the same data. We have used two \ different data sets, namely A2 (for the logistic) and A3 (for the quadratic \ fit). In this case, we use ", StyleBox["FitComp", FontWeight->"Bold"], " on each fitted function separately, and then compare the total error for \ the two functions." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(FitComp[A3, {PolyFitFunc[A3, 2]}]\)], "Input"], Cell[BoxData[ InterpretationBox[ RowBox[{ InterpretationBox[ RowBox[{\("f"\), "\[InvisibleSpace]", InterpretationBox[\(\[Null]\_1\), Subscript[ 1], Editable->False], "\[InvisibleSpace]", \("(x)"\)}], SequenceForm[ "f", Subscript[ 1], "(x)"], Editable->False], "\[InvisibleSpace]", \(" = "\), "\[InvisibleSpace]", \(\(20.2291666666667024`\[InvisibleSpace]\) + 0.0434523809523805759`\ x + 0.00245238095238065056`\ x\^2\)}], SequenceForm[ SequenceForm[ "f", Subscript[ 1], "(x)"], " = ", Plus[ 20.229166666666703, Times[ .043452380952380576, x], Times[ .0024523809523806506, Power[ x, 2]]]], Editable->False]], "Print"], Cell[BoxData[ InterpretationBox[ RowBox[{\("Total Error for "\), "\[InvisibleSpace]", InterpretationBox[ RowBox[{\("f"\), 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Thus, we conclude that the logistic function is a better fit. \ Note that the function has to be adjusted for the adjustment in output values \ done previously by adding back 20. Thus, our best fitting function for the \ original data is\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(LogisticFitFunc[A2] + 20\)], "Input"], Cell[BoxData[ \(20 + 6.41889634693810684` \/\(1 + 2.58312248271390298`*^97\ E\^\(\(-0.112903835761244408`\)\ x\)\)\)], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Making Predictions and Answering Questions", "Subsection"], Cell[CellGroupData[{ Cell["\<\ Question 1: What will the median age of the first marriage be in the year \ 2010? \ \>", "Subsubsection"], Cell[TextData[ "To answer this question means that we need to evaluate the best fitting \ function given above at x = 2010. To substitute the value 2010 for the \ independent variable x, we use /. {x\[Rule] 2010}. "], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(LogisticFitFunc[A2] + 20 /. {x -> 2010}\)], "Input"], Cell[OutputFormData["\<\ 25.99013026365002\ \>", "\<\ 25.9901\ \>"], "Output"] }, Open ]], Cell["\<\ Note that with this prediction for the year 2010 we are outside the range of \ input values for the given data. Thus, we have to keep in mind that this \ prediction is only true if \ta) we managed to choose the correct function type \tb) the future continues just like the past. \ \>", "Text"], Cell[TextData[{ "Just for comparison, let's see what the quadratic model would have \ predicted. Note that we have to translate 2010 into years after 1960, as that \ is the meaning of the data to which we fit the function. To modify 2010, we \ need to substract 1960:\n\t 2010\[LeftRightArrow]50. \nWe substitute the \ value 50 for the independent variable ", StyleBox["x", FontSlant->"Italic"], " in the fitted polynomial:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(PolyFitFunc[A3, 2] /. {x -> 50}\)], "Input"], Cell[OutputFormData["\<\ 28.53273809523735\ \>", "\<\ 28.5327\ \>"], "Output"] }, Open ]], Cell["\<\ The quadratic model gives a higher value. Since both models had a relatively \ small total error, both could be potentially true. We can only see which one \ is the more appropriate model once the year 2010 is reached and we know the \ actual median age of first marriage. In the mean time, the model should be \ checked whenever new data becomes available. If the model predictions match \ the actual data, then the model is quite reasonable. If model's prediction \ differs significantly from the data, then the additional data needs to be \ incorporated and a new model has to be fitted. \ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["\<\ Question 2: When will the median age of first marriage reach 30? \ \>", "Subsubsection"], Cell[TextData[{ "To answer this question, we need to solve the following equations for ", StyleBox["x:\n\t\t\n\t\t", FontSlant->"Italic"], " LogisticFitFunc[A2]+20=30 or PolyFitFunc[A3]=30." }], "Text"], Cell[TextData[{ "To get an idea as to what to expect as an answer, we can graph the fitted \ function for a larger set of input values by using ", StyleBox["xplot\[ShortRightArrow]{xmin, xmax}", FontWeight->"Bold"], ". This draws the graph for input values between ", StyleBox["xmin", FontWeight->"Bold"], " and ", StyleBox["xmax", FontWeight->"Bold"], ", and may produce output values that are outside the default range \ determined by the data. To see a larger range of output values, use", StyleBox[" yplot\[ShortRightArrow]{ymin, ymax}", FontWeight->"Bold"], ", and the range for output value is enlarge to the interval ", StyleBox["ymin", FontWeight->"Bold"], " to ", StyleBox["ymax", FontWeight->"Bold"], ".