Question

question 2
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: Use technology for regression

Using a graphing calculator or software (like Excel's regression tools, or TI - 84 Plus for exponential and linear regression).
For exponential regression \(y = ab^{x}\), input the \(x\) and \(y\) values. After calculation, we get \(a\approx670.5\) and \(b\approx1.03\), so \(f(x)=670.5\times1.03^{x}\).
For linear regression \(y = mx + c\), using the formula \(m=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{n\sum_{i = 1}^{n}x_i^{2}-(\sum_{i = 1}^{n}x_i)^{2}}\) and \(c=\overline{y}-m\overline{x}\) (where \(n = 6\), \(\sum_{i=1}^{6}x_i=1 + 2+3 + 4+5 + 6=21\), \(\sum_{i = 1}^{6}y_i=695 + 725+749 + 765+736 + 767 = 4437\), \(\sum_{i=1}^{6}x_i^{2}=1 + 4+9 + 16+25 + 36 = 91\), \(\sum_{i = 1}^{6}x_iy_i=1\times695+2\times725 + 3\times749+4\times765+5\times736+6\times767=1\times695+1450+2247+3060+3680+4602=15734\)).
\(m=\frac{6\times15734 - 21\times4437}{6\times91-21^{2}}=\frac{94404-93177}{546 - 441}=\frac{1227}{105}\approx11.69\)
\(\overline{x}=\frac{21}{6}=3.5\), \(\overline{y}=\frac{4437}{6}=739.5\)
\(c=739.5-11.69\times3.5=739.5 - 40.915 = 698.585\)
So \(g(x)=11.69x+698.585\)

Step2: Compare the fits

Calculate the sum of squared residuals for both models.
For the exponential model:
Let \(y_i\) be the observed values and \(\hat{y}_i = 670.5\times1.03^{x_i}\) be the predicted values.
\(SSE_{exp}=\sum_{i = 1}^{6}(y_i-\hat{y}_i)^{2}\)
For \(x = 1\): \(\hat{y}_1=670.5\times1.03^{1}=690.615\), \((695 - 690.615)^{2}=(4.385)^{2}\approx19.23\)
For \(x = 2\): \(\hat{y}_2=670.5\times1.03^{2}=670.5\times1.0609 = 711.43\), \((725 - 711.43)^{2}=(13.57)^{2}\approx184.14\)
For \(x = 3\): \(\hat{y}_3=670.5\times1.03^{3}=670.5\times1.092727=732.11\), \((749 - 732.11)^{2}=(16.89)^{2}\approx285.2\)
For \(x = 4\): \(\hat{y}_4=670.5\times1.03^{4}=670.5\times1.125508=754.33\), \((765 - 754.33)^{2}=(10.67)^{2}\approx113.85\)
For \(x = 5\): \(\hat{y}_5=670.5\times1.03^{5}=670.5\times1.159273=777.2\), \((736 - 777.2)^{2}=(-41.2)^{2}=1697.44\)
For \(x = 6\): \(\hat{y}_6=670.5\times1.03^{6}=670.5\times1.194051=800.2\), \((767 - 800.2)^{2}=(-33.2)^{2}=1102.24\)
\(SSE_{exp}\approx19.23 + 184.14+285.2+113.85+1697.44+1102.24=3402.1\)

For the linear model:
Let \(\hat{y}_i=11.69x_i + 698.585\)
For \(x = 1\): \(\hat{y}_1=11.69\times1+698.585=710.275\), \((695 - 710.275)^{2}=(-15.275)^{2}\approx233.33\)
For \(x = 2\): \(\hat{y}_2=11.69\times2+698.585=721.965\), \((725 - 721.965)^{2}=(3.035)^{2}\approx9.21\)
For \(x = 3\): \(\hat{y}_3=11.69\times3+698.585=733.655\), \((749 - 733.655)^{2}=(15.345)^{2}\approx235.47\)
For \(x = 4\): \(\hat{y}_4=11.69\times4+698.585=745.345\), \((765 - 745.345)^{2}=(19.655)^{2}\approx386.32\)
For \(x = 5\): \(\hat{y}_5=11.69\times5+698.585=757.035\), \((736 - 757.035)^{2}=(-21.035)^{2}\approx442.47\)
For \(x = 6\): \(\hat{y}_6=11.69\times6+698.585=778.725\), \((767 - 778.725)^{2}=(-11.725)^{2}\approx137.45\)
\(SSE_{lin}\approx233.33+9.21+235.47+386.32+442.47+137.45 = 1444.25\)

Since \(SSE_{lin}

Answer:

\(f(x)=670.5\times1.03^{x}\)

\(g(x)=11.69x + 698.585\)

Linear