Question

use regression to find an exponential equation that best fits the data above. the equation has form y = ab^x where: a = b = round your answers to 2 decimal places. question help: post to forum submit question jump to answer

Explanation:

Step1: Transform the exponential model

Take the natural - logarithm of $y = ab^{x}$, we get $\ln(y)=\ln(a)+x\ln(b)$. Let $Y = \ln(y)$, $A=\ln(a)$ and $B = \ln(b)$, then the model becomes a linear model $Y=A + Bx$.

Step2: Calculate the necessary sums

Assume we have $n$ data points $(x_i,y_i)$. First, calculate $\sum_{i = 1}^{n}x_i$, $\sum_{i = 1}^{n}Y_i=\sum_{i = 1}^{n}\ln(y_i)$, $\sum_{i = 1}^{n}x_i^2$, $\sum_{i = 1}^{n}x_iY_i$. Since the $x$ - values are not given in the problem, assume $x$ is the index of the data points starting from $x = 1,2,\cdots,6$.
For $y_1 = 1016$, $\ln(y_1)=\ln(1016)\approx6.92$; for $y_2 = 1814$, $\ln(y_2)=\ln(1814)\approx7.50$; for $y_3 = 3017$, $\ln(y_3)=\ln(3017)\approx8.01$; for $y_4 = 4968$, $\ln(y_4)=\ln(4968)\approx8.51$; for $y_5 = 8094$, $\ln(y_5)=\ln(8094)\approx9.00$; for $y_6 = 13598$, $\ln(y_6)=\ln(13598)\approx9.52$.
$\sum_{i = 1}^{6}x_i=1 + 2+3 + 4+5 + 6=\frac{6\times(6 + 1)}{2}=21$, $\sum_{i = 1}^{6}Y_i=6.92+7.50+8.01+8.51+9.00+9.52 = 49.46$, $\sum_{i = 1}^{6}x_i^2=1^2+2^2+3^2+4^2+5^2+6^2=\frac{6\times(6 + 1)\times(2\times6 + 1)}{6}=91$, $\sum_{i = 1}^{6}x_iY_i=1\times6.92+2\times7.50+3\times8.01+4\times8.51+5\times9.00+6\times9.52=6.92 + 15.00+24.03+34.04+45.00+57.12 = 182.11$.

Step3: Calculate $B$ and $A$ using the least - squares formula for linear regression

The formula for $B$ in linear regression $Y = A + Bx$ is $B=\frac{n\sum_{i = 1}^{n}x_iY_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}Y_i}{n\sum_{i = 1}^{n}x_i^2-(\sum_{i = 1}^{n}x_i)^2}$.
$B=\frac{6\times182.11-21\times49.46}{6\times91 - 21^2}=\frac{1092.66-1038.66}{546 - 441}=\frac{54}{105}\approx0.5143$.
The formula for $A$ is $A=\overline{Y}-B\overline{x}$, where $\overline{x}=\frac{\sum_{i = 1}^{n}x_i}{n}=\frac{21}{6}=3.5$ and $\overline{Y}=\frac{\sum_{i = 1}^{n}Y_i}{n}=\frac{49.46}{6}\approx8.2433$.
$A = 8.2433-0.5143\times3.5=8.2433 - 1.80005=6.44325$.

Step4: Calculate $a$ and $b$

Since $A=\ln(a)$, then $a = e^A$. So $a = e^{6.44325}\approx628.57$.
Since $B=\ln(b)$, then $b = e^B$. So $b = e^{0.5143}\approx1.67$.

Answer:

$a = 628.57$
$b = 1.67$