\n\nLet's first look at the logistic function and draw the graph from \ 1900 to 2050. 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Graphics \[SkeletonIndicator]\), False, Editable->False]], "Output"] }, Open ]], Cell[TextData[{ "From the graph we see that the logistic levels off at approximately 26.5. \ This is not surprising since the fitted logistic function has the value 6.41 \ in the numerator. Recall that this value corresponds to ", StyleBox["L", FontSlant->"Italic"], ", which is the largest value that could ever be achieved. With the added \ amount of 20 years, the largest the median age could ever be under this model \ is 26.41. Thus, the answer to our question is NEVER.\n\nThe situation is \ different for the quadratic function. Here we see that there is an input \ value for which the median age has reached 30. From the graph we can get an \ estimate, somewhere around 55. This would be in the year 1960+55 = 2015. To \ get a more exact value, we can use the built-in function ", StyleBox["Solve", FontWeight->"Bold"], " to solve the equation PolyFitFunc[A3]=30.\n\nHere is how the built-in \ function ", StyleBox["Solve", FontWeight->"Bold"], " works:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\(?Solve\)\)], "Input"], Cell[BoxData[ \("Solve[eqns, vars] attempts to solve an equation or set of equations \ for the variables vars. Solve[eqns, vars, elims] attempts to solve the \ equations for vars, eliminating the variables elims."\)], "Print"] }, Open ]], Cell[TextData[{ "The important thing to remember is to use == in the equation within ", StyleBox["Solve", FontWeight->"Bold"], " (otherwise ", StyleBox["Mathematica", FontSlant->"Italic"], " thinks you want to make an assignment)." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Solve[PolyFitFunc[A3, 2] == 30, x]\)], "Input"], Cell[BoxData[ \({{x \[Rule] \(-72.5986078168882009`\)}, { x \[Rule] 54.8801612149444206`}}\)], "Output"] }, Open ]], Cell[TextData[{ "There are two answers, namely ", StyleBox["x", FontSlant->"Italic"], " = -72.5986 and ", StyleBox["x", FontSlant->"Italic"], " = 54.8802, which is a result of the symmetry of quadratic functions. If \ we translate the answers into actual years (by rounding to full years and \ adding 1960), we see that the median age of first marriage is 30 years in \ both 1887 and 2015:\n\n\t\t1887 \[LongLeftRightArrow] -73 ( - 73 + 1960 \ = 1887) \n\t\t2015 \[LongLeftRightArrow] 55\t ( 55 + 1960 = 2015)\n\n\ This answer tells us two things: \n1) We can expect the median age of first \ marriage to be 30 in the year 2015.\t\n2) Since it is unlikely that the \ median age of marriage was higher in earlies times, the model may be only \ applicable for years correponding to the increasing part of the parabola \ (i.e., to the right of the vertex). From the graph we can estimate the input \ value corresponding to the vertex to be -10 (corresponds to 1960-10 = 1950). \ We can actually use the values for the constants ", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b,", FontSlant->"Italic"], " and ", StyleBox["c", FontSlant->"Italic"], " to compute the input value of the vertex. Recall that the ", StyleBox["x", FontSlant->"Italic"], "-coordinate of the vertex for a quadratic function of the form ", StyleBox["f(x)=a", FontSlant->"Italic"], Cell[BoxData[ FormBox[ RowBox[{\(x\^2\), "+", StyleBox["bx", FontSlant->"Italic"], "+", "c"}], TraditionalForm]]], " can be computed as\n\n\t\t", Cell[BoxData[ \(TraditionalForm\`\(-b\)\/\(2 a\)\)], FontSize->14], StyleBox["=", FontSize->14], Cell[BoxData[ FormBox[ StyleBox[\(\(-0.0434524\)\/\(2 \((0.00245238)\)\)\), FontSize->16], TraditionalForm]], FontSize->14], StyleBox["= ", FontSize->14], "-8.85923", StyleBox["\n\t\t", FontSize->14], "\nThus, the model should be applicable starting from 1960 - 9 = 1951. We \ can also try to get additional data for earlier years to further validate the \ model. " }], "Text"] }, Open ]] }, Open ]] }, Open ]] }, FrontEndVersion->"Microsoft Windows 3.0", ScreenRectangle->{{0, 800}, {0, 544}}, WindowSize->{658, 445}, WindowMargins->{{2, Automatic}, {Automatic, 5}}, PrintingCopies->1, PrintingStartingPageNumber->19, PrintingPageRange->{Automatic, Automatic}, PageHeaders->{{Cell[ TextData[ "M-", { CounterBox[ "Page"]}], "PageNumber"], Inherited, Cell[ TextData[ "Silvia Heubach"], "Header"]}, {Cell[ TextData[ "Introduction to Modeling"], "Header"], Inherited, Cell[ TextData[ "M-", { CounterBox[ "Page"]}], "PageNumber"]}}, PrintingOptions->{"FacingPages"->True} ] (*********************************************************************** Cached data follows. If you edit this Notebook file directly, not using Mathematica, you must remove the line containing CacheID at the top of the file. 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