Exercise 1.2.29

Question

The table shows (lifetime) peptic ulcer rates (per 100 population) for various family incomes as reported by the National Health Interview Survey.
    \begin{enumerate}[label=(\alph*)]
        \item Make a scatter plot of these data and decide whether a linear model is appropriate.
        \item Find and graph a linear model using the first and last data points.
        \item Find and graph the regression line.
        \item Use the linear model in part (c) to estimate the ulcer rate for people with an income of \$25,000.
        \item According to the model, how likely is someone with an income of \$80,000 to suffer from peptic ulcers?
        \item Do you think it would be reasonable to apply the model to someone with an income of \$200,000?
    \end{enumerate}

Khagan Gulusoy · Accepted Answer

\begin{enumerate}[label=(\alph*)]

\item Judging from the graph, the linear model is appropriate for this case. 
    \begin{figure}[H] \centering
    \begin{tikzpicture}
    \begin{axis}[axis lines=left, xmin=0, xmax=65000, ymax=15, ymin=0, xlabel={Income, \$}, ylabel={Ulcer rate (per 100 population)}, x label style = {at={(ticklabel cs:0.9)},anchor=south,above=10mm}, y label style={anchor=north, above=3mm}]
         \addplot[mark=*, mark options={scale=1, fill=black}](4000,14.1);
         \addplot[mark=*, mark options={scale=1, fill=black}](6000,13);
         \addplot[mark=*, mark options={scale=1, fill=black}](8000,13.4);
         \addplot[mark=*, mark options={scale=1, fill=black}](12000,12.5);
         \addplot[mark=*, mark options={scale=1, fill=black}](16000,12);
         \addplot[mark=*, mark options={scale=1, fill=black}](20000,12.4);
         \addplot[mark=*, mark options={scale=1, fill=black}](30000,10.5);
         \addplot[mark=*, mark options={scale=1, fill=black}](45000,9.4);
         \addplot[mark=*, mark options={scale=1, fill=black}](60000,8.2);
    \end{axis}    
    \end{tikzpicture}
    \end{figure}

\item The linear function $f$ that passes through points $(4000,14.1)$ and $(60000,8.2)$ can be found by the equation
    $$\forall x\in [0, 139808]:\quad \frac{x-4000}{60000-4000}=\frac{f(x)-14.1}{8.2-14.1}.$$
    $$\frac{x-4000}{56000}=\frac{f(x)-14.1}{-5.9}.$$
    $$-5.9x+23600=56000f(x)-789600.$$
    $$f(x)\approx 14.5214 + 0.000105357 x.$$
    \begin{figure}[H] \centering
    \begin{tikzpicture}
    \begin{axis}[axis lines=left, xmin=0, xmax=65000, ymax=15, ymin=0, xlabel={Income, \$}, ylabel={Ulcer rate (per 100 population)}, x label style = {at={(ticklabel cs:0.9)},anchor=south,above=10mm}, y label style={anchor=north, above=3mm}]
         \addplot[mark=*, mark options={scale=1, fill=black}](4000,14.1);
         \addplot[domain=0:65000, samples=2, mark=none, color=blue, ultra thick] {14.5214 - 0.000105357*x};
         \addplot[mark=*, mark options={scale=1, fill=black}](60000,8.2);
         \addplot[mark=*, mark options={scale=1, fill=black}](6000,13);
         \addplot[mark=*, mark options={scale=1, fill=black}](8000,13.4);
         \addplot[mark=*, mark options={scale=1, fill=black}](12000,12.5);
         \addplot[mark=*, mark options={scale=1, fill=black}](16000,12);
         \addplot[mark=*, mark options={scale=1, fill=black}](20000,12.4);
         \addplot[mark=*, mark options={scale=1, fill=black}](30000,10.5);
         \addplot[mark=*, mark options={scale=1, fill=black}](45000,9.4);
    \end{axis}    
    \end{tikzpicture}
    \end{figure}

\item Using Python, we calculate the regression function
    \begin{minted}{python}
    import numpy as np
    X = np.array([4000, 6000, 8000, 12000, 16000, 20000, 30000, 45000, 60000])
    Y = np.array([14.1, 13, 13.4, 12.5, 12, 12.4, 10.5, 9.4, 8.2])

result = np.polyfit(X, Y, 1)  # 1 indicates linear regression
    slope = result[0]
    intercept = result[1]

print("Slope (m):", slope)
    print("Intercept (b):", intercept)
    \end{minted}
    (Alternatively, we can use \href{https://www.wolframalpha.com/input?i=linear+regression+%284000%2C14.1%29+%2860000%2C8.2%29+%286000%2C13%29+%288000%2C13.4%29+%2812000%2C12.5%29+%2816000%2C12%29+%2820000%2C12.4%29+%2830000%2C10.5%29+%2845000%2C9.4%29}{WolframAlpha}). In other words, we get the regression formula:
    \begin{equation*}
        T:[0, 139808) 	o \mathbb{R}, \quad T(x) = -0.00009978545618789518\cdot x+13.950764077085212.
    \end{equation*}
    The regression line then looks as follows: \begin{figure}[H] \centering
    \begin{tikzpicture}
    \begin{axis}[axis lines=left, xmin=0, xmax=65000, ymax=15, ymin=0, xlabel={Income, \$}, ylabel={Ulcer rate (per 100 population)}, x label style = {at={(ticklabel cs:0.9)},anchor=south,above=10mm}, y label style={anchor=north, above=3mm}]
         \addplot[mark=*, mark options={scale=1, fill=black}](4000,14.1);
         \addplot[domain=0:65000, samples=2, mark=none, color=blue, ultra thick] {-9.978545618789518e-05*x+13.950764077085212};
         \addplot[mark=*, mark options={scale=1, fill=black}](60000,8.2);
         \addplot[mark=*, mark options={scale=1, fill=black}](6000,13);
         \addplot[mark=*, mark options={scale=1, fill=black}](8000,13.4);
         \addplot[mark=*, mark options={scale=1, fill=black}](12000,12.5);
         \addplot[mark=*, mark options={scale=1, fill=black}](16000,12);
         \addplot[mark=*, mark options={scale=1, fill=black}](20000,12.4);
         \addplot[mark=*, mark options={scale=1, fill=black}](30000,10.5);
         \addplot[mark=*, mark options={scale=1, fill=black}](45000,9.4);
    \end{axis}    
    \end{tikzpicture}
    \end{figure}

\item For people with an income of \$25,000 the ulcer rate is:
    $$-0.00009978545618789518\cdot 25000+13.950764077085212=11.4561276723878325\approx 11.5.$$

\item For people with an income of \$80,000 the ulcer rate is:
    $$-0.00009978545618789518\cdot 80000+13.950764077085212=5.9679275820535976\approx 6.$$

\item No, that's impossible as the rate would turn negative. The model only seems to make sense for medium-income levels. After \$100,000-\$150,000 income, the ulcer level probably does not depend on income anymore since all of the healthcare needs are fully covered after that. %It's highly unlikely for someone earning \$200,000 to develop an ulcer.
    
\end{enumerate}

Exercise 1.2.29

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Exercise 1.2.29

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