Question

question 3
use exponential regression to find an exponential function that best fits this data,
$f(x)=$
use linear regression to find an linear function that best fits this data,
$g(x)=$
of these three, which equation best fits the data?
exponential
linear
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Explanation:

Step1: Input data into calculator

Using a graphing calculator or statistical software, input the \(x\) - values (\(1,2,3,4,5,6\)) and \(y\) - values (\(767,739,742,752,835,814\)).

Step2: Perform exponential regression

For exponential regression, the general form is \(y = ab^{x}\). Using the calculator's exponential regression function, we get \(a\approx752.7\) and \(b\approx1.017\). So \(f(x)=752.7\times(1.017)^{x}\).

Step3: Perform linear regression

For linear regression, the general form is \(y = mx + c\). Using the calculator's linear regression function, \(m=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}\), \(\bar{x}=\frac{1 + 2+3+4+5+6}{6}=3.5\), \(\bar{y}=\frac{767 + 739+742+752+835+814}{6}\approx771.83\). After calculation, \(m\approx11.9\) and \(c = 729.1\). So \(g(x)=11.9x+729.1\).

Step4: Compare the fits

To compare which model fits better, we can look at the \(R^{2}\) (coefficient of determination) values. The closer \(R^{2}\) is to \(1\), the better the fit. For the given data, when calculated (using calculator's output for \(R^{2}\) in regression), the exponential model has a higher \(R^{2}\) value.

Answer:

\(f(x)=752.7\times(1.017)^{x}\); \(g(x)=11.9x + 729.1\); Exponential